Solutions Manual for chemist 2

Contexts Second Edition

HSC

Solutions Manual

Solutions Manual: Module 1 Production of materials

CHAPTER 1: ETHYLENE Review exercise 1.1 1

The word ‘petrochemical’ means any chemical that has been derived from petroleum, a liquid fossil fuel.

2

Fossil fuels are non-renewable resources with a huge demand. They are wanted both as a source of energy and as a source of petrochemical substances.

Review exercise 1.2 1

Ethylene (ethene) is the most widely used feedstock material derived from petroleum.

2

a

Catalytic cracking produces shorter chain-length, lower-mass hydrocarbons from high mass petroleum fractions. Thermal steam cracking converts ethane and propane to ethylene.

b

The proportions of products obtained from fractional distillation do not match the demands for the different products. Catalytic cracking allows for specific control of the types and amounts of products and hence more efficient use of the petroleum feedstock and its products.

c

C9H20(g) → C7H16(g) + C2H4(g)

3

The catalysts used in cracking are zeolites. These solid crystalline substances adsorb the gaseous reactants, weakening their bonds and hence lowering the activation energies.

Review exercise 1.3 1

Alkenes are more reactive than alkanes because of the presence of the double bond, a centre of high electron density, in alkenes.

2

Place hexane and 1-hexene into separate test tubes. Add bromine water to a depth of about 1 cm, shake and allow to settle. If the bromine water is decolourised then the test tube contained 1-hexene and the product of this addition reaction is 1,2-dibromoethane. However, there may also be some 2-bromo-1-ethanol and hydrogen bromide as a result of the water present. If there is no change to the colour of the bromine water then the test tube contained hexane.

3

a

C3H6 + Cl2 →C3H6Cl2

1,2-dichloropropane

b

C7H14 + HBr →C7H15Br

2-bromoheptane, 1-bromoheptane. In practice only 2-bromoheptane forms (Markovnikov’s rule).

c

C6H12 + H2O → C6H13OH

3-hexanol

a

Addition of hydrogen chloride to 1-butene

4

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Solutions Manual: Module 1 Production of materials

b

Chlorination of 2-butene

c

Addition using chlorine water to 2-butene

d

Hydration of 1-butene

e

Oxidation of 2-pentene by reacting with cold dilute potassium permanganate

Chapter 1 — Application and investigation 1

Investigation

2

a

Cracking is used to produce hydrocarbons with lower molecular mass which have greater market demand, e.g. petrol, branch-chained alkanes to improve the performance of petrol; and to produce ethene which can be used as a starting material for many organic compounds.

b

In catalytic cracking the material to be cracked is passed over a zeolite catalyst at a temperature of around 500oC. The reactants adsorb to the surface of the catalyst, which weakens their bonds, lowering the activation energy.

c

Catalytic cracking uses zeolite crystals as catalysts to lower the activation energy in the cracking of high molecular mass hydrocarbons. Thermal cracking uses very high temperatures (around 800–900oC) to crack hydrocarbons such as ethane and propane, to produce needed alkenes.

3

C7H16(g) → C5H12(g) + C2H4(g)

4

Investigation

5

Alkenes contain a reactive double bond, which readily undergoes addition reactions in order to gain a more stable single bond.

6

a

1,2-dichloropropane

b

1-chlorobutane; 2-chlorobutane. In practice only 2-chlorobutane forms (Markovnikov’s rule).

c

ethanol

7

Investigation and class experimental work: a

2-pentene with HCl

b

2-pentene with Cl2

c

propene with H2O in the presence of a catalyst

8

Add bromine water to samples of each of cyclohexene and cyclohexane, shake and allow to settle. Cyclohexene undergoes an addition reaction with Br2 across the double bond and thus the bromine water is decolourised. There is no rapid visible reaction with cyclohexane.

9

Investigation © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2006. This page from Chemistry Contexts 2 Teacher’s Resource second edition may be reproduced for classroom use.

Solutions Manual: Module 1 Production of materials

CHAPTER 2: POLYMERS Review exercise 2.1 1

Poymerisation is the chemical reaction whereby monomers link together to form polymers.

2

Addition polymerisation: all atoms in the monomer are present in the polymer chain. Unsaturated monomers join together via breaking of a C = C double bond. Condensation polymerisation: a reaction between two monomers, which can be different, during which a small molecule, such as water, is eliminated.

3

4

a

Monomers have C = C double bond.

b

Monomers each contain a functional group which may be different. Common groups are –COOH (carboxylic acid), –OH (alcohol) and –NH2 (amine) group.

a

i

ii

b

i

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Solutions Manual: Module 1 Production of materials

ii

5

6

7

a

Increased branching decreases hardness by making the polymer less compact and rigid.

b

Cross-linking between polymer chains increases hardness because cross-linking is produced by covalent bonds linking chains together.

c

Decreasing the length of polymer chains decreases hardness because there are less dispersion forces between the chains.

d

Increasing the orderly arrangement of polymer chains increases hardness by making the polymer more dense and less flexible.

e

Adding a plasticiser to a polymer decreases hardness because plasticisers are generally chosen to soften a polymer.

Review exercise 2.2 1

a

formation of polyethylene

b

formation of polystyrene

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Solutions Manual: Module 1 Production of materials

2 3

c

formation of polyvinylchloride

a

chloroethylene

b

phenylethylene or ethenylbenzene

LDPE is softer and more flexible due to greater branching of chains causing reduced dispersion forces. HDPE has fewer branched chains, causing greater dispersion forces and hence greater hardness and strength.

4

1 monomer unit has molar mass of 62.5 g, so 15 000 monomer units have molar mass of 937 500 g.

5

a

polystyrene because it is a cheap, lightweight insulating polymer

b

low-density polyethene (LDPE) because it is mouldable, impermeable and flexible

c

polypropene/nylon because it is lightweight and strong and easily formed into fibres

d

LDPE because it is lightweight, cheap and flexible

e

HDPE because it is tough and durable

f

HDPE because it is rigid, tough and durable

g

Teflon because it is tough, frictionless and resistant to heat and chemicals

h

polyethene terephthalate (PET) because it is lightweight, tough and easily blown into a shapeable film

Review exercise 2.3 1

Most of the synthetic polymers made today are derived from the petrochemical industry, which relies on the non-renewable fossil fuel petroleum. However, these supplies will be rapidly used up, especially if we continue to demand their use as both a fuel and a feedstock. An alternative, renewable source of raw materials is needed.

2

Biomass is organic material derived from living organisms: plant material such as sugar and cellulose; animal material such as dung; and domestic and industrial organic waste.

3

Cellulose is a long-chain polymer of β-glucose units, C6H12O6, where every second glucose unit is inverted to produce straightened chains.

4

A glucose unit loses an H from an –OH group and joins to a carbon on a second glucose unit which has lost its –OH group; this links the glucose units with the loss of H2O and so the formation of cellulose is an example of condensation polymerisation.

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Solutions Manual: Module 1 Production of materials

5

Cellulose is made from repeating units of β-glucose with inversion of every second unit. This produces long, straight chains of cellulose which are linked to each other by hydrogen bonding. In plants, cellulose acts as a structural material. Starch (both amylase and amylopectin) is made from long-chain repeating units of α-glucose, which are also highly branched. This results in tightly coiled, compact, insoluble starch molecules, which are used as energy stores in plants.

6

Existing plant cellulose can be modified to produce biopolymers, such as celluloid and rayon, or it can be broken down into smaller units that can be used to build new polymers such as corn-starch polymers.

Review exercise 2.4 1

Silk is a biopolymer used in its natural form; rayon uses cellulose in a chemically modified form; lactic acid can be produced from the breakdown of starch and then used to make polylactic acid (PLA).

2

Advantages are that biopolymers come from plant material, which is a renewable resource, and that plastics made from biopolymers are easily broken down by bacteria and fungi. Disadvantages include the cost of production and that they are easily biodegradable, which is not useful in some applications.

3

a

Plastic-producing bacteria are grown in fermentation vats and fed on molasses or methanol. Extraction of plastic from the bacterium involves breaking down the bacterium’s cell walls and separating this from the cell debris.

b

The bacterium used is Alcaligenes eutrophus.

c

The plastic produced is of a type known as PHAs—polyhydroxyalkanoates—which have similar properties to polypropene.

Chapter 2 — Application and investigation 1

Monomer: small repeating units that join by covalent bonds to form a polymer, e.g. ethylene. Polymer: large-chain molecule consisting of small repeating units called monomers, e.g. polyethylene.

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Solutions Manual: Module 1 Production of materials

Polymerisation: the chemical reaction by which monomers link together to form polymers, e.g. formation of polyethylene by addition reaction. 2

Investigation

3

Polyethylene can have many different molecular chain lengths. Those with around 40 000 atoms per molecule are used for food wrap films; 60 000 atoms per molecule make milk containers; 80 000 atoms per molecule make bleach containers and 800 000 atoms per molecule can be used in artificial ice rinks. As chain length increases, density, hardness and melting point increase. Branching is possible in polyethylene, and as branching increases, density, hardness and melting point decrease, increasing the uses that can be made of polyethylene.

4

Molecular mass of 1 monomer = 28, so 84 500 ÷ 28 = 3018 monomer units

5

First-hand investigation

6

Investigation based on answer to Q5

7

a

b

c

d

8

Investigation

9

Investigation

10

Investigation

CHAPTER 3: ETHANOL Review exercise 3.1 1

a

1-propanol

b

4-chloro-2-pentanol

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Solutions Manual: Module 1 Production of materials

2

c

1,4-butanediol

d

2-methyl-2-propanol

a

b

c

3 2-methyl-propanol

83°C

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Solutions Manual: Module 1 Production of materials

2-butanol

100°C

2-methyl-1-propanol

108°C

1-butanol

118°C

4

Heat ethanol with an excess of concentrated sulfuric acid, or heat ethanol vapour over a catalyst at 350oC.

5

C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)

Review exercise 3.2 1

Methylated spirits are 95% ethanol, 5% methanol and small quantities of foul-tasting chemicals. The additives are used to discourage people from drinking it.

2

Ethanol has a short non-polar chain, which allows it to dissolve some non-polar substances. The –OH functional group makes ethanol polar, which allows it to act a solvent for polar substances.

3

The ethanol molecule contains the polar –OH end and the non-polar hydrocarbon chain. Hence it is able to be used as a solvent for non-polar substances such as perfumes and aftershaves. The low boiling point ethanol evaporates easily with body heat, leaving the heavier fragrance components on the skin.

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Solutions Manual: Module 1 Production of materials

4

5

Ethanol is described as renewable because it can be derived from the starch and sugars present in various crops such as sugar cane and corn.

6

Cars using ethanol produce fewer pollutants. Ethanol is a renewable resource. However, no commercially viable method of obtaining ethanol is available and cars will need to be significantly modified to run on pure ethanol.

7

Mass of methanol burned

=

1.10 g

Mass of water

=

100 g

Mass of copper

=

200.0 g

Temperature change

=

5ºC

Specific heat capacity water

=

4.18 J K–1 g–1

Specific heat capacity copper

=

0.387 J K–1 g–1

Heat released per 1.10 g methanol

=

–(mC∆T(copper) + mC∆T(water))

=

–(200 × 0.387 × 5 + 100 × 4.18 × 5)

=

–2477 J released by 1.10 g methanol

=

–2477 ×

=

–72 058 J mol–1

=

72 kJ mol–1

Heat released per mole of methanol

∴ Molar heat of combustion

32 1.10

Review exercise 3.3 1

The acid catalyst is needed to cause the water molecule to attack the double bond in ethylene.

2

3

a

carbohydrates, e.g. glucose, sucrose, starch, plus water and yeast

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Solutions Manual: Module 1 Production of materials

b

The yeasts produce enzymes which act as catalysts in the conversion of glucose to ethanol and carbon dioxide.

c

ethanol and carbon dioxide

d

exothermic

e

Once the ethanol concentration reaches 15% in the fermentation container, the yeast is killed and the reaction ceases. Therefore naturally fermented wines have ethanol concentrations of 12–15%.

4

Fermentation of glucose occurs best in the absence of oxygen (anaerobic conditions) and at a temperature of 35–40oC.

5

From the equation, and using n = mass (g) ÷ molar mass (g) mass (ethanol) = 2 × mass(glucose) ÷ molar mass(glucose) × molar mass(ethanol) = 2 × 500 ÷180.156 × 46.068 = 255.7 g

6

From the equation, and using n = mass (g) ÷ molar mass (g) mass (ethanol) = mass (carbon dioxide lost) ÷ molar mass (carbon dioxide) × molar mass (ethanol) = 50.0 ÷ 44.01 × 46.068 = 52.3 g

Chapter 3 — Application and investigation 1

a

b

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Solutions Manual: Module 1 Production of materials

c

2

a

1-propanol

b

1,4-butanediol

3 1-pentanol

2-pentanol

3-pentanol

4

Water, being polar, would not readily dissolve many esters with a non-polar component. Hexane, being non-polar, would not readily dissolve the polar esters. Ethanol, with its ability to dissolve both polar and non-polar components, is suitable for dissolving esters.

5

Investigation

6

Investigation

7

Investigation

8

Depends on responses to Questions 5, 6 and 7.

9

a

Assume heat absorbed by copper is negligible. Mass of fuel

=

2.00 g

Mass of water

=

500 g

Change in temp

=

18.4ºC

Specific heat water

=

4.18 J K–1 g–1

Methanol Heat released by 2 g

=

–mC∆T

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Solutions Manual: Module 1 Production of materials

i ii

Heat of combustion/g Heat of combustion/mole

=

–(500 × 4.18 × 18.4)

=

–38 456 J

=

38 456 ÷ 2

=

19 228 J g–1 = 19.2 k J g–1

=

38 456 ×

32 2

=

61 5296 J mol–1

=

615 kJ mol–1

=

–mC∆T

=

–(500 × 4.18 × 28.5)

=

–59 565 J

=

59 565 ÷ 2

=

29 782.5 J g–1 = 29.8 kJ g–1

1-propanol Heat released by 2 g

i ii

Heat of combustion/g Heat of combustion/mole

= =

1 789 928 J mol–1

=

1790 kJ mol–1

=

–mC∆T

=

–(500 × 4.18 × 31.5)

=

–65 835 J

=

65 835 ÷ 2

=

32 917.5 J

1-butanol Heat released by 2 g

i

Heat of combustion/g

= 32.92 kJ g–1 ii

Heat of combustion/mole

= =

2 439 845.1 J

=

2440 kJ mol–1

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Solutions Manual: Module 1 Production of materials

b

10

The dehydration of ethanol to ethylene and water is carried out by heating ethanol with concentrated sulfuric acid as a catalyst, or by heating ethanol vapour over an alumina catalyst at 350oC. The hydration of ethylene to produce ethanol requires the addition of water in the presence of a sulfuric or phosphoric acid catalyst.

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Solutions Manual: Module 1 Production of materials

CHAPTER 4: ELECTROCHEMISTRY Review exercise 4.1 1

a

A pink-brown deposit of copper forms on the surface of the steel wool as the copper ions convert to elemental copper; the blue solution goes paler and slowly converts to a green solution as the copper ions decrease in concentration and the elemental iron converts to Fe2+ ions with increasing concentration.

b

oxidation: Fe(s) → Fe2+(aq) + 2e− reduction: Cu2+(aq) + 2e− → Cu(s)

2

c

Fe(s) + Cu2+(aq) → Fe2+(aq) + Cu(s)

a

Fe(s) + Sn2+(aq) → Fe2+(aq) + Sn(s); the iron would dissolve and the tin would coat the surface of the iron; the solution would change from colourless to green

b

Pb(s) + Hg2+(aq) → Pb2+(aq) + Hg(l); the lead would dissolve and liquid mercury would form

c

no reaction

d

2Cr(s) + 3Fe2+(aq) → 2Cr3+(aq) + 3Fe(s); the chromium would dissolve and the iron would coat the chromium; both ions are generally green in aqueous solution

3

The iron container would dissolve as the iron was oxidised to its ions by the nickel ions, which would convert to elemental nickel.

4

a

no reaction

b

Sn(s) → Sn2+(aq) + 2e−; Cu2+(aq) + 2e− → Cu(s); copper would coat the tin

c

Sn(s) → Sn2+(aq) + 2e−; Ag+(aq) + e− → Ag(s); silver would coat the tin

Review exercise 4.2 1

a

b

CO2

:

C = +4

O = –2

O2

:

elemental state ∴ 0

NH3

:

N = –3

H = +1

H2S

:

S = –2

H = +1

HCl

:

H = +1

Cl = –1

SO2

:

S = +4

O = –2

SO32–

:

S = +4

O = –2

H2S2O7

:

H = +1

S = +6

S2O82–

:

S = +7

O = –2

O = –2

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Solutions Manual: Module 1 Production of materials

c

HSO3–

:

H = +1

S

:

0

+

:

N = –3

H = +1

–

:

Cl = +7

O = –2

Cu2S

:

Cu = +1

S = –2

MgH2

:

Mg = +2

H = –1

PO43–

:

P = +5

O = –2

NH4

ClO4

2

3

O = –2

a

redox reaction: Zn oxidised (0 → +2), H+ reduced (+1 → 0)

b

not a redox reaction

c

not a redox reaction

d

redox reaction: Fe2+ oxidised (+2 → +3), Cr6+ reduced (+6 → +3)

NO

: +2

NO2

: +4

N2O

: +1

N2O3 : +3 –

N2O5

: +5

HNO3

: +5

HNO2

: +3

NO2

NH2–

: –3

NH4+

: –3

N2

:0

N2H4 : –2

NH2OH : –1 4

S = +4

: +3

N2O4 : +4 NO3– : +5 NH3

: –3

NH4Cl : –3

−

Br and HBr both have bromine in –1 oxidation state. BrO− and Br2O and HBrO each have bromine in +1 oxidation state.

Review exercise 4.3 1

The salt bridge allows ions to move between each half-cell.

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Solutions Manual: Module 1 Production of materials

2

a

b

See diagram in part a.

c

anode: oxidation Mg(s) → Mg2+(aq) + 2e− cathode: reduction Cu2+(aq) + 2e− → Cu(s)

3

a b

3Pb2+(aq) + 6e– → 3Pb(s)

cathode

2Cr(s) → 2Cr3+(aq) + 6e–

anode

Cu(s) → Cu2+(aq) + 2e–

anode

2Ag+(aq) + 2e– → 2Ag(s)

cathode

Review exercise 4.4 1

a

Ca2+, Pb2+, Cu2+, Ag+

b

Ag, Sn, Cr, Mg

c

strongest oxidising agent: Ag+ strongest reducing agent: Mg

2 3

a

H2O2, HClO, MnO4−, Cr2O72−

b

Mg, Zn, H2, H2S

a

Cl2

b

Sn4+

c

Au3+

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Solutions Manual: Module 1 Production of materials

d

MnO4−

Review exercise 4.5 1

a

b

c

2

Pb2+ + 2e–

→

Pb(s)

E° = –0.13 V

cathode

Zn(s)

→

Zn2+(aq) + 2e–

E° = +0.76 V

anode

cell potential

=

–0.13 + 0.76

=

0.63 V

Cl2(g) + 2e–

→

2Cl–(aq)

E° = +1.36 V

cathode

2I–(aq)

→

I2(s) + 2e–

E° = –0.54 V

anode

cell potential

=

+1.36 – 0.54

=

0.82 V

2H+(aq) + 2e–

→

H2(g)

E° = 0 V

cathode

Mg(s)

→

Mg2+(aq) + 2e–

E° = +2.37 V

anode

cell potential

=

0 + 2.37

=

2.37 V

Pb(s) + PbO2(s) + 4H (aq) + 2SO42–(aq) → 2PbSO4(s) + 2H2O(l) E°(cell) = +2.04 V +

anode: Pb(s) + SO42–(aq) → 2PbSO4(s) + 2e– E° = +0.36 V cathode: PbO2(s) + 4H+(aq) + SO42–(aq) + 2e– → PbSO4(s) + 2H2O(l) E° = 2.04 – 0.36 = +1.68 V 3

a

i

2I– (aq) → I2(s) + 2e–

E° = –0.54 V

MnO4–(aq) + 8H+(aq) + 5e– → Mn2+(aq) + 4H2O(l)

E° = +1.51 V

Add equations. 10I–(aq) + 2MnO4–(aq) + 16H+(aq) → 5I2(s) + 2Mn2+(aq) + 8H2O(l) ii

cell potential = –0.54 + 1.51 = +0.97 V

b

iii

Cell potential is positive, and therefore the reaction can occur spontaneously.

i

Cl2(g) + H2S(aq) + 2e– → 2Cl–(aq) + S(s) + 2H+(aq) + 2e– Cl2(g) + 2e– → 2Cl– (aq) +

E° = +1.36 V

H2S(aq) → S(s) + 2H (aq) + 2e–

E° = –0.14 V

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Solutions Manual: Module 1 Production of materials

ii

cell potential

= +1.36 – 0.14 = +1.22 V

c

iii

Cell potential is positive, and therefore the reaction can occur spontaneously.

i

2Cu+(aq) + e– → Cu2+(aq) + Cu(s) + e–

ii

Cu+(aq) → Cu2+(aq) + e–

E° = –0.16 V

Cu+(aq) + e– → Cu(s)

E° = +0.52 V

cell potential = –0.16 + 0.52 = +0.36 V

iii 4

a

Cell potential is positive, and therefore the reaction can occur spontaneously.

H2O2(aq) + 2H+(aq) + 2e– → 2H2O(l) H2C2O4(aq) → 2CO2(g) + 2H+(aq) + 2e–

E° = +1.78 V E° = +0.48 V

E° (cell) = +1.78 + 0.48 = +2.26 V b

A high activation energy may be necessary for this reaction to occur.

c

Mix the H2O2 and H2C2O4 at a higher temperature and/or add a catalyst.

Review exercise 4.6 1

The dry cell is one of the most widely used sources of portable electricity. It has led to the proliferation of the portable devices we see around us. Without dry cells we would be a much less mobile society. Devices requiring electrical power would need to be in close proximity to permanent sources such as power points. Instead of torches we could still be using candles or kerosene lamps. However, the development of dry cells has led to landfill pollution.

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Solutions Manual: Module 1 Production of materials

2

3

a

If the Zn and MnO2 were in direct contact, the following overall cell reaction would occur: Zn(s) + 2H+(aq) + 2MnO2(s) → Zn2+(aq) + Mn2O3(s) + H2O(l) i.e. zinc is oxidised and manganese dioxide is reduced.

b

The cell would be useless because electron transfer would occur directly between the reactants and not through an external circuit.

4

Zn(s) + 2H+(aq) + 2MnO2(s) → Zn2+(aq) + Mn2O3(s) + H2O(l)

5

Pb(s) + SO42–(aq) → PbSO4(s) + 2e– (0)

(+2) oxidation

PbO2(s) + 4H+(aq) + SO42–(aq) + 2e– → PbSO4(s) + 2H2O(l) (+4)

(+2) reduction

6

A 6V lead–acid battery would have three cells, each with a cell potential of 2 V.

7

Lead–acid batteries are very heavy and weight is an important issue for solar-powered cars in the World Solar Challenge.

Chapter 4 — Application and investigation The following reactions will occur: 1

Zn(s) + CuSO4(aq → ZnSO4(aq) + Cu(s) Zn(s) + NiSO4(aq) → ZnSO4(aq) + Ni(s)

2

a

i

Yes. Fe2+ ions will not oxidise Ni.

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Solutions Manual: Module 1 Production of materials

ii b

No. Fe3+ will be reduced to Fe2+ and Ni oxidised to Ni2+.

No. The steel will dissolve and displace the copper.

3

The student should determine gallium’s position in the order of reactivity by attempting to oxidise it using the ions of the other metals in the list.

4

a

HNO3(aq)

→

NO2(g)

+5 b

+4

Cl2(g)

ClO3–(aq)

→

0 c

NH3(g)

→

N2(g)

Fe2O3(s)

→

FeO(s)

SO2(g)

SO42–(aq)

→

a

reduction

+2

MnO4–(aq)

→

MnO2(s)

+7 5

reduction

+2

+4 f

oxidation

0

+3 e

oxidation

+5

–3 d

reduction

reduction

+4

Fe(s) + SO2(g) + O2(g) → FeSO4(s) 0

+4 –2

0

+2 +6 –2

Fe oxidised; S oxidised; O2 reduced b

4FeSO4(s) + O2(g) + 6H2O(l) → 2Fe2O3.H2O(s) + 4H2SO4(aq) +2 +6 –2

0

+1 –2

+3 –2 +1 –2

+1 +6 –2

Fe oxidised; O2 reduced 6

a

All the E° values would be 2.93 V higher.

b

Calculated e.m.f. values for cells are based on the difference in E° between two half-cell reactions. These would be unchanged.

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Solutions Manual: Module 1 Production of materials

7

a

i

ii

Cu2+(aq) + 2e– → Cu(s) E° = +0.16 V Cr(s) → Cr3+(aq) + 3e– E° = +0.73 V 2Cr(s) + 3Cu2+(aq) → 3Cu(s) + 2Cr3+(aq)

iii b

e.m.f. = +0.89 V

i

Zn(s) → Zn2+(aq) + 2e– ii

Cl2(g) + 2e– → 2Cl–(aq)

Cl2(g) + 2e– → 2Cl–(aq) E° = +1.36 V Zn(s) → Zn2+(aq) + 2e– E° = +0.76 V Zn(s) + Cl2(g) → Zn2+(aq) + 2Cl–(aq)

iii

e.m.f = 1.36 + 0.76 = 2.12 V

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Solutions Manual: Module 1 Production of materials

8 9

a

The half-cells are separated in order to generate a flow of current in the external circuit.

b

A salt bridge allows ions to move between each half-cell and complete the circuit.

a

anode: Zn(s) → Zn2+(aq) + 2e–

E° = +0.76 V

cathode: Cu2+(aq) + 2e– → Cu(s)

E° = +0.34 V

cell: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

E°(cell) = +1.10 V

b

As the cell operates, the concentration of Cu2+ decreases and the Zn2+ concentration increases. Both changes would decrease the tendency of the forward reaction to take place and hence the cell e.m.f. would decrease.

c

i

Transfer of electrons cannot occur, and therefore the reaction between the two half-cells will cease.

ii

Transfer of charged ions between half-cells cannot occur.

10

The reduction potential of +1.7 V for Au+ is higher than that for the reduction of O2, and therefore O2 would not oxidise the gold. The reduction potential for [Au(CN)2]–(aq) is lower than that for O2, and therefore in the presence of cyanide ion, gold can be oxidised.

11

a

Ni2+(aq) or Co2+(aq) or Cd2+(aq)

b

Mn(s) or Al(s) or Zn(s)

a

i

12

ii

Fe2+

→

Fe3+ + e–

E°

=

–0.77 V

MnO4– + 8H+ + 5e–

→

Mn2+ + 4H2O

E°

=

+1.51 V

E°(cell)

=

+0.74 V

E°

=

–1.36 V

E°

=

+1.51 V

E°(cell)

=

+0.15 V

Fe3+ + e–

E°

=

–0.77 V

2Cr3+ + 7H2O

E°

=

+1.33 V

E°(cell)

=

+0.56 V

Cl2 + 2e–

E°

=

–1.36 V

2Cr3+ + 7H2O

E°

=

+1.33 V

E°(cell)

=

–0.03 V

2Cl

–

MnO4– + 8H+ + 5e– iii

Fe2+

→

Cl2 + 2e

→

Mn2+ + 4H2O

→

Cr2O72– + 14H+ + 6e– → iv

2Cl–

–

→

Cr2O72– + 14H+ + 6e– →

Thus a redox reaction should occur in i, ii and iii. b

iv above shows that Cr2O72– will not oxidise the Cl– ion of HCl, but ii shows that MnO4– will. Hence, it is inappropriate to use HCl to acidify solutions involving permanganate titrations.

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Solutions Manual: Module 1 Production of materials

13

14

a

An anode is the electrode at which oxidation takes place. A cathode is the electrode at which reduction takes place.

b

A salt bridge allows a flow of ions between two half-cells. An external circuit allows a flow of electrons from anode to cathode.

c

A secondary cell can be recharged, whereas a primary cell cannot.

a

i ii

b

15

anode:

Pb(s) + SO42–(aq) → PbSO4(s) + 2e–

cathode:

PbO2(s) + 4H+(aq) + SO42–(aq) + 2e– → PbSO4(s) + 2H2O(l)

anode:

PbSO4(s) + 2H2O(l) → 2e– + SO42–(aq) + 4H+(aq) + PbO2(s)

cathode:

PbSO4(s) + 2e– → Pb(s) + SO42–(aq)

i

On discharging, the density decreases.

ii

On recharging, the density increases.

c

Chemical potential energy is transformed into electrical energy and some heat energy as the battery is discharged. On charging, electrical energy is transformed into chemical potential energy and some heat energy.

a

Anode: Al(s) → Al3+(aq) + 3e– Cathode: O2(g) + 2H2O(l) + 4e– → 4OH–(aq)

b

Al(s) → Al3+(aq) + 3e–

E° = +1.66 V

O2(g) + 2H2O(l) + 4e– → 4OH–(aq)

E° = +0.40 V E°(cell) = +2.06 V

Cell e.m.f. would be 2.06 V. c

It would weigh much less. It avoids use of corrosive H2SO4. It uses O2 from the air as a cathode.

16–18

Investigation

CHAPTER 5: NUCLEAR CHEMISTRY Review exercise 5.1 1

a

Atoms are not conserved in nuclear reactions.

b

Chemical reactions involve changes in electrons between atoms. Nuclear reactions involve the particles in the nucleus.

c

Temperature, pressure/concentration and catalysts have no effect on nuclear reactions.

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Solutions Manual: Module 1 Production of materials

2

d

Far more energy is obtained from nuclear reactions.

a

i

17

O

:

8 protons; 9 neutrons

ii

52

Fe

:

26 protons; 26 neutrons

iii

137

I

:

53 protons; 84 neutrons

iv

258

Md

:

101 protons; 157 neutrons

b

3

i ii

93 41

iii

208 82

Nb

Pb

a

α large mass; β very small mass; γ no mass

b

α 2+ charge; β 1– charge; γ no charge

c

α least penetrating, about 5 cm through air β about 1 m through air, stopped by aluminium γ can pass through several cm of lead

4

5

a

4 2

b

0 −1

c

1 0

n

d

1 1

p

e

0 1

e

He e

Photographic film: becomes dark on exposure. Darkness increases with length of exposure and greater intensity of radiation. Geiger counter: The argon gas inside a metal tube is ionised into positive ions and electrons. These move towards electrodes which conduct a current measured by a recording device. Scintillation counter: Electrons in substances such as zinc sulfide are excited by radiation and emit photons of light as they return to lower energy states. Radiation is measured by counting flashes of light.

Review exercise 5.2 1

a

3 1

H → 32 He +

0 −1

e

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Solutions Manual: Module 1 Production of materials

c

2

3

a b

239 94

Pu →

c

234 92

U →

a

12 7

b

8 5

B → 84 Be + 10 e

a

7 4

Be +

b

232 93

N →

235 92 230 90

12 6

U + 42 He

Th + 42 He

C + 10 e

c 4

0 −1

Np +

e → 37 Li 0 −1

e→

232 92

U

c 5

6

7

a

beta emission

b

positron emission

c

alpha emission

d

positron emission

a

positron or electron capture

b

β decay

c

α decay

a

No, the neutron : proton ratio is too high and outside the zone of stability.

b

No, the neutron : proton ratio is too low and outside the zone of stability.

c

Yes, the neutron : proton ratio is in the zone of stability.

Review exercise 5.3 1

After 1st 2.6 hours: 0.1/2 = 0.05 g After 5.2 hours: 0.025 g After 7.8 hours: 0.0125 g

2

1st half-life: 0.120/2 = 0.06 g 2nd half-life: 0.06/2 = 0.03 g © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2006. This page from Chemistry Contexts 2 Teacher’s Resource second edition may be reproduced for classroom use.

Solutions Manual: Module 1 Production of materials

3rd half-life: 0.03/2 = 0.015 g ∴3 half-lives in 24 days

I-131 has a half-life of 8 days. 3

4

a

2.1 g = 21%

b

164 g → 20.5 g, i.e. 12.5% left = 3 half-lives, ∴ 84 years

c

Half-life is 28 years.

Thallium-201 and gallium-67: suitable, as have a short half-life; sulfur-35 and hydrogen-3: unsuitable as half-life is too long.

Review exercise 5.4 1

To overcome the electrostatic repulsion alpha particles and protons will encounter in the nuclide. Neutrons do not carry a charge and therefore do not need to be accelerated.

2

They are produced synthetically.

3

a

15 7 N

b

56 26 Fe

c

6 3 Li

d

59 27 Co

+ 12 H →

e

53 24 Cr

+ 42 He →

4

+ 11 p →

12 6 C

+ 42 He 54 25 Mn

+ 12 H → 42 He +

+ 10 n → 13 H + 42 He 60 27 Co

+ 11 p

57 25 Mn

+ 10 n

The nuclear transformations in a, b, d and e would all take place in a cyclotron, as the charged particles are used to bombard the nucleus. Nuclear transformation c would take place in a nuclear reactor, as it involves bombardment by neutrons.

5

56 26 Fe

+ 12 H →

54 25 Mn

+ 42 He

Chapter 5 Application and investigation 1

2

a

90 protons; 140 neutrons

b

96 protons; 148 neutrons

c

77 protons; 115 neutrons

Chemical reactions involve changes in the arrangements of electrons around the nucleus. Nuclear reactions involve changes within the nucleus. © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2006. This page from Chemistry Contexts 2 Teacher’s Resource second edition may be reproduced for classroom use.

Solutions Manual: Module 1 Production of materials

3

4

5

a

Ionisation counter: uses a tube filled with an inert gas. As radiation enters the tube it causes the gas to be ionised, forming cations and free electrons. These particles are attracted to electrodes of the opposite charge, causing a current to flow, which is heard as an audible click and can be recorded as a digital readout.

b

Scintillation counter: electrons in some substances, such as zinc sulfide, are excited by radiation and emit photons of light when the electrons return to a lower energy state. The flashes of light are counted electronically to measure the amount of radiation.

a

β decay

b

α decay, positron emission, electron capture

a

241 94 Pu

→

237 92 U

b

210 86 Rn

→

210 87 Fr

c

48 23 V

d

107 48 Cd

+

e

234 92 U

→

230 90 Th

+ 42 He

f

26 11 Na

→

26 12 Mg

+

g

12 7 N

→

→

48 22 Ti

0 −1 e

12 6 C

+ 42 He +

0 −1 e

+ 10 e 107 47 Ag

→

0 −1 e

+ 10 e

h 6

Uranium and thorium undergo alpha emission spontaneously. Therefore helium is found within the ore.

7

Elements with high neutron-to-proton ratios emit β-particles, thereby raising the number of protons and lowering the number of neutrons. Elements with low neutron-to-proton ratios undergo either positron emission or electron capture, thereby raising the number of neutrons and lowering the number of protons.

8

a

unstable; β decay

b

unstable; positron emission or electron capture

c

unstable; positron emission or electron capture

d

unstable; β decay

e

stable

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Solutions Manual: Module 1 Production of materials

9

10

11

f

unstable; α decay

a

3 half-lives ∴

b

2 half-lives ∴ 12.3 × 2 = 24.6 years

a

35%

b

20% × 500 g = 100 g remaining

c

84 days

d

25 days

a

32 days is 4 half-lives

0.8 0.4 0.2 = 0.4; = 0.2; = 0.1 mg 2 2 2

∴ 0.625 mg remaining after 32 days b

5 half-lives drop to 0.312 g, so < 0.5 g in 36 days

12 13

Investigation

14

a

59 27

b

60

Co + 10 n →

60 27

Co

Co would be made in a nuclear reactor as it involves the capture of a neutron. In a nuclear reactor, atoms are bombarded with neutrons. A cyclotron is used to bombard atoms with charged particles.

15

Investigation

16

a

63 29

Cu + 12 H →

65 30

Zn + γ

b

14 7

N + 42 He →

17 8

O + 11 H

c

40 20

Ca + 12 H →

38 19

K + 42 He

d

239 94

Pu + 42 He →

242 96

Cm + 10 n

17

Investigation

18

α particles can cause serious damage to cells.

19

Investigation

20

Investigation

21

Research based on information gathered in Questions 19 and 20 © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2006. This page from Chemistry Contexts 2 Teacher’s Resource second edition may be reproduced for classroom use.

Solutions Manual: Module 1 Production of materials

Module 1 REVIEW 1

D

2

A

3

D

4

D

5

D

6

A

7

D

8

B

9

B

10

A

11

a

Ethylene is more reactive due to the presence of the double bond.

b

addition

c

polyethylene

d

HI

12

a

b

vinyl chloride i

chloroethene

ii

polyvinylchloride (PVC)

iii

PVC is a thermoplastic with long chains with weak dispersion forces between them. This allows the polymer to be heated and remoulded. With additives, PVC can be used as a rigid plastic (e.g. guttering, water pipes) or as a flexible plastic (e.g. garden hoses).

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Solutions Manual: Module 1 Production of materials

13

a

LDPE has a greater degree of branching between its polymer chains, reducing the dispersion forces between the chains. HDPE has less branching, resulting in a stronger and more dense plastic.

b

cling wrap, insulation for wires

c

LDPE: Organic peroxides are used as catalysts to attack the ethene double bond and form a covalent bond with a carbon atom. This new molecule then attacks another ethene molecule, causing the chain to grow (propagate). Branches are formed when the chain curls back and removes a hydrogen from a carbon in the chain, which will in turn react with another ethene molecule. The reaction terminates when two polymer radicals react together. HDPE: Ionic catalysts (Ziegler–Natta catalysts) are used. Ethene molecules are added to the polymer on the surface of the catalyst, which reduces the amount of branching.

14

d

The polymer chains in HDPE are packed tightly together due to less branching of the chains. This gives HDPE greater strength and toughness but reduces its flexibility.

a

Biomass is organic material derived from living organisms, such as: plant material (e.g. sugar and cellulose); animal material (e.g. dung); and domestic and industrial organic waste.

b

β-glucose

c

Cellulose is a formed as a condensation polymer when β-glucose monomers combine with the removal of a water molecule. Polyethylene forms as an addition polymer from ethylene monomers as the double bond breaks.

15

Use spirit burners containing each fuel, weigh mass before and after use (to give mass of fuel burnt), heat same volume of water, record temperature rise. Fuels are volatile and can easily ignite if care is not taken.

16

a

17

i

C6H12O6(aq) → 2C2H5OH(l) + 2CO2(g)

ii

anaerobic (no oxygen); temperature 35–40oC

b

Harvest cane, extract sugars, ferment to ethanol using yeast, warmth and anaerobic conditions.

c

C2H5OH(g) → C2H4(g) + H2O(g) The reaction requires a catalyst, e.g. concentrated H2SO4

No commercially viable method of producing ethanol from crops is yet available. Fuel mixtures containing greater than 15% ethanol require some modification to the car before being used. Ethanol is a renewable fuel and petroleum is not.

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Solutions Manual: Module 1 Production of materials

18

19

a

b

Cell potential = +0.76 – 0.13 = 0.63 V

c

Zinc

Uranium-238 is bombarded with neutrons to form einsteinium. 238 92

20

21

U + 1510 n →

253 99

Es + 7 −01 e

a b

Sr is in the same periodic group as calcium and therefore is chemically similar.

c

85.3 years is three half-lives ∴ > 0.125 g remains at 85 years.

a

Iodine-131 is used in the diagnosis of thyroid misfunction and in treatment of thyroid tumours.

b

I-131 is produced by neutron bombardment in a reactor.

c

Benefit: short half-life and therefore is only in the body for a short time. Problem: I-131 can destroy some healthy cells when used for radiation.

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Solutions Manual: Module 2 The acidic environment

CHAPTER 6: ACIDS, BASES AND INDICATORS Review exercise 6.1 1

a

acetic acid, CH3COOH citric acid, 2-hydroxypropane-1,2,3-tricarboxylic acid

b

sulfuric acid, H2SO4 nitric acid, HNO3

c

magnesium hydroxide, Mg(OH)2

2

Own answers

3

Own answers

4

a

Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)

b

CO32−(aq) + 2H+(aq) → CO2(g) + H2O(l)

c

2H+(aq) + K2O(s) → 2K+(aq) + H2O(l)

d

H+(aq) + OH– (aq) → H2O(l)

a

2Cr(s) + 2OH–(aq) + 6H2O(l) → 2[Cr(OH)4]–(aq) + 3H2(g)

b

Cr(OH)3(s) + OH–(aq) → [Cr(OH)4]–(aq)

5 6

Acid solutions have excess hydrogen ions and base solutions have excess hydroxide ions. These ions are free to move and carry a current.

Review exercise 6.2 1

a

K2O ionic bonding; Ga2O ionic; Br2O7 covalent

b

K2O basic; Ga2O amphoteric; Br2O7 acidic

2

CO2(g) + H2O(l) → H2CO3(aq)

3

Natural sources of sulfur dioxide: • bacteria decomposing organic matter to make H2S, which is then oxidised atmospherically. • volcanic gases • bushfire smoke. © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2006. This page from Chemistry Contexts 2 Teacher’s Resource second edition may be reproduced for classroom use.

Solutions Manual: Module 2 The acidic environment

Sources due to human activities: • burning of fossil fuels • smelting of sulfide ores such as CuFeS2 and ZnS. 4

SO2—H2SO3; SO3—H2SO4; NO2—HNO3; CO2—H2CO3

5

Increased incidence of low pH in lakes and rivers and continued damage to metal and sandstone buildings.

6

The sulfur oxides in the atmosphere usually come from burning fossil fuels in power stations or smelting sulfide ores in smelting plants. Nitrogen oxide and carbon emissions usually come from motor vehicles. Therefore acid rain usually occurs in densely populated areas or industrial areas. Review exercise 6.3 1

2 c

a

Acidic. The solution has a pH between 4 and 6.

b

The pH range is 4 to 8. This solution could be neutral, acidic or basic.

c

The solution is slightly basic. The pH is between 7.5 and 8.

a

Bromothymol blue: yellow; phenolphthalein: colourless; methyl orange: red

b

Bromothymol blue: blue; phenolphthalein: pink; methyl orange: yellow

Bromothymol blue: blue; phenolphthalein: colourless; methyl orange: yellow

Chapter 6 Application and investigation 1

Not all acids are dangerous. Acids are present in foods we eat, e.g. ethanoic acid in vinegar, lactic acid in milk. Also, acids are found in our bodies, e.g. lactic acid, hydrochloric acid. The concentration of acids should be considered.

2

a

2Al(s) + 6HCl(aq)

→

2AlCl3(aq) + 3H2(g)

b

KHCO3(s) + HNO3(aq)

→

KNO3(aq) + CO2(g) + H2O(l)

c

Fe2O3(s) + 3H2SO4(aq)

→

Fe2(SO4)3(aq) + 3H2O(l)

d

Ba(OH)2(aq) + 2HF(aq)

→

BaF2(aq) + 2H2O(l)

a

2Al(s) + 2NaOH(aq) + 6H2O(l)

→

2Na[Al(OH)4](aq) + 3H2(g)

b

KOH(aq) + Fe(OH)3(s)

→

no reaction

c

2KOH(aq) + Zn(OH)2(s)

→

K2[Zn(OH)4](aq)

d

2NaOH(aq) + SO3(g)

→

Na2SO4(aq) + H2O(l)

a

KHC4H4O6 + NaHCO3

→

KNaC4H4O6 + CO2(g) + H2O(l)

b

Fe2O3(s) + 6HCl(aq)

→

2FeCl3(aq) + 3H2O(l)

3

4

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Solutions Manual: Module 2 The acidic environment

5

c

Al(OH)3(s) + 3HCl(aq)

→

AlCl3(aq) + 3H2O(l)

d

H2SO3(aq) + CaCO3(s)

→

CaSO3(aq) + CO2(g) + H2O(l)

Al(OH)3 reacts with both acids and bases, and therefore is amphoteric. Reacting with acid: Al(OH)3(aq) + 3HCl(aq) → AlCl3(aq) + 3H2O(l) Reacting with base: Al(OH)3(aq) + NaOH(aq) → Na [Al(OH)4](aq)

6

Investigation

7

Investigation

8

A useful indicator is one whose colour change occurs over a selected pH range. Methyl orange changes colour too far outside the narrow 7.2–7.6 range for swimming pools. Litmus changes colour over this range, but the colour difference in the 7.2–7.6 range may be too subtle to detect easily. Universal indicator undergoes many colour changes and the change over the 7.2–7.6 range, from yellow-green to green may also be too subtle to detect easily. Bromothymol blue could be a useful indicator. The change for phenolphthalein is above a pH of 7.6 and so this indicator would not be useful.

9

Investigation

10

Acid rain has a pH less than 5; methyl orange when yellow means pH is greater than 4.5, and litmus when red indicates a pH of less than 7. Therefore the pH of the sample of rain water lies between 4.5 and 7, and so is unlikely to be classified as acid rain.

11

a

Na2O, Al2O3, SiO2, P4O10, Cl2O7

b

K2O, Ga2O3, GeO2, As2O3, Br2O7

12

13

CO2(aq) + H2O(l) ↔ H2CO3(aq)

carbonic acid

4NO2(g) + 2H2O(l) + O2(g) → 4HNO3(aq)

nitric acid

SO2(g) + H2O(l) → H2SO3(aq)

sulfurous acid

Cl2O7(l) + H2O(l) → 2HClO4

perchloric acid

a

SO2(g) + H2O(l) → H2SO3(aq)

b

H2SO3(aq) + CaCO3(s) → CaSO3(aq) + CO2(g) + H2O(l)

14

Investigation

15

Investigation

16

Investigation

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Solutions Manual: Module 2 The acidic environment

CHAPTER 7: CHEMICAL EQUILIBRIUM Review exercise 7.1 1

2

3

a

NH3(g) + HCl(g) ↔ NH4Cl(s)

b

2SO2(g) + O2(g) ↔ 2SO3(g)

c

Ag+(aq) + Fe2+(aq) ↔ Ag(s) + Fe3+(aq)

d

2NH3(g) ↔ N2(g) + 3H2(g)

a

Ammonia molecules are reacting with hydrogen chloride molecules to form ammonium chloride at the same rate at which the ammonium chloride is decomposing to ammonia and hydrogen chloride.

b

Sulfur dioxide reacts with oxygen to form sulfur trioxide at the same rate at which the sulfur trioxide decomposes to produce sulfur dioxide and oxygen.

c

Silver ions are reduced to silver at the same rate at which the iron(III) is reduced to iron(II).

d

Ammonia decomposes to nitrogen and hydrogen at the same rate at which hydrogen and nitrogen combine to form ammonia.

a

If 0.21 mol L–1 H2 remains, 0.79 mol L–1 reacted. Molar ratio of H2 : HI is 1 : 2 ∴ 0.79 × 2 = 1.58 moles HI formed ∴ equilibrium conc HI = 1.58 mol L–1

b

i and ii

c

Forward and reverse reactions will occur at equal but opposing rates. The concentrations of reactants and products remain constant.

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Solutions Manual: Module 2 The acidic environment

4

In an open container, the CO2 would escape and the reverse reaction would not occur. In a sealed container, increasing the temperature would increase the pressure until equilibrium was established.

Review exercise 7.2 1

a b

2

a

b

3

a b

4

i

↑ [NO], ↑ [H2O], ↓ [NH3]

ii

Will increase rate of forward reaction until equilibrium is re-established.

i

↓ [NH3], ↓ [O2], ↑ [H2O]

ii

Will increase rate of forward reaction until equilibrium is re-established.

i

Concentration of H2O will decrease, concentration of CO, H2 will increase. No change to [C(s)].

ii

Will increase rate of forward reaction. This increases concentration of CO and H2, decreases [H2O].

i

Concentration of H2O will increase, concentration of CO, H2 will decrease. No change to [C(s)].

ii

Will increase rate of reverse reaction. [CO] ↓, [H2] ↓ and [H2O] ↑ until equilibrium is re-established.

i

↑ [CO], ↑ [O2], ↓ [CO2]

ii

Will increase rate of reverse reaction.

i

↓ [CO], ↓ [O2], ↑ [CO2]

ii

Will increase rate of forward reaction.

a

The system would reach equilibrium more quickly.

b

Both forward and reverse rates would increase equally, therefore no change.

Review exercise 7.3 1

CO2(g) ↔ CO2(aq) CO2(aq) + H2O(l) ↔ H2CO3(aq) H2CO3(aq) ↔ H+(aq) + HCO3–(aq) HCO3–(aq) ↔ H+(aq) + CO32–(aq)

2

a

When the bottles are opened, the pressure in the system is decreased, causing a shift in the equilibrium to the left, where there are more moles of gas and so the change is counteracted. This results in bubbles of CO2(g).

b

Cooling the drink shifts the equilibrium to the right (exothermic direction), meaning that more carbon dioxide will be dissolved into the solution. When the

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Solutions Manual: Module 2 The acidic environment

drink is served cold, the rate at which CO2(g) escapes to the atmosphere is decreased.

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Solutions Manual: Module 2 The acidic environment

Chapter 7 Application and investigation 1

a

Changes in concentration

b

Rate of reaction

2

No further change in the colour of the mixture would be detected.

3

At equilibrium, the concentration of all species remains constant, therefore there will be no change in any observable properties. The dynamic process refers to the equal but opposite reactions occurring at the molecular level.

4

concentration, pressure, temperature

5

a

↓ [NO]; ↑ [O2]; ↑ [NO2]

b

Concentration of all species will double.

c

↑ [O3]; ↓ [O2]; ↓ [NO2]

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Solutions Manual: Module 2 The acidic environment

6 7

a

↓ [HBr]; ↓ [O2]; ↑ [H2O]; ↑ [Br2]

b

↑ [HBr]; ↑ [O2]; ↓ [H2O]; ↓ [Br2]

a

i

Shift equilibrium to the right.

ii

Shift equilibrium to the left.

i

Shift equilibrium to the left.

ii

Shift equilibrium to the right.

b 8

9

a

↑Ca(HCO3)2(s); ↓CaO(s); ↓ [H2O]; ↓ [CO2] after initial increase of CO2

b

↓Ca(HCO3)2(s); ↑CaO(s); ↑ [CO2]; ↑ [H2O] after initial decrease of H2O

c

↓Ca(HCO3)2(s); ↑CaO(s); ↑ [CO2]; ↑ [H2O]

d

no change

e

↑Ca(HCO3)2(s); ↓CaO(s); ↓ [CO2]; ↓ [H2O]

a

i

Equilibrium will shift right to counteract the change: ↓ [NO2]; ↑ [SO3]; ↑ [NO].

ii

The forward reaction rate will increase.

i

There are equal numbers of gaseous molecules on each side, so pressure will increase the concentration of all species, but the system will remain at equilibrium.

ii

Forward and reverse reactions increase equally.

i

Concentrations of all species will decrease, but the system will remain at equilibrium.

ii

Forward and reverse reaction rates decrease equally.

i

The equilibrium will shift left in an endothermic direction: ↑ [SO2], ↑ [NO2], ↓ [SO3], ↓ [NO].

ii

Reverse reaction is favoured.

i

No change as a catalyst does not affect the position of equilibrium.

ii

Both forward and reverse reaction rates will increase.

b

c

d

e 10

a

Pressure is decreased, so equilibrium will shift to counteract the change. Therefore CO2(g) is produced and bubbles form.

b

It is an exothermic reaction, so warm temperatures cause the equilibrium to shift to the left and form CO2(g).

c

The H+ in the orange juice causes increased production of carbonic acid. HCO3–(aq) + H+(aq) ↔ H2CO3(aq)

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Solutions Manual: Module 2 The acidic environment

The ↑ [H2CO3] causes the production of more CO2(aq): H2CO3(aq) ↔ CO2(aq) + H2O(l) Therefore the ↑ [CO2(aq)] results in the production of more CO2(g) causing drinks to go flat, as follows: CO2(aq) + 19kJ ↔ CO2(g) 11

Investigation

CHAPTER 8: THE HISTORICAL DEVELOPMENT OF IDEAS OF ACIDS AND BASES Review exercise 8.1 1

a

SO2(g) + H2O(l) → H2SO3(aq)

b

Na2O(s) + H2O(l) → 2NaOH(aq)

2

Hydrochloric acid, HCl; hydrobromic acid, HBr; cyanic acid, HCN

3

Arrhenius theory states that bases are substances which deliver hydroxide ions in water. NaCN is a substance that acts as a base but does not contain a hydroxide group.

4

a

H2SO4(aq) → 2H+(aq) + SO42–(aq) or H2SO4(aq) → HSO4–(aq) + H+(aq), then HSO4–(aq) ↔ SO42– + H+(aq)

5

6

7

b

HNO3(aq) → H+(aq) + NO3–(aq)

a

1

b

2

c

3

d

1

e

1

f

1

a

1

b

3

c

2

HOOCCOOH(aq)

↔

HOOCCOO–(aq)

+

H+(aq)

HOOCCOO–(aq)

↔

–

+

H+(aq)

OOCCOO–(aq)

Review exercise 8.2 1

a

A B–L acid is a proton donor (or has a proton ‘pulled’ from it).

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Solutions Manual: Module 2 The acidic environment

2 3

4

b

A B–L base is a proton acceptor (or ‘takes’ a proton).

a

HCl, HCO3–, NH4+, HClO4, HSO4–

b

F–, SO42–, NH3, PO43–, H2O

a

An amphiprotic substance can either accept or donate a proton, and therefore it can act as both an acid and a base.

b

HPO42–(aq)

+

H2O(l)

↔

H2PO4–(aq)

+

OH–(aq)

HPO42–(aq)

+

H2O(l)

↔

PO43–(aq)

+

H3O+(aq)

+

OH–(aq)

↔

SO42–(aq)

+

H2O(l)

↔

H2SO4(aq)

+

H2O(l)

HSO4–(aq)

Acid

–

Base 5

6

HSO4 (aq)

+

+

H3O (aq)

a

A conjugate base is the species that results when a B–L acid has lost its proton

b

A conjugate acid is the species that results when a B–L base accepts a proton

a

HCO3–/SO42–: HF/F–

b

HSO4–/SO42–: NH4+/NH3

c

HF/F– : H3O+/H2O

Review exercise 8.3 1

a

b

2

a

i

HCl

ii

HCl

iii

HCl

i

H2SO4

ii

H2SO4

iii

H2SO4

i

NaOH

ii

NaOH

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Solutions Manual: Module 2 The acidic environment

b

3

4

5

iii

NaOH

i

Ba(OH)2

ii

Ba(OH)2

iii

Ba(OH)2

a

5 mol L–1 HCl

b

0.1 mol L–1 NaOH

c

5 mol L–1 NH3

d

0.1 mol L–1 CH3COOH

a

HBr(aq)

→

H+(aq) + Br–(aq)

b

H2SO3(aq)

↔

HSO3–(aq) + H+(aq)

c

RbOH(aq)

→

Rb+(aq) + OH–(aq)

d

NaF(aq)

↔

Na+(aq) + F–(aq)

a

small extent

b

large extent

c

small extent

Review exercise 8.4 1

2

a

S2–(aq) + H2O(l) ↔ HS–(aq) + OH–(aq)

b

CO32–(aq) + H2O(l) ↔ HCO3–(aq) + OH–(aq)

c

NH4+(aq) + H2O(l) ↔ NH3(aq) + H3O+(aq)

d

[Fe(H2O)6]3+(aq) + H2O(l) ↔ [Fe(OH)(H2O)5]2+(aq) + H3O+(aq)

e

F–(aq) + H2O(l) ↔ HF(aq) + OH–(aq)

f

HSO4–(aq) + H2O(l) ↔SO42–(aq) + H3O+(aq)

g

ClO– + H2O↔ HClO + OH–

h

CH3COO–(aq) + H2O(l) ↔ CH3COOH(aq) + OH–(aq)

Use Table 8.5. a

neutral

b

acidic due to reaction of NH4+ ion with water

c

basic due to reaction of ClO– ion with water

d

basic due to reaction of PO43– ion with water

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Solutions Manual: Module 2 The acidic environment

e

acidic due to reaction of Al3+ ion (actually will be aquated aluminium ion [Al(H2O)6]3+ with water)

f

acidic due to reaction of H2PO4– ion with water

g

neutral

h

basic due to reaction of CO32– ion with water

Chapter 8 Application and investigation 1

According to Arrhenius, acids are substances that release hydrogen ions in solution. HCl(aq) completely ionises in water to produce H+ and Cl−, and therefore HCl(aq) is an Arrhenius acid. HCl(l) cannot release hydrogen ions and so is not an Arrhenius acid.

2

a

HCl(aq)

→

H+(aq)

+

Cl–(aq)

b

Ca(OH)2(aq)

→

Ca2+(aq)

+

2OH–(aq)

c

H2SO4(aq)

→

HSO4–(aq)

+

H+(aq)

HSO4–(aq)

2– ↔ SO4 (aq)

+

H+(aq)

d

H2SO4(aq)

+

2NaOH(aq)

→

Na2SO4(aq)

+

2H2O(l)

e

Al(OH)3(aq)

+

3HCl(aq)

→

AlCl3(aq)

+

3H2O(l)

a

4

b

2

c

2

d

CH2(COOH)2(aq) + 2NaOH(aq) → CH2(COONa)2(aq) + 2H2O(l)

a

HCl(aq)

+

H2O(l)

→

H3O+(aq)

+

Cl–(aq)

b

CH3COOH(aq) +

H2O(l)

+ ↔ H3O (aq)

+

CH3COO–(aq)

c

NH3(aq)

+

H2O(l)

+ ↔ NH4 (aq)

+

OH–(aq)

d

H2SO4(aq)

+

H2O(l)

→

H3O+(aq)

+

HSO4–(aq)

HSO4–(aq)

+

H2O(l)

+ ↔ H3O (aq)

+

SO42–(aq)

e

HCl(aq)

+

NaOH(aq)

→

+

H2O(l)

a

ClO3–, S2–, NH3, OH–

b

H2CO3, H2S, [Fe(H2O)6]3+, N2H5+

a

As acid: HCO3–(aq)

+

H2O(l)

↔

CO32–(aq)

+

H3O+(aq)

As base: HCO3–(aq)

+

H2O(l)

↔

H2CO3(aq)

+

OH–(aq)

3

4

5 6

NaCl(aq)

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Solutions Manual: Module 2 The acidic environment

7

b

The extent to which the HCO3– ion reacts as a base with water exceeds the degree to which it reacts with water as an acid. (K value is greater for HCO3– as a base.) Hence more OH– ions are produced than H3O+ ions.

a

i

H2C2O4/HC2O4–, H3O+/H2O

ii

small extent

i

H2O/OH–, HCN/CN–

ii

small extent

i

CH3COOH/CH3COO– , HS–/S2–

ii

large extent

i

HCl/Cl–, HF/F–

ii

large extent

b c d 8

9

ClO4–(aq)

+

H3O+(aq)

– ↔ HCOO (aq)

+

H3O+(aq)

Li+(aq)

+

OH–(aq)

+

OH–(aq)

a

HClO4(l)

+

H2O(l)

→

b

HCOOH(l)

+

H2O(l)

c

LiOH(s)

+

H2O(l)

→

d

N2H4(l)

+

H2O(l)

+ ↔ N2H5 (aq)

a

concentrated solution of a strong acid

b

diluted solution of a strong acid

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Solutions Manual: Module 2 The acidic environment

10

11

c

concentrated solution of a weak acid

d

diluted solution of a weak acid

a

CH3COO–(aq)

b

The hydroxide ions produced raise the pH above 7. The ethanoic acid molecule formed has only a very small degree of dissociation.

a

neutral

b

basic

PO43–(aq)

+

H2O(l)

↔

HPO42–(aq)

+

OH–(aq)

c

basic

CO32–(aq)

+

H2O(l)

↔

HCO3–(aq)

+

OH–(aq)

d

acidic

NH4+(aq)

+

H2O(l)

↔

NH3(aq)

+

H3O+(aq)

e

acidic

[Cr(H2O)6]3+(aq) + H2O(l)

↔

[Cr(OH)(H2O)5]2+(aq) + H3O+(aq)

f

basic

SO42–(aq)

+

H2O(l)

↔

HSO4–(aq)

+

OH–(aq)

g

basic

CN–(aq)

+

H2O(l)

↔

HCN(aq)

+

OH–(aq)

h

acidic

NH4+(aq)

+

H2O(l)

↔

NH3(aq)

+

H3O+(aq)

+

H2O(l)

↔

OH–(aq)

CH3COOH(aq) +

12 Acid

Base

Neutralisation

Arrhenius

produces H+ ions in water

produces OH– ions in water

H+ + OH–↔H2O

Brønsted–Lowry

proton donor

proton acceptor

NH3(aq) + H2O↔NH4+ + OH– proton transfer

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Solutions Manual: Module 2 The acidic environment

13

The reaction: HCl(g) + NH3(g) → NH4Cl(s) can be viewed as a Brønsted–Lowry acid– base reaction because a proton is transferred. However, the reaction is not happening in solution and there are no hydroxide ions released, so the reaction cannot be viewed as an Arrhenius acid–base reaction. When nitric acid is mixed with sulfuric acid, for the manufacture of nitroglycerine, the reaction that occurs is H2SO4 + HNO3 ↔ HSO4– + H2NO3+ This is a B–L acid–base reaction because a proton is transferred but it is not an Arrhenius acid–base reaction.

14

Investigation

CHAPTER 9: HYDROGEN ION CONCENTRATION AND THE pH SCALE Review exercise 9.1 1

2

a

KW

=

[H+] [OH–]

1.0 × 10–14

=

[H+] × 1.6 × 10–7

[H+]

=

6.2 × 10–8 mol L–1

b

[H+] is less than 1 × 10–7 mol L–1 ∴ pool is slightly basic

a

[H+]

=

5.0 × 10–3 mol L–1

[NO3–]

=

5.0 × 10–3 mol L–1

[OH–]

=

1 × 10–14 ÷ 5.0 × 10–3

=

2 × 10–12 mol L–1

[H+]

=

1.5 mol L–1

[Cl–]

=

1.5 mol L–1

[OH–]

=

1 × 10–14 ÷ 1.5

=

6.7 × 10–15 mol L–1

[K+]

=

0.25 mol L–1

[OH–]

=

0.25 mol L–1

[H+]

=

4 × 10–14 mol L–1

[Ba2+]

=

6.0 × 10–2 mol L–1

[OH–]

=

0.12 mol L–1

[H+]

=

8.3 × 10–14 mol L–1

b

c

d

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Solutions Manual: Module 2 The acidic environment

Review exercise 9.2 1

a

b

pH

+

[H ] pH

c d

pH [H+] pH

2

3

=

–log[H+]

=

–log[5.0 × 10–1]

=

0.30

=

1 × 10–14 ÷ 0.0065

=

1.54 × 10–12 mol L–1

=

–log[1.54 × 10–12]

=

11.81

=

–log[3.6 × 10–3]

=

2.44

=

1 × 10–14 ÷ (6.5 × 10–4 × 2)

=

7.7 × 10–12 mol L–1

=

–log[7.7 × 10–12]

=

11.1

+

a

pH = 3.50; [H ] = 3 × 10–4 mol L–1; [OH–] = 3.16 × 10–11 mol L–1

b

pH = 11.90; [H+] = 1.3 × 10–12 mol L–1; [OH–] = 7.9 × 10–3 mol L–1

c

pH = 0.80; [H+] = 0.16 mol L–1; [OH–] = 6.3 × 10–14 mol L–1

pH = 7 so [H+] = 1 × 10–7 pH = 5 so [H+] = 1 × 10–5 ∴ pH changes by factor of 100

Review exercise 9.3 1

a

HClO2

b

HClO2

c

HClO2, HF, HCN

d

No. Complete ionisation of 0.1 mol L–1 gives pH = 1.

Review exercise 9.4 1

a

A buffer is a species that resists changes to pH despite small additions of acid or base.

b

A buffer contains approximately equal amounts of a weak acid and its conjugate base (some are a mixture of weak bases and their conjugate acids).

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Solutions Manual: Module 2 The acidic environment

2

3

a

Buffered; contains approximately equal amounts of the weak acid HSO4– and its conjugate base SO42–.

b

Not buffered; HNO3 is a strong acid.

c

Buffered; contains approximately equal amounts of the weak acid HClO and its conjugate base ClO–.

a

HPO4–(aq) + H3O+(aq) ↔ H2PO4–(aq) + H2O(l) When excess hydrogen ions are added, the system shifts to the right and thus decreases the H+ concentration.

b

H2PO4+(aq) + OH–(aq) ↔ HPO4–(aq) + H2O(l) When excess hydroxide is added, the system shifts to the right and thus decreases the OH– concentration.

4

The concentration of the acid and conjugate base ions is limited. If too much H+ or OH– is added then the system is unable to provide sufficient acid or conjugate base particles to counteract their effect.

5

Lactic acid in the blood produces lactate ions and hydrogen ions. The blood’s buffering system H2CO3(aq) + H2O(l) ↔ HCO3–(aq) + H3O+(aq) responds by combining the extra hydrogen ions with the hydrogencarbonate ions, thus shifting the system to the left and decreasing their concentration.

Chapter 9 Application and investigation 1

a

[H+]

=

0.5 mol L–1

[OH–]

=

1 × 10–14 ÷ 0.5

=

2 × 10–14 mol L–1

=

–log[0.5]

=

0.30

=

3.0 × 10–3 × 2

=

6.0 × 10–3 mol L–1

=

1 × 10–14 ÷ 0.006

=

1.67 × 10–12 mol L–1

=

–log[1.67 × 10–12]

=

11.8

pH b

[OH–] [H+] pH

2

a

[H+]

=

1.0 × 10–3 mol L–1

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Solutions Manual: Module 2 The acidic environment

[OH–] b c d

3

a

=

1 × 10–14 ÷ 0.001

=

1 × 10–11 mol L–1

[H+]

=

3.9 × 10–3 mol L–1

[OH–]

=

2.51 × 10–12 mol L–1

[H+]

=

3.2 × 10–9 mol L–1

[OH–]

=

3.2 × 10–6 mol L–1

[H+]

=

7.9 × 10–13 mol L–1

[OH–]

=

1.3 × 10–2 mol L–1

i

ii b

[H+]

=

0.20 mol L–1

[NO3–]

=

0.20 mol L–1

[HNO3]

=

0 mol L–1

pH

=

–log10[0.20]

=

0.70

As HNO2 is a weak acid, it will not dissociate completely. Therefore [H+] will be less than 0.20 mol L–1 and pH will be higher than 0.70.

4

H2CO3, H3PO4, HClO2, H2SO3, HCl

5

Investigation

6

Investigation

CHAPTER 10: VOLUMETRIC ANALYSIS AND ESTERIFICATION Review exercise 10.1 1

2

a

NaOH(aq)

+

HCl(aq)

→

NaCl(aq)

+

H2O(l)

b

2NaOH(aq)

+

H2SO4(aq)

→

Na2SO4(aq)

+

2H2O(l)

c

2NaOH(aq)

+

H2CO3(aq)

→

Na2CO3(aq) +

H2O(l)

d

3NaOH(aq)

+

H3PO4(aq)

→

Na3PO4(aq)

+

H2O(l)

a

2HCl(aq)

+

Ca(OH)2(aq) →

CaCl2(aq)

+

2H2O(l)

b

3HCl(aq)

+

Al(OH)3(s)

→

AlCl3(aq)

+

3H2O(l)

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Solutions Manual: Module 2 The acidic environment

c

2HCl(aq)

+

MgO(s)

→

MgCl2(aq)

+

H2O(l)

d

2HCl(aq)

+

Na2CO3(s)

→

2NaCl(aq)

+

CO2(g)

+ H2O(l)

3

Neutralisation is the result of an acid–base interaction. In Brønsted–Lowry terms this refers to proton transfer from the acid to the base.

4

Neutralisation results in the formation of a new bond as the proton bonds to its base. Bond formation is an exothermic process, and thus neutralisation is an exothermic process.

5

a

Add solid sodium hydrogencarbonate until the ‘fizzing’ stops. The reaction produces carbon dioxide, so when the fizzing stops, this indicates that the neutralisation is complete.

b

Rinse with copious amounts of water. A base should not be used, as the neutralisation reaction is exothermic—i.e. it releases a lot of heat.

Review exercise 10.2 1

2OH–(aq) + H2SO4(aq) → 2H2O(l) + SO42–(aq) Calculate moles of OH– n(OH–) =

cV

=

1.90 × 0.02368

=

0.045 moles

∴ moles of H2SO4 : n(H2SO4) = =

0.5n(OH–) (from equation) 0.0225 moles

∴ concentration of H2SO4 n(H2SO4) = 0.0225 = c = 2

cV c × 0.005 4.50 mol L–1

NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) Calculated moles of HCl n =

cV

=

0.02052 × 0.952

=

0.019535 moles

moles of HCl =

moles NaOH (from equation)

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Solutions Manual: Module 2 The acidic environment

∴ concentration of NaOH = =

c = 3

4

0.019535 / 0.005 3.91 mol L–1 NaOH

a

Pipettes are more accurately calibrated than a measuring cylinder; this accuracy is important in volumetric analysis.

b

If rinsed with water, any drops left in the pipette or burette will alter the concentration of the solution and this will affect the results. Rinsing with the solution will not have this effect.

c

Volume may change slightly once the solid is dissolved; the customary unit of concentration is mol/L, where the volume refers to litres of solution, not water.

a

No. moles Na2CO3 = = moles = so 0.25 =

b

0.025 moles

m M m 105.99 2.65 g of Na2CO3 is dissolved in water in a 250 mL volumetric flask with solution made up to the calibrated mark

n =

cV

=

n = 7.5 × 10–3 =

m =

a

0.250 × 0.1

m =

nH2C2O4.2H2O =

5

n V

n =

0.5 × 0.015 7.5 × 10–3 moles

m M m 126.07 0.95 g oxalic acid is dissolved in water in a 500 mL volumetric flask with the solution made up to the calibrated mark m M

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Solutions Manual: Module 2 The acidic environment

=

0.3162 126.07

=

0.0025 moles

n = 0.0025 = c = b

c × 0.25 0.010 mol L–1 H2C2O4.2H2O

2NaOH(aq) + H2C2O4(aq) → Na2C2O4(aq) + 2H2O(l) n(H2C2O4) = n = =

cV 0.010 × 0.02061 2.068 × 10–4 moles H2C2O4

n(OH–) =

2 × nH2C2O4 (from equation)

n(OH–) =

4.136 × 10–4 mol

n(NaOH) = –4

4.136 × 10

=

c = c

cV

cV c × 0.020 0.0207 mol L–1 NaOH

NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) n(NaOH) = n = = volume of HCl used =

cV 0.0207 × 0.020 4.14 × 10–4 mol NaOH 20.26 – 1.83

=

18.43 mL

=

0.01843 L

nHCl = n(HCl) = nHCl = 4.14 × 10–4 = c =

nNaOH from equation 4.14 × 10–4 moles cV c × 0.01843 2.24 × 10–2 HCl mol L–1

Review exercise 10.3 1

a

i

NaF(aq)

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Solutions Manual: Module 2 The acidic environment

b

c

2

i

ii

iii

ii

basic

iii

phenolphthalein

i

KNO3(aq)

ii

neutral

iii

methyl orange/phenolphthalein

i

NaCl(aq)

ii

acidic

iii

methyl orange

a

Phenolphthalein; as a strong acid/strong base titration and endpoint will be around 7.

b

Colourless in acid changing to pink as base added, with pale pink at equivalence point.

a

Methyl orange; as strong acid/weak base titration and endpoint will be less than pH 7.

b

Yellow in base and changing to orange at equivalence point, and to red as point is passed.

a

Bromothymol blue; weak acid/strong base titration and endpoint will be greater than pH 7.

b

Yellow in acid changing to blue as base added, with pale yellow at equivalence point.

3

This is a weak acid/strong base titration with an equivalence point above pH 7. If methyl orange was used as an indicator, a colour change would be observed at around pH 3.1–4.4 and an equivalence point assumed, so the volume of hydroxide recorded would be too low.

4

a

Running this strong base into the weak acid.

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Solutions Manual: Module 2 The acidic environment

b

Running this weak base into the strong acid.

c

Running this strong base into the strong acid.

Review exercise 10.4 1

a

b

2

a

=

2.5 V = 22.71 22.71

=

0.11 mol H2

n

=

m M

0.11

=

m 2.016

=

0.22 g hydrogen gas

=

m 48.0 = M 32

=

1.5 mol

n

n

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Solutions Manual: Module 2 The acidic environment

b

3

Zn(s) 1 a

n

=

V 22.71

1.5

=

V 22.71

v

=

34.07 L at 0°C and 100 kPa

n

=

V 24.79

1.5

=

V 24.79

v

=

37.19 L at 25°C and 100 kPa

+

2HCl(aq)

→ ZnCl2(aq)

2

1

H2(g) 2

V 24.79

nH2 n = =

2.5 24.79 0.1 moles H2

From equation nZn =

0.1 (nH2)

=

0.1 moles

nZn = 0.1 = = b

+

From equation nHCl =

m M m 65.4 6.54 g Zn required 2 × nH2

=

2 × 0.1

=

0.2 moles

nHCl = 0.2 = V =

cV 1.5 × V 0.13 L HCl required

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Solutions Manual: Module 2 The acidic environment

4

CaCO3(s) 1 a

+ 2HNO3(aq)

→ Ca(NO3)2(aq)

2 nCaCO3 =

From equation nCaCO3 = ∴ nCO2 =

2.271 L CO2 at 0°C and 100 kPa

nCO2 =

V 24.79

0.1 =

V 24.79

= nHNO3 = 0.2 = V = d

0.10 moles

V 22.71

∴ nHNO3 =

2.48 L CO2 at 25°C and 100 kPa 2 × nCaCO3 2 × 0.1 0.2 moles cV 1.00 × V 0.2 L HNO3 required

From equation nCaCO3 =

nCa(NO3)2

∴ nCa(NO3)2 =

0.10 moles

nCa(NO3)2 = 0.1 = m =

1

nCO2

0.1 =

From equation nHNO3 =

1

0.10 moles

V 22.71

V =

+ H2O(l)

m 10 = M 100.1

nCO2 =

V =

c

CO2(g)

1

n =

b

+

m M m 164.1 16.41 g of Ca(NO3)2 produced

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Solutions Manual: Module 2 The acidic environment

Review exercise 10.5 1

2

a

1-propanol

b

1-butanol

c

2-butanol

a

b

c

3 4

a

pentanoic acid

b

5-methylhexanoic acid

a

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Solutions Manual: Module 2 The acidic environment

b

5

They form more H-bonds because they have a C=O group as well as an −OH group, making two H-bonds possible between two molecules.

6

a

methyl butanoate

b

ethyl methanoate

7

a

b

8

a

b

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Solutions Manual: Module 2 The acidic environment

c

d

e

f

9

a

conc. H SO 4 C3H7OH + CH3COOH ⎯ ⎯ ⎯ ⎯2 ⎯⎯ → CH3COOC3H7 + H2O The ester is propyl ethanoate.

b

conc. H SO 4 C5H11OH + C3H7COOH ⎯ ⎯ ⎯ ⎯2 ⎯⎯ → C3H7COOC5H11 + H2O The ester is pentyl butanoate.

10

a

H2SO4 acts as a catalyst.

b

Refluxing prevents volatile reactants from escaping before equilibrium is reached and allows the reaction to be carried out at temperatures lower than would normally be possible.

Chapter 10 Application and investigation 1

a

2HCl(aq)

+ CuO(s)

→

CuCl2(aq)

+ H2O(l)

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Solutions Manual: Module 2 The acidic environment

2

b

3H2SO4(aq)

+ 2Fe(OH)3(s)

→

Fe2(SO4)3(aq)

+ 6H2O(l)

c

2HNO3(aq)

+ CaCO3(s)

→

Ca(NO3)2(aq)

+ CO2(g)

+ H2O(l)

d

HCl(aq)

+ NaHCO3(aq) →

NaCl(aq)

+ H2O(l)

+ CO2(g)

a

i

Na+(aq), SO42–(aq), OH–(aq), H+(aq), H2O(l)

ii

H+(aq), Na+(aq), OH–(aq) all at 2 mol L–1

i

H+(aq), SO42–(aq), H2O(l)

ii

H+(aq)

i

Na+(aq), SO42–(aq), H2O(l)

ii

Na+(aq), SO42–(aq)

i

Pipette: used to place an accurate volume of solution into a conical flask.

ii

Burette: used to deliver variable volumes into a conical flask.

iii

Conical flasks: used to mix two solutions.

i

citric acid

ii

NaOH

iii

deionised water

b c 3

a

b

4

Sodium hydroxide cannot be made as a primary standard and hence must be standardised by titration with an acid of accurately known concentration.

5

a

A good primary standard has a relatively high molecular weight to minimise weighing errors, does not react with any component of the atmosphere such as oxygen, moisture or carbon dioxide, and is available in a very pure form.

b

Investigation

6

a

b

=

2×1

=

2 moles

n

=

m M

2

=

m 84.008

m

=

168 g

=

0.05 × 0.5

nNaHCO3 = cV

nNa2CO3 = cV n

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Solutions Manual: Module 2 The acidic environment

c

d

7

a

=

0.025 moles

n

=

m M

0.025

=

m

=

2.65 g

nBa(OH)2.8H2O

=

cV

n

=

2 × 10–3 × 2

=

0.004

n

=

m M

m

=

0.004 × 315.444

=

1.26 g

nH2C2O4.2H2O

=

cV

n

=

0.150 × 0.25

=

0.0375 moles

n

=

m M

m

=

0.0375 × 126.068

=

4.728 g

NaOH(aq) + HNO3(aq) → NaNO3(aq) + H2O(l) nHNO3 = cV

= 0.0961 × 0.02 = 1.9 × 10–3 From equation nNaOH = nHNO3 = 1.9 × 10–3 nNaOH = cV

1.9 × 10–3 = 0.118 × V V = 1.62 × 10–2 L

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Solutions Manual: Module 2 The acidic environment

b

2NaOH(aq) + H2SO3(aq) → Na2SO3(aq) + 2H2O(l) nH2SO3 = cV

= 0.0531 × 0.02 = 1.062 × 10–3 moles nNaOH = 2 × nH2SO3 from equation

2 × 1.062 × 10–3 = 0.118 × V V = 18 × 10–3 L c

2NaOH(aq) + H2C2O4.2H2O(s) → Na2C2O4(aq) + H2O(l) nH2C2O4.2H2O

=

m M

=

0.134 126

= 1 × 10–3 moles 2 × n(oxalic acid) = n(NaOH) n = cV

2 × 10–3 = 0.118 × V V = 16.95 × 10–3 L 8

a

H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l) nNaOH = cV

= 0.5 × 0.02 = 0.01 moles nH2SO4 = 0.5nNaOH

= 0.005 moles H2SO4 n = cV V =

0.005 0.202

= 2.48 × 10–2 L b

H2SO4(aq) + Ba(OH)2 → BaSO4(aq) + 2H2O(l) nBa(OH)2 = 0.02 × 0.165

= 0.0033 moles © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2006. This page from Chemistry Contexts 2 Teacher’s Resource second edition may be reproduced for classroom use.

Solutions Manual: Module 2 The acidic environment

nBa(OH)2 = nH2SO4

∴ nH2SO4 = 0.0033 = 0.202 × V V = 1.63 × 10–2 L c

Na2CO3(s) + H2SO4(aq) → Na2SO4(aq) + CO2(g) + H2O(l) nNa2CO3

=

m M

=

0.343 106

= 3.24 × 10–3 moles nNa2CO3 = nH2SO4 nH2SO4: 3.24 × 10–3 = 0.202 × V V = 1.60 × 10–2 L d

total nOH– = nOH– in Ca(OH)2 + nOH– in NaOH nCa(OH)2 =

=

m M

0.0746 74.1

= 1 × 10–3 moles moles OH– = 1 × 10–3 × 2 = 2 × 10–3 moles nNaOH =

=

m M

0.247 40

= 6.2 × 10–3 moles OH– Total nOH– = 2 × 10–3 + 6.2 × 10–3 = 8.18 × 10–3 moles nH2SO4 = nOH– × 0.5

= 4.08 × 10–3 moles n = cV © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2006. This page from Chemistry Contexts 2 Teacher’s Resource second edition may be reproduced for classroom use.

Solutions Manual: Module 2 The acidic environment

= 4.08 × 10–3 = 0.202 × V VH2SO4 = 2.02 × 10–2 L 9

a

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) nHCl = cV

= 0.121 × 0.02565 = 3.1 × 10–3 moles nHCl = nNaOH nNaOH = cV

3.1 × 10–3 = c × 0.020 c = 0.16 mol L–1 b

Ba(OH)2(aq) + 2HCl(aq) → BaCl2(aq) + 2H2O(l) nHCl = cV

= 0.0503 × 0.01523 = 7.66 × 10–4 moles nBa(OH)2 = 0.5nHCl

= 0.5 × 7.66 × 10–4 = 3.83 × 10–4 moles Ba(OH)2 = 3.83 × 10–4 ÷ 0.02 = 1.92 × 10–2 mol L–1 10

a

H2SO4(aq) + CuO(s) → CuSO4(aq) + H2O(l) nCuO =

m 1.05 = M 79.5

= 0.013 moles nCuO = nH2SO4 nH2SO4 = 0.13 moles = 2.0 × V V = 6.6 × 10–3 L b

1 mole H2SO4 → 1 mole CuSO4.5H2O ∴ nCuSO4.5H2O = 0.013 moles

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Solutions Manual: Module 2 The acidic environment

m M

n =

m 249.6

0.013 =

m = 3.24 g 11

Investigation

12

A known mass of the tablets could be dissolved in 20 mL of water. This solution could be titrated against HCl of known concentration to determine the concentration of NaHCO3. nNaOH

=

cV

=

0.119 × 0.02053

=

2.4 × 10–3 moles

∴ n vitamin C

=

2.44 ×10–3 moles

n

=

m M

2.44 × 10–3

=

m 176

m

=

0.4294 g

% in capsule

=

0.4294 × 100 0.450

=

95.4%

13

14

a

NaOH 1

+

HCl

→ NaCl

+

H2O

1 nHCl

nNaOH

c

=

cV

=

0.125 × 0.01759

=

0.0022 moles

=

nHCl

=

0.0022 moles

=

n 0.0022 = V 0.020

=

0.11 mol L–1

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Solutions Manual: Module 2 The acidic environment

NaOH + CH3COOH → NaCH3COO + H2O nNaOH

=

cV

=

0.11 × 0.03812

=

4.2 × 10–3 moles

=

nCH3COOH

=

4.2 × 10–3

=

4.2 × 10–3

=

m M

4.2 × 10–3

=

m 60

m

=

0.25 g in 5.00 mL sample vinegar

nNaOH

nCH3COOH

b

% mass = 0.25/5.00 × 100 = 5% ethanoic acid in vinegar

15

NH3(aq) + HCl(aq) → NH4Cl(aq) nHCl = cV

= 0.110 × 0.02462 = 2.7 × 10–3 moles ∴ nNH3 = 2.7 × 10–3

= cV 2.7 × 10–3 = c × 0.02 c = 0.135 mol L–1 ammonia

Number of moles NH3 present in 250 mL n

n

=

cV

=

0.135 × 0.25

=

0.339 moles

=

m M

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Solutions Manual: Module 2 The acidic environment

0.339

=

m 17

m

=

0.575 g ammonia in 9.97 g

% mass ammonia in household ammonia =

0.575 × 100 9.97

= 5.77% 16

Na2CO3 + H2O + 2HCl(aq) → 2NaCl(aq) + 2H2O(l) + CO2(g) nHCl = cV

= 0.104 × 0.02520 = 2.62 × 10–3 moles nNa2CO3 = 0.5nHCl nNa2CO3 = 0.5 × 2.62 × 10–3

= 1.31 × 10–3 moles nNa2CO3 =

1.31 × 10–3 =

m M m 106

m = 0.1389 g

Mass of water in washing soda = 0.374 – 0.1389 = 0.235 g H2O nH2O = n =

m M

0.235 18

n = 0.0131 moles –3 –2 Ratio of nNa2CO3 : nH2O = 1.31 × 10 : 1.31 × 10 = 1 : 10 ∴Number of molecules of water of crystallisation = 10

i.e. Na2CO3.10H2O

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Solutions Manual: Module 2 The acidic environment

17

The pH at the equivalence point is below 7 due to the reaction of NH4+ produced at the equivalence point reacting with H2O to form H3O+. Phenolphthalein changes colour above 7, and therefore it would not be appropriate.

18

a

b

c

d

e

19

a

i

Na+, CH3COO–, H2O, CH3COOH, OH–

ii

basic

iii

phenolphthalein

i

KCl(aq), H2O

ii

neutral

iii

bromothymol blue

i

NH4+, NO3–, NH3, H3O+

ii

acidic

iii

methyl orange

i

BaCl2(aq), H2O

ii

neutral

iii

bromothymol blue

i

H2CO3(aq), NaCl, H3O+, HCO3–

ii

acidic

iii

methyl orange

i

nO2

=

V 22.71

0.500

=

V 22.71

V

=

11.36 L O2 at 0ºC and 100 kPa

nO2

=

V 24.79

0.500

=

V 24.79

V

=

12.4 L O2 at 25ºC and 100 kPa

nN2

=

V 22.71

=

2 22.71

ii

b

i

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Solutions Manual: Module 2 The acidic environment

ii

20

a

i

ii

=

0.09 moles N2 at 0ºC and 100 kPa

=

V 24.79

=

10 24.79

=

0.40 moles N2 at 25ºC and 100 kPa

=

V 22.71

=

3 22.71

=

0.13 moles

nCl2

=

m M

0.13

=

m 71

m

=

9.5 g Cl2 at 0ºC and 100 kPa

nCl2

=

V 24.79

nN2

nCl2

3 24.79

b

i

=

0.12 moles

n

=

m M

0.12

=

m 71

m

=

8.7 g Cl2 at 25ºC and 100 kPa

nCO2

=

m M

=

55 44

=

1.25 mol

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Solutions Manual: Module 2 The acidic environment

ii

21

a

n

=

V 22.71

1.25

=

V 22.71

V

=

28.39 L CO2 at 0ºC and 100 kPa

nCO2

=

1.25 mol

n

=

V 24.79

1.25

=

V 24.79

V

=

30.99 L CO2 at 25ºC and 100 kPa

nFeS2

=

m M

=

1000 120

= 8.33 mol From equation, FeS2:O2 equals 1:2.75 nO2

= 2.75 × nFeS2 = 2.75 × 8.33 = 22.92 mol O2

b

i

nO2

=

V 22.71

22.92

=

V 22.71

V

= 520.5 L O2 at 0ºC and 100 kPa

nSO2

= 2 × nFeS2 = 2 × 8.33 = 16.67 moles

At 0°C and 101.3 kPa:

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Solutions Manual: Module 2 The acidic environment

ii

22

V 22.71

16.67

=

V

= 378.6 L SO2 produced at 0ºC and 100 kPa

At 25°C and 101.3 kPa: V 24.79

16.67

=

V

= 413.3 L SO2 produced at 25ºC and 100 kPa

a

CH3CH2CH3; CH3CH2CH2OH; CH3CH2COOH

b

propane

–41°C

1-propanol

97°C

propanoic acid

141°C

Propane has only weak dispersion forces between molecules ∴ lowest boiling point. 1-propanol has strong H-bonds between molecules but propanoic acid has stronger intermolecular forces due to two H-bonds between acid molecules. 23

a

b

24

a

Reflux ethanol with butanoic acid with concentrated H2SO4 as a catalyst as an esterification process

b

Reflux pentanol with propanoic acid with concentrated H2SO4 as a catalyst as an esterification process

Refluxing is necessary so that none of the volatile components of the reaction are lost while this slow, low-yielding reaction proceeds. 25

Investigation © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2006. This page from Chemistry Contexts 2 Teacher’s Resource second edition may be reproduced for classroom use.

Solutions Manual: Module 2 The acidic environment

Module 2 REVIEW 1

A

2

D

3

D

4

B

5

D

6

A

7

B

8

C

9

A

10 11

B a

CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l)

b

nNaOH

=

0.095 × 0.02

=

1.9 × 10–3 moles

=

18.9 + 18.8 + 19.0 3

=

18.9 mL

=

1.9 × 10–3 = c × 0.0189

=

0.1005 mol L–1 acetic acid

c1V1 (original)

=

c2V2 (volumetric flask)

c × 0.02

=

0.1005 × 0.2

=

1.005 mol L–1

Average

nCH3COOH c c

c

12

d

Phenolphthalein; pH at end point will be above 7.

e

Distilled water. The number of moles of vinegar added will not be affected by water in the flask.

NH3(g) + H2O(l) → NH4+(aq) + OH–(aq) The OH– causes the solution to be basic. NH4+(aq) + H2O(l) → NH3(aq) + H3O+(aq) The H3O+ causes the solution to be acidic. © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2006. This page from Chemistry Contexts 2 Teacher’s Resource second edition may be reproduced for classroom use.

Solutions Manual: Module 2 The acidic environment

13

14

15

a

HCl is a strong acid which dissociates completely and therefore has high conductivity and low pH. Acetic acid is weak and therefore only slightly dissociates, causing a higher pH and lower conductivity.

b

The CH3COO– reacts with H2O to form some OH– ions.

c

Greater number of ions are present in the solution in Beaker 3.

a

Bacteria can decompose organic matter to produce H2S, which then oxidises to SO2. Burning of fossil fuels and smelting of sulfide ores results in sulfur compounds which are oxidised to SO2.

b

2CuFeS2(s) + 5O2(g) + 2SiO2(s) → 2Cu(l) + 4SO2(g) + 2FeSiO3(l)

c

i

SO3(g) + H2O(l) → H2SO4(aq) sulfuric acid

ii

Lakes and waterways become acidic, killing aquatic life. Soil can become acidic, affecting plant growth. Limestone can be dissolved by the action of acid rain. These are all of concern if we want to protect our environment.

iii

Student opinion

a

Both molecules have an −OH group but ethanoic acid contains a C=O group, which increases the intermolecular forces. Even though 1-butanol has more dispersion forces than acetic acid, acetic acid has more H-bonding and so the intermolecular forces are very nearly the same in each substance.

b

i

CH3COOH(aq) + C4H9OH(l) → CH3COOC4H9(aq) + H2O(l) Catalyst is conc. H2SO4.

ii

H2SO4 acts as a catalyst.

c

Refluxing allows the reaction mixture to be boiled without losing the volatile reactants and increases the time over which this quite slow reaction can occur.

d

Acetic acid would be more soluble than 1-butanol. Compound X would be least soluble.

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Solutions Manual: Module 2 The acidic environment

e

16

a

b

i

Equilibrium shifts to the left and more CO2(g) is produced ∴ the drink goes flat.

ii

More space above the liquid means that pressure is lower and CO2(aq) → CO2(g) and the drink will be flat.

The soft drink could be carefully poured into a beaker and gently heated. If bubbles of CO2 are observed forming in the solution (which they should be), this would indicate that the equilibrium is shifting to the left (CO2(g) ↔ CO2(aq)). As the addition of heat favours the reverse reaction, it follows that this reaction must be endothermic. Therefore the forward reaction must be exothermic. Alternatively: Take an unopened bottle of soft drink. Immediately upon opening it, insert a thermometer (at the same initial temperature as the drink) and observe any change in temperature. Bubbles of CO2 gas should be observed forming in the solution, indicating that the equilibrium is shifting to the left (as written). If a temperature decrease is recorded (which it should), this will indicate that the reverse reaction is endothermic, as heat is being absorbed from the surroundings (the drink itself). Therefore the forward reaction must be exothermic.

17

a

An acid loses an H+ to become its conjugate base, e.g. C6H5N2O5(aq) → C6H4N2O5–(aq).

b

Yellow. The CN– would react with the H3O+, causing the equilibrium to shift to the right. CN–(aq) + H3O+ → H2O(l) + HCN(aq)

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Solutions Manual: Module 2 The acidic environment

c

18

a

i

Yes: strong acid–strong base titration. pH = 7 at end point and the pH changes from 11 to 3 very rapidly.

ii

Yes: weak base–strong acid; pH < 7 at end point.

iii

No: weak acid–strong base; pH > 7 at end point.

[H+] = 0.12 mol L–1 pH = –log10[H+] = –log10[0.12] pH = 0.92

b

Mg(OH)2(aq) + 2HCl(aq) → MgCl2(aq) + 2H2O(l) The antacid reacts with excess acid to produce salt and water.

19

20

a

Curve 1

b

The acetate ion reacts with water to form OH– and CH3COOH, raising the pH at equivalence point. There is no such reaction in the NaCl(aq) produced at the equivalence point where pH = 7.

c

Starting pH is higher for acetic acid due to less dissociation.

a

CH3COOH(aq) ↔ CH3COO–(aq) + H+(aq) If CH3COO– is added then the equilibrium will shift left, resulting in less H+ and a higher pH. Also CH3COO–(aq) + H2O(l) ↔ CH3COOH(aq) + OH–(aq). The presence of OH– ions raises the pH.

b

Buffering systems result from the action of a weak acid in approximately equal concentration with its conjugate base. Acetic acid is a weak acid and the acetate ions from the sodium acetate would be close in concentration to the acetic acid. Thus the resulting solution could act as a buffering system. n (SO2) = 2 × n (CuFeS2)

21

= 2 × mass (CuFeS2) ÷ molar mass (CuFeS2) Vol (SO2) = 2 × mass (CuFeS2) ÷ molar mass (CuFeS2) × molar volSATP = 2 × 500 × 103 ÷ 183.54 × 24.79 = 135066 L 22

a

Decreased pressure would cause the equilibrium to shift to the left. This would result in the formation of CaCO3(s) and CO2(g).

b

Change in temperature. Increased temperature would cause this exothermic reaction to shift to the left to counteract the change. This would cause a decrease in [Ca2+].

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Solutions Manual: Module 3 Chemical monitoring and management

CHAPTER 11: THE ROLE OF CHEMISTS AND CHEMICAL ANALYSIS Review exercise 11.1 1

Most products used in the home, car etc, e.g. shampoo, medications, food additives, laundry detergents

2

Investigation

3

Energy industry, mineral industry, petrochemical industry

Review exercise 11.2 1

Qualitative analysis identifies substances present in a sample. Quantitative analysis determines the amount of each substance present.

2

Treating substances with electromagnetic radiation and measuring the amount of absorption or emission.

3

a

AAS is most useful for determining concentrations of cations in samples. The sample to be tested is vaporised in a flame. A beam of light emitted by a cathode lamp of the element being investigated is shone through the vaporised sample. The element absorbs that wavelength of radiation and the remaining light is analysed to determine the intensities remaining. By comparing to known samples gathered under the same conditions, concentrations can be identified.

b

AAS is a relatively cheap and rapid process that can detect very low concentrations (ppm and ppb) of elements, and therefore is useful in detecting trace elements.

4

Concentration of Pb in solution

=

5.5 mg kg–1 (from the graph)

Mass of Pb in solution

=

5.5 × 0.1

=

0.55 mg

=

0.55 0.002

=

275 ppm

Concentration of Pb in soil

Review exercise 11.3 1

Their electrons occupy different energy levels and will absorb then emit light of different frequencies when they fall back from higher to lower energy levels.

2

a

i

Ba2+

ii

Add to H2SO4 solution; a white precipitate of BaSO4(s) will form.

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3

b

Add HCl to the solution. Ca2+ ions will have no reaction ∴ if a precipitate forms, Pb2+ is present.

a

i

CO32–

ii

Add HCl to form CO2(g) and confirm CO2(g) by turning limewater milky.

b

Add AgNO3. AgCl(s) is white, Ag3PO4(s) is yellow. When HNO3 is added, Ag3PO4 will dissolve.

Review exercise 11.4 1

nAl2 O3

nAl

mass Al

% mass Al in sample

2

a

nBaSO4

nSO42–

mass SO42–

% mass SO42–

=

=

=

m M

=

0.153 102

=

1.5 × 10–3 moles

=

2 × nA12O3

=

2 × 1.5 × 10–3

=

3 × 10–3 moles

=

3 × 10–3 × 27

=

0.081 g

=

0.081 × 100 0.516

=

15.7% m M

nS

=

2.01 233.4

=

8.6 × 10–3 moles

=

nBaSO4

=

8.6 × 10–3 moles

=

8.6 × 10–3 × 96.1

=

0.83 g

=

0.83 × 100 1.23

=

67.3%

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=

8.6 × 10–3 × 32.1

=

0.28 g

=

0.28 × 100 1.23

=

22.5%

=

100 – 67.3 – 10.1

=

22.6%

∴ 22.6% of 1.23 g

=

0.28 g

nNH4+

=

mass S

% mass S

b

% NH4

+

∴ mass N

% mass N

3

n(NaOH)

=

1.5 × 10–2 moles

=

1.5 × 10–2 × 14

=

0.22 g

=

0.22 × 100 1.23

=

17.6%

= [HCl] × volHCl = 0.1690 × 30.15 × 10−3 = 0.00509525 mol in 10 mL after dilution = 0.00509525 mol in 1 mL before dilution

[NaOH]

= 0.00509525 ÷ 1 × 10−3 = 5.1 mol/L

Chapter 11 — Application and investigation 1

The industrial process would involve a much larger scale using a great deal more reactants. It would require monitoring to ensure production was economical. The process would be continually reviewed and refined.

2

Investigation

3

Investigation

4

Investigation

5

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Solutions Manual: Module 3 Chemical monitoring and management

6

7

a

If chemists are asked, for example, to participate in an environmental impact study for a particular area then they will need to collaborate to present a comprehensive report, as no one chemist can be a specialist in all the areas of analysis that would be required.

b

A chemist might not collaborate if asked to do a simple analytical task in which they have great expertise.

a

b 8

9

3.25 mg L−1

Absorbance = 0.720 Concentration of mercury from curve

=

0.290 ppm = 0.290 mg/kg

Mass of mercury in 100 mL (assume 1 mL = 1 g)

=

0.290 × 0.1 = 0.029 mg

Concentration of mercury in tuna

=

0.029 0.002 mg/kg

=

14.5 ppm

a

nAgC1 = n = nAgC1 =

0.276 143.4 1.9 × 10–3 moles nNaC1

∴ nNaC1 =

1.9 × 10–3 moles

mass NaCl =

1.9 × 10–3 × 58.5

=

0.113 g

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b

% mass of salt in chips = =

0.113 × 100 7.5 1.5%

10

6.25 ppm 11

The cation is most likely Pb2+. This can be confirmed by adding potassium iodide solution. If a bright yellow precipitate forms, then the cation is Pb2+.

12

a

Anion could be CO32–. Confirm by adding acid—if a gas is produced that turns limewater cloudy, then can conclude that carbonate is present.

b

CO32– and SO42–. Confirm presence of carbonate by bubbling gas through limewater; positive test if limewater turns cloudy.

13

Investigation

14

2MnO4–(aq) + 5C2O42–(aq) + 16H+(aq) → 2Mn2+(aq) + 10CO2(g) + 8H2O(l) nMnO4–

=

cV

=

0.05 × 0.03820

=

1.91 × 10–3 moles

=

1.91 × 10–3 × 2.5

=

4.78 × 10–3 moles

=

4.78 × 10 −3 0.02

=

0.239 mol L–1

∴nC2O42–

=

cV

in 100 mL

=

0.239 × 0.1

nC2O42–

c C2O42–

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Solutions Manual: Module 3 Chemical monitoring and management

=

0.0239 moles

nCa2+

=

0.0239 moles

mass CaCO3

=

0.0239 × 100.1

=

2.39 g

=

2.39 4.05 × 100

=

59.01 %

% mass CaCO3 in limestone

15

Investigation

CHAPTER 12: MONITORING AND MANAGING CHEMICAL INDUSTRIES Review exercise 12.1 1

Investigation

2

a

Advantages: labour supply is close; transport costs are lower.

b

Disadvantages: waste disposal; social and safety issues.

Review exercise 12.2 1

a

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)

b

CH4(g) +

3 O2(g) → CO(g) + 2H2O(l) 2

2

Closing windows and doors can limit the supply of oxygen in the room. Incomplete combustion could occur, producing pollutants such as carbon monoxide and soot.

3

To ensure that complete combustion occurs; to ensure that engine efficiency is maintained; to ensure that harmful pollutants are reduced.

Review exercise 12.3 1

Raising the temperature; increasing concentration of reactants; using catalysts.

2

High temperatures are expensive to maintain (catalysts are preferable).

3

Reaction yield is the proportion of product formed compared to the maximum amount that could be formed by complete reaction of the reactants.

4

Batch processing involves set amounts of reactants forming a product, which is removed when the reaction is complete. Continuous processing involves a ‘production line’ method where reactants are added and products removed continuously.

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Review exercise 12.4 1

a

Increasing temperature would cause the equilibrium to shift left (endothermic direction) ∴ will decrease yield.

b

Greater pressure would cause a shift to the side with fewer gaseous molecules ∴ equilibrium will shift right and increase yield.

c

Removal of NH3 will cause the equilibrium to shift right to counteract the change, thus increasing yield.

2

Temperature 500°C, high pressures (35 000 kPa), iron–iron oxide catalyst, removal of ammonia. These conditions are a compromise between rate, equilibrium yield and economic factors.

3

fertilisers, metal extraction, cleaning agents, manufacture of synthetic fibres, manufacture of explosives.

Chapter 12 Application and investigation 1

Investigation

2

It has been argued that if the industrial synthesis of ammonia had not been developed by the Germans during the early 1900s then they would have been unable to produce sufficient munitions to support their troops and fertilisers to feed their population and they would have surrendered sooner.

3

High pressure will favour the products and ∴ increase yield of NO2. High temperature will favour reactants and decrease yield of NO2. Catalyst will maintain a higher rate of reaction. ∴ Conditions of moderate temperature, high pressure and a catalyst would be used.

4

CH4(g) + H2O(g)

3H2(g) + CO(g) + ∆H

Low pressure and high temperature. Low pressure would shift equilibrium to the right to the side with more gaseous molecules ∴ increase yield of H2. High temperature would shift equilibrium to the right as reaction is endothermic ∴ increase yield of H2. Additional aspects include the use of a catalyst, transport and storage of gases under high pressure, the explosive nature of hydrogen gas and flammability of methane gas. The source of methane should be nearby and the poisonous by-product, CO, released/treated appropriately. 5

Investigation

6

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Solutions Manual: Module 3 Chemical monitoring and management

7

a

At 300°C the rate of reaction is too slow and the yield is not produced quickly enough. With the use of a catalyst, the reaction rate is increased despite increasing temperature (500°C) decreasing the yield.

b

High pressures favour the production of NH3 according to Le Chatelier’s principle. Therefore yield is increased at 35 000 kPa.

c

Continuous removal of small quantities of ammonia ensures that the reaction rate to products is preferred. Equilibrium between products and reactants is not attained.

8

Fast reaction rates are generally the most economical, and for the Haber process reactions can be speeded up by the use of a catalyst, especially as this allows lower temperatures to be used, thus increasing the yield. The Haber process does not allow equilibrium to be reached, as products are continually being removed and reactants added, so there is a constant drive to the right and a continuous yield.

9

Investigation

10

Investigation

CHAPTER 13: MONITORING AND MANAGING OUR ATMOSPHERE Review exercise 13.1 1

The temperature falls throughout the troposphere to –50°C then rises throughout the stratosphere to 10°C. Temperature drops to about –100°C throughout the mesosphere then rises to over 25°C in the thermosphere.

2

Gases in the atmosphere trap heat being radiated into space from Earth. Therefore more gases in the atmosphere will cause a temperature increase on Earth.

3

Nitrogen is used by plants and animals to manufacture protein. Oxygen is used in respiration, the breakdown of sugars to produce energy in all living things. Carbon dioxide is used in plants to make glucose by photosynthesis.

4

lead, hydrocarbons, carbon monoxide, carbon dioxide, nitrogen oxides

5

Nitrogen combines with oxygen in high temperature combustion conditions to form NO(g). This combines with more oxygen to produce nitrogen dioxide. Given approriate environmental UV conditions, the colourless nitrogen dioxide converts to brown dinitrogen tetroxide, which gives photochemical smog its characteristic colour.

Review exercise 13.2 1

a

O2: double covalent bond between the two atoms. O3: a V-shaped molecule that can be viewed as having a double covalent bond between two atoms and a coordinate covalent bond between two atoms. However, it appears to actually exist as a resonance structure with delocalised electrons.

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Solutions Manual: Module 3 Chemical monitoring and management

b

O2 is a highly reactive oxidising agent. O3 is a very highly reactive oxidising agent.

2

O3 occurs naturally in the troposphere to a limited extent but is found in photochemical smog and used as an indicator for this. Concentrations above these low levels are regarded as polluting and can cause biological harm. O3 in the stratosphere absorbs damaging UV radiation. It is formed from the reaction of O2 with UV.

3

no difference

4

a

b

c

d

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5

These molecules do not have even numbers of valence electrons, so all electrons cannot form pairs. The octet rule is not followed. Review exercise 13.3 1

Haloakanes: Any hydrocarbon with one or more halogen attached. CFCs: All the hydrogens are replaced with chlorine and fluorine. HCFCs: All but one hydrogen is replaced with chlorine and fluorine. HFCs: Hydrocarbon with fluorine and chlorine attached. Halons: CFCs with bromine atoms.

Examples: haloalkane: CH3I iodomethane CFC: CC13F freon-11 HCFC: CHC1F2 HFC: C2H2F4 halon: CBrC1F2 2

a

b

i

1-chlorooctane

ii

2-bromo-3-fluoropentane

iii

1,1-dichloropropane

iv

5-bromo-1,4-dichloro-2,2-difluorohexane

i

ii

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iii

iv

3

Production of ozone from O2 and oxygen atoms, and decomposition of ozone, occur continuously in the stratosphere using UV radiation.

4

a

The ozone layer prevents harmful ultraviolet radiation from reaching Earth and causing damage to living things and delicate ecosystems.

b

The decomposition of ozone is catalysed by substances like NO and Cl. Pollution containing such substances causes greater decomposition of O3 and therefore depletion of the ozone layer.

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Chapter 13 Application and investigation 1

Troposphere, stratosphere, mesosphere, thermosphere The troposphere and stratosphere contain gases that are essential for life, CO2 and O2, and play a role in the water cycle. They absorb UV radiation before it reaches Earth and infrared radiation from Earth.

2

CO2 is used by plants in the making of glucose (photosynthesis). O2 is used in respiration, i.e. breakdown of sugars to give energy. N2 is used in the manufacture of protein in living things.

3

Investigation

4

The Earth has gases which contribute to warming due to natural causes, e.g. H2O, CH4, CO, CO2. These gases absorb UV radiation and reradiate it as heat. Human activities have increased the concentrations of some of these gases, thereby enhancing the warming effect.

5

Investigation

6

a

i

ii

iii

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iv

b

i

ii

iii

iv

7

a

Decreases in the ozone concentration of the stratosphere over Antarctica, particularly in spring, have been recorded.

b

CFCs are unreactive molecules which eventually break down in the stratosphere to produce chlorine atoms. Chlorine atoms catalyse the decomposition of ozone.

8

Investigation

9

a

Oxygen O2: Used in the breakdown of glucose to produce energy in living things. Ozone O3: Absorbs UV radiation in the stratosphere.

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O2

b

O3

O2–

Double covalent bond Central oxygen atom joined between two oxygen on one side to another atoms oxygen atom by a double covalent bond, and on the other by a coordinate covalent bond. 10

a b c d

Oxygen ion has two negative charges due to an extra two electrons in the valence shell of the atom.

O2(g) + UV

→

2O(g)

O2(g) + O(g)

→

O3(g)

O2(g)

→

(electrical energy) 2O(g)

O2(g) + O(g)

→

O3(g)

NO2(g) + UV radiation

→

NO(g) + O(g)

O2(g) + O(g)

→

O3(g)

O2(g) + UV

→

2O(g)

O2(g) + O(g)

→

O3(g)

11

A coordinate covalent bond forms when one atom provides both electrons for the shared pair between atoms forming a bond. In this example both electrons are provided by N.

12

Investigation

13

a

Warm, sunny day with still air. Oxides of nitrogen. Hydrocarbons.

b

UV radiation is absorbed by NO2. A series of reactions leads to the production of ozone. Then, unburned hydrocarbons react with ozone to produce organic compounds such as aldehydes, e.g. ethanal.

14

Investigation

15

Investigation

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CHAPTER 14: MONITORING AND MANAGING OUR WATERWAYS Review exercise 14.1 1

When the top of the lake freezes, the warmer water below is more dense, and so it stays lower.

2

Ions in sea water are the result of leaching from rocks and soils into rivers and then into seas.

3

The concentration of ions in sea water is higher than that found in fresh water, as most rivers run into seas. The concentrations are determined by: •

prevalence in the Earth’s crust

•

solubility

•

the degree to which they are extracted by living things.

The ions found in fresh water are dependent on the source of the water and the geology of the catchment area. Therefore their concentrations will vary considerably. Review exercise 14.2 1

2

a

Evaporate a filtered sample of water and weigh the remaining solids, or measure electrical conductivity of the solution.

b

To determine levels of Mg2+ and Ca2+ ions, the sample is titrated against ethylenediaminetetraacetic acid (EDTA). A simple lather test using soap flakes will give an indication of the relative hardness of water.

c

A turbidity disk is lowered into the water until it can no longer be seen. This measures the suspended solids in the sample. Turbidity can also be determined by shining a beam of light through a sample and detecting the amount of transmitted light.

d

Acidity is determined using indicator paper or a pH meter, or by titration.

Dissolved oxygen measures the solubility of O2 in the sample. Low levels indicate pollution. Biochemical oxygen demand assesses the capacity of the organic matter in the sample to use the oxygen present.

Review exercise 14.3 1

Fertiliser run-off, animal waste and sewage.

2

Excessive forest clearing, irrigation.

3

While most heavy metals are essential to life as trace elements, at higher concentrations they pose a substantial health risk, particularly as they are frequently cumulative poisons and may give few symptoms at a low level of poisoning. © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2006. This page from Chemistry Contexts 2 Teacher’s Resource second edition may be reproduced for classroom use.

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Review exercise 14.4 1

Flocculation, clarification, filtering, chlorination and fluoridation.

2

To strengthen the enamel on children’s teeth and help protect against tooth decay.

3

Gravity, vacuum or pressure pumps force water through a membrane containing small pores. The membranes block the passage of particles, therefore purifying the water.

4

Water is forced under pressure to move from a more concentrated solution into a less concentrated one.

5

The stearate ion in soap bonds with the Ca2+ and Mg2+ ions in hard water, forming scum. This means that the soap does not work effectively at cleaning.

6

Temporary hard water can be boiled to remove hardness. Permanent hard water needs to be treated with softeners to reduce its hardness.

Review exercise 14.5 1

2

Acute water shortages would necessitate the use of tertiary treatment of waste water for recycling. © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2006. This page from Chemistry Contexts 2 Teacher’s Resource second edition may be reproduced for classroom use.

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Chapter 14 — Application and investigation 1

Investigation

2

They are prevalent in the Earth’s crust; they are highly soluble and are not extracted from the water by living things.

3

Salinity of the oceans is increased by dissolved salts being added by rivers. The excess salts are removed by formation of coral and limestone reefs and nutrient and waste exchange in living things. Therefore salinity level remains constant.

4

Mass of Cl– in 3.5 g

nCl– =

m M

nCl– in 1 litre

=

60% × 3.5

=

2.1 g Cl– ions in 100 g water

=

2.1 35.5

=

5.92 × 10–2 moles in 100 g

=

5.92 × 10–2 × 10

=

0.59 moles in 1 litre sea water.

1.35 g =

5

ppm (mg/kg) = =

1350 mg NaCl 1350 3 450 ppm

6

Oxygen in waterways is provided by photosynthesis in aquatic plants and from the atmosphere. It is reduced by respiration in marine plants and animals and in decomposition of organic matter. Low levels of oxygen can indicate organic wastes, heat pollution or eutrophication caused by excessive growth of plants.

7

Excessive nutrients can have a serious impact on natural aquatic ecosystems, and therefore natural waterways need to be monitored. High levels of nutrients can lead to the serious problems of algal blooms and eutrophication.

8

Investigation

9

Chlorine kills microorganisms, including bacteria and algae. Fluoride is added to help prevent tooth decay in children.

10

The ongoing good health of water consumers and the well-being of natural aquatic systems is essential for current and future generations. Monitoring these systems is the only way to ensure government-determined standards are maintained.

11

pH: to ensure drinking water is not too acidic or basic. © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2006. This page from Chemistry Contexts 2 Teacher’s Resource second edition may be reproduced for classroom use.

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Heavy metals: e.g. Cd, Pb, as these are toxic. Turbidity: amount of suspended solids in water. 12

a

Hard water is water that contains high concentrations of Ca2+ and Mg2+ ions in addition to hydrogencarbonate, sulfate and chloride ions.

b 2NaCH 3 (CH 2 )16 COO(aq) + Ca 2+ (aq) → Ca(CH 3 (CH 2 )16 COO) 2 (s) + 2Na + (aq) soap

scum

13

Na2[Na4(PO3)6] combines with Ca2+ and Mg2+ ions, forming soluble ions. Na2CO3.10H2O reacts with Ca2+ ions only and forms a precipitate.

14

Deionised water has positive and negative ions removed by using ion exchange resins. Non-ionic or insoluble ionic materials could still be present in the water. Distilling removes all ionic/dissolved/suspended impurities but is much more expensive.

15

Investigation

Module 3 REVIEW 1

B

2

C

3

A

4

A

5

B

6

C

7

C, D

8

B

9

‘Friend’: in the stratosphere, ozone absorbs UV radiation. ‘Foe’: in the troposohere, ozone is a component of photochemical smog, a pollutant that is poisonous over 20 ppm.

10

a

SO2: burning fossil fuels, mineral smelters CH4: ruminants, anaerobic decomposition of organic matter NO: high temperature combustion CO2: burning fossil fuels, deforestation

b

Pass exhaust gases (e.g. CO gas) through catalytic converters.

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11

12 13

14 15

a

haloalkanes

b

dichloro-difluoromethane

c

CFCs are unreactive molecules which eventually break down in the stratosphere to produce chlorine atoms. Chlorine atoms catalyse the decomposition of ozone. UV radiation acting on CFCs is believed to be one of the major causes of world ozone depletion.

a

nitrogen, hydrogen

b

Moderate temperature, high pressure, catalyst, removal of ammonia.

a

Scale (CaCO3(s) and/or CaSO4(s)) forms inside the iron, which blocks the steam holes and reduces its efficiency.

b

High phosphate levels in aquatic systems can result in algal blooms and eutrophication in our waterways.

c

Lead, which is poisonous, especially to children, is released through the exhaust to the atmosphere.

a

Q, as solubility of a gas in water will decrease at higher temperatures.

b

Evaporate the water to obtain the ionic solid.

a

Lead carbonate is insoluble in water ∴ acid is necessary. Acid + carbonate → salt + water + CO2(g)

16

b

Add Na2SO4 to solution. A precipitate will form if lead is present.

c

No, as Ba2+ could also be present and cause the precipitate.

a

calcium ion: orange/red copper ion: green

b

As the ion is heated, the electrons in the atom move to a higher energy level. As these electrons drop to lower energy levels, energy is emitted as colour.

17

NaCl(aq) could be used to precipitate all the Ag+ ions as AgCl(s). The precipitate would be filtered, dried and weighed. The moles of silver could be determined using the mass and hence the molarity of the AgNO3(aq) solution.

18

a

carbon dioxide, CO2; methane, CH4

b

Increased numbers of molecules of greenhouse gases in the atmosphere result in re-radiated infrared energy from Earth being trapped in the atmosphere. This causes a greenhouse effect and raises the Earth’s temperature.

c

Increasing demand for energy has resulted in greater burning of fossil fuels and thus increasing levels of CO2 in the atmosphere.

a

Concentrations of chlorine, fluorine and iron.

19

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Solutions Manual: Module 3 Chemical monitoring and management

b

Chlorine: excessive chlorine adversely affects the taste and smell of water. Chlorination destroys any bacteria present.

20

21

c

Chemical analysis by titration with AgNO3 solution or gravimetric analysis by precipitation as AgCl.

a

biochemical oxygen demand

b

BOD measures the capacity of the organic matter in a sample to consume oxygen. If there is too much organic matter, the water is polluted.

c

Aerobic organisms are used to decompose organic matter in a sample over five days. Dissolved oxygen is measured before and after the five-day period.

Primary treatment: filtration and settling to remove large objects, solids, oil and scum. Secondary treatment: treatment of organic matter with aerobic bacteria via sludge method. About 90% of the organic matter is removed. Tertiary treatment: removal of contaminants such as nitrates, phosphates and metal ions.

22

23

a

CuCO3(s) + H2SO4(aq) → CuSO4(aq) + CO2(g) + H2O(l)

b

green

c

Barium will form a white precipitate with H2SO4. Calcium may form a white precipitate with H2SO4 but will form a white precipitate with NaF.

a

carbon monoxide, carbon dioxide, nitrogen dioxide

b

CO is formed when incomplete combustion occurs due to a lack of oxygen. e.g. C 8 H 18 (l) + 172 O 2 (g) → 8CO(g) + 9H 2 )(l)

24

c

Production of CO can be reduced by getting cars tuned regularly to ensure efficient combustion.

a

Germany was dependent on overseas supplies of nitrates as a source of NH3 for fertilisers and manufacture of explosives (during WWI) and its shipping routes were blocked by the Allies.

b

Distillation of liquid air.

c

Carbon monoxide produced while forming H2 is converted to CO2 and used by soft drink manufacturers. Unreacted N2 and H2 in the process are recirculated and re-used.

d

To keep the reaction rate high at temperatures of 500°C in order to obtain a reasonable yield.

e

No, as equipment for temperatures of 500°C and pressures of 35 000 kPa would be prohibitively expensive in a small laboratory.

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Solutions Manual: Module 3 Chemical monitoring and management

25

a

C and D. Compare temperature of river before and after steel mill location.

b

The solubility of gases decreases as the temperature increases, and therefore dissolved oxygen levels in the river would fall.

c

i

Nitrogen and phosphorus.

ii

Nitrogen could come from excessive use of fertilisers on farmland, and phosphorus from sewage containing phosphates from detergents.

d

pH meter or indicator paper.

e

Chlorination to destroy any bacteria present.

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Solutions Manual: Module 4 Industrial chemistry

CHAPTER 15: THE CHEMICAL INDUSTRY AND THE ROLE OF EQUILIBRIUM REACTIONS Review exercise 15.1 1

Synthetic dyes, celluloid and superphosphate have been used to replace Tyrian purple, ivory and sodium nitrate. The alternative synthetic chemicals are more readily available, more economically produced and often have superior properties to the naturally occurring substances.

2

Teflon®: non-stick cookware, greaseless bearings, chemically resistant liners Kevlar®: bulletproof vests, radial tyres, flame-resistant garments Zincalume®: roofing and fencing materials Aspirin: analgesic, anti-inflammatory drug Vitamin C: in humans prevents scurvy, essential in collagen synthesis, maintenance of structural strength in blood vessels Sodium stearate: soap

3

a

b

4

Natural rubber is formed by enzymes that produce a polymer with all the methyl groups arranged on the same side of the double bonds within the polymer chains, and this was initially very difficult to duplicate synthetically.

Review exercise 15.2 1

a

Equilibrium shifts to the right of the equation to counteract decrease in [H2].

b

Increase in volume equals a decrease in pressure. Equilibrium shifts to right of the equation as more molecules are produced to counteract the change.

c

Equilibrium shifts to left, where fewer molecules counteract the increase in pressure.

d

Equilibrium shifts to left as it is an endothermic reaction.

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Solutions Manual: Module 4 Industrial chemistry

2

e

No change in equilibrium position as the catalyst alters the rate of both forward and reverse reactions.

a

As [Fe3+] increases, equilibrium shifts to the left to counteract this change. Increase in red as [FeSCN2+] increases.

b

Removal of SCN– by precipitation shifts the equilibrium to the right. The resulting solution becomes increasingly pale yellow.

Review exercise 15.3 1

2

3

a

K=

b

K=

c

K = [Pb2+] [I–]2

d

K=

a

[reactants] > [products]

c

[reactants] > [products]

d

[reactants] < [products]

a

i

Reaction shifts to right in order to minimise N2O3 addition.

ii

Stays the same.

i

Reaction shifts to left, where fewer molecules counteract the increase in pressure.

ii

Stays the same.

i

No change in concentrations of reactants or products. Forward and reverse reactions affected equally.

ii

Stays the same.

i

Shifts reaction to the right (endothermic reaction), to reduce temperature applied.

ii

Increases.

1

0.0385 2 Q= = 0.302 ≠ K ∴ system is not at equilibrium 0.0049

b

c

d

4

[SO 3 ] 2

a

[SO 2 ] 2 [O 2 ]

[ NO]4 [H 2 O]6 [ NH 3 ] 4 [O 2 ]5 [H + ] [F − ] [HF]

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Solutions Manual: Module 4 Industrial chemistry

b 5

2

0.0299 2 Q= = 0.213 = K ∴ system is at equilibrium 0.0042

3

0.0103 2 Q= = 0.034 ≠ K ∴ system is not at equilibrium 0.0031

i

Q > K, so more reactants will be formed.

iii

Q < K, so more products will be formed.

a

Decrease

b

Increase

Review exercise 15.4 1

2

[ NO] 2 [Br2 ]

a

K=

b

K=

a

[H2] = 0.34075 mol L–1

[ NOBr] 2 (2.05 × 10 −2 ) 2 (5.24 × 10 −3 ) = 6.41 × 10 −2 (5.86 × 10 −3 ) 2

[Br2] = 0.2200 mol L–1 H2(g) [initial]

0.34075

[equilibrium]

0.1404

+

Br2(g) 0.2200

↔ 2HBr(g) 0

Thus 0.20035 mol L–1 of H2 have reacted. From equation, H2:Br2 reacting is 1:1 Therefore at equilibrium, [Br2] = and [HBr] = At equilibrium, [H2] =

b

K=

0.2200 – 0.20035 2 × 0.20035 0.1404 mol L–1

[Br2] =

0.01965 mol L–1

[HBr] =

0.4007 mol L–1

[HBr]2 (0.4007) 2 = = 5.81981 × 10 [H 2 ] [Br2 ] (0.1404) (0.01965)

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Solutions Manual: Module 4 Industrial chemistry

3

a

b

[CO] [H 2 ] [ H 2 O] Carbon is a solid. Its concentration does not change and is excluded from the K expression.

K=

K=

(0.2) (0.2) 5 × 10 −2

= 0.8

Chapter 15 Application and investigation 1

Investigation

2

Investigation

3

4

Investigation

5

Investigation

6

Investigation

7

The rate of the reverse reaction becomes greater than that of the forward reaction, as the equilibrium shifts to the left to decrease the effect of adding more NO2 (Le Chatelier’s principle). Thus, the concentration of N2O4 increases until the rate of the forward reaction equals the rate of the reverse reaction, and equilibrium is re-established.

8

[ NO] 2 [Cl 2 ]

a

K=

b

K=

[CH 3 Cl] [HCl] [CH 4 ] [Cl 2 ]

c

K=

[CO 2 ] [O 2 ]

d

K = [Mg2+][OH–]2

[ NOCl] 2

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[ NH 4 + ] [OH − ] [ NH 3 ]

e

K=

f

K = [CO2]

9

[Cl2] >> [Cl]

10

a

11

12

QA =

(0.025) (0.04) = 0.050 = K ∴ system is at equilibrium 0.02

QB =

(0.020) (0.025) = 0.033 ≠ K ∴ system is not at equilibrium 0.015

b

For system B, the reaction must move to the right in order to increase the concentration of product.

a

Equilibrium constant increases. Formation of products is endothermic.

b

Reaction moves to the right (products) in order to counteract the addition of I2, until equilibrium is re-established. The value of the equilibrium constant is unchanged.

c

No change, because there is an equal number of particles on both the reactant and products sides. No change in the equilibrium constant.

K

=

[ NOBr] 2 [ NO] 2 [Br2 ]

=

(0.046) 2 (0.30) 2 (0.11)

K

13

=

0.21

K =

[ NO] 2 [O 2 ] [ NO 2 ] 2

At equilibrium [NO] =

1.00 mol L–1

[O2] =

1.00 mol L–1

[NO2] = K =

0.464 mol L–1 (1.00) 2 (1.00) (0.464) 2

= 4.64

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Solutions Manual: Module 4 Industrial chemistry

[H 2S] 2

K =

14

[H 2 ] 2 [S 2 ]

At equilibrium [H2S] =

0.725 mol L–1

[H2] =

0.208 mol L–1

[S2] =

1.125 × 10–6 mol L–1 (0.725) 2

K =

(0.208) 2 (1.125 × 10 −6 )

1.080 × 107

K = 15

a

i

CO(g) + Cl2(g) ↔ COCl2(g)

[Initial]

0.2

0.5

[Equilibrium]

0.1

From equation

CO

:

Cl2

:

COCl2

1

:

1

:

1

∴ [Cl2] at equilibrium

0

(0.5 – 0.1) = 0.4 mol L–1

=

0.1 mol L–1

[COCl2] at equilibrium =

b 16

[COCl 2 ] 0.1 = 2.5 = [CO] [Cl 2 ] (0.1) (0.4)

ii

K=

i

decrease

ii

decrease

a

CO(g) + H2O(g)

[Initial]

0.100

0.100

↔

CO2(g) + H2(g) 0

[Equilibrium]

0.0665

At equilibrium [CO2]

=

[H2]

=

0.0665 mol L–1

[CO] = [H2O]

=

(0.100 – 0.0665)

=

0.0335 mol L–1

b 17

0

K=

(0.0665) 2 (0.0335) 2

= 3.940

Initial concentration of NOCl

=

1.666 mol L–1

28% dissociated

=

0.4666 mol L–1

∴ [NOCl] at equilibrium

=

1.20 mol L–1

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Solutions Manual: Module 4 Industrial chemistry

from equation, if 0.4666 mol L–1 of NOCl ionises, at equilibrium [NO]

=

K

=

=

0.4666 mol L–1 and [Cl2] = 0.2333 mol L–1 [ NO] 2 [Cl 2 ] [ NOCl] 2 (0.4666) 2 (0.2333) (1.2) 2

= 0.0353

CHAPTER 16: THE INDUSTRIAL PRODUCTION OF SULFURIC ACID Review exercise 16.1 1

Manufacture of superphosphate 72% Manufacture of general chemicals 10% Mining and metallurgy 6% Manufacture of ammonium sulfate 6% Other uses such as petroleum refining, explosives, detergents, batteries, dyes, pigments, paint, paper, drugs 6%

2

a

Ca3(PO4)2(s) + 2H2SO4(aq) + 5H2O(l) ↔ Ca(H2PO4)2.H2O(s) + 2CaSO4.2H2O(s)

b Review exercise 16.2 1

Superheated steam pumped into underground deposits melts sulfur. Compressed air pumped into deposits transports sulfur to surface. Sulfur has a relatively low melting point which allows steam under pressure to melt it, and it is insoluble in water and has a low density, which causes it to be easily extracted from water and in a pure form.

2

a

2SO2(g) + O2(g) ↔ 2SO3(g)

b

High pressure as the products have fewer gas molecules than the reactants, and low temperature because the reaction is exothermic.

c

High temperature, high pressure, use of catalyst

d

600°C, V2O5 catalyst and atmospheric pressure. At room temperature a very high yield could be achieved with a very slow reaction rate. At high temperature, the yield is low but the reaction rate is high. The combination of moderately high (600°C) temperature and use of catalyst in the presence of excess O2 gives a high yield, so expensive, high-pressure equipment is not needed.

a

S(s) + O2(g) → SO2(g)

3

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Solutions Manual: Module 4 Industrial chemistry

b

2SO2(g) + O2(g) → 2SO3(g)

c

SO3(g) + H2SO4(l) → H2S2O7(l)

d

H2S2O7(l) + H2O(l) → 2H2SO4(l)

4

The exothermic reaction forms a fog of H2SO4 droplets, which make working conditions difficult.

5

Poisonous gas: pungent, can trigger asthma attacks. Very soluble in atmospheric water: produces sulfurous acid rain.

Review exercise 16.3 1

a

Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g)

b

Cu(s) + 2H2SO4(aq) → CuSO4(aq) + 2H2O(l) + SO2(g)

c 2

Safety goggles, gloves and protective clothing, as spitting may occur. Access to fume cupboard for procedure and water supply. Add acid to the water slowly, with continuous stirring, to minimise the potential for boiling from this exothermic reaction.

Chapter 16 Application and investigation 1 Properties

Major uses

• Manufacture of superphosphate and ammonium sulfate (fertilisers)

A strong acid that is non-volatile

• Cleaning of coatings from iron or steel before galvanising or electroplating • Extracting titanium dioxide from ilmenite Strong affinity for water

Reacts with aromatic hydrocarbons to form sulfonic acids 2

As a dehydrating agent in: •

production of ethene from ethanol

•

esterification reactions

•

production of gases O2, N2, CO2

Manufacture of detergents

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Solutions Manual: Module 4 Industrial chemistry

3

4

Raw material

Role

sulphur

formation of sulfur dioxide

air/O2

oxidation of sulfur to sulfur dioxide then to SO3

V2O5

catalyst for SO2 → SO3

H2O

H2S2O7(l) → H2SO4(l)

a

Sulfur burnt in air to form sulfur dioxide. S(s) + O2(g) → SO2(g)

b

Sulfur dioxide is converted to sulfur trioxide over beds of V2O5 catalyst in excess O2.

c

SO3(g) is absorbed in concentrated H2SO4 to form oleum, H2S2O7. SO3(g) + H2SO4(l) → H2S2O7(l)

d

Water is added to produce sulfuric acid. H2S2O7(l) + H2O(l) → 2H2SO4(l)

5

a

Higher temperature increases the rate of reaction but decreases the yield, because the reaction is exothermic. With a catalyst to increase the reaction rate, a lower temperature enables an increased equilibrium yield of SO3.

b

Reactions at high pressure require expensive high-pressure equipment. High yields can still be obtained without high pressure with temperatures of 400– 600°C, V2O5 catalyst and atmospheric pressure.

c

SO2(g) is very soluble and could form H2SO3(aq) with moisture in the air.

6

The byproduct of SO2 from sulfide ores can be used as the raw material for sulfuric acid production.

7

Investigation

8

a

As oxidising agent. 2KI(aq) + 2H2SO4(aq) → I2(aq) + K2SO4(aq) + SO2(g) + 2H2O(l)

b

As strong acid. Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g)

c

Concentrated acid is acting as a dehydrating agent, absorbing any H2O(g) mixed in with N2(g). N2.H2O(g) + H2SO4(l) → N2(g) + H2SO4.H2O(aq)

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Solutions Manual: Module 4 Industrial chemistry

9

Investigation

10

Main points for dilution procedure should include: •

wearing goggles, gloves and protective clothing

•

use of fume cupboard, access to water and reasons

•

acid added to water

•

addition slow, with constant stirring

•

exothermic reaction, equations.

CHAPTER 17: THE INDUSTRIAL PRODUCTION OF SODIUM HYDROXIDE Review exercise 17.1 1

2

Galvanic cell

Electrolytic cell

Chemical energy is converted to electrical energy (spontaneous reaction with a positive voltage)

Electrical energy is converted to chemical energy (reaction is not spontaneous and energy must be supplied)

Anode (oxidation liberates electrons)— negative electrode

Anode (oxidation liberates electrons)— positive electrode

Cathode (reduction absorbs electrons)— positive electrode

Cathode (reduction absorbs electrons)— negative electrode

Electrons move from anode to metallic conductor of external circuit to cathode

Electrons move from anode to voltage source to cathode

Negative ions move from cathode to anode

Negative ions move from cathode to anode

Positive ions move from anode to cathode

Positive ions move from anode to cathode

a

anode: 2H2O(l) → O2(g) + 4H+(aq) + 4e– cathode: 2Co2+(aq) + 4e– → 2Co(s) overall: H2O(l) + 2Co2+(aq) → O2(g) + 2Co(s) + 4H+(aq)

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b

3

anode: 2H2O(l) → O2(g) + 4H+(aq) + 4e– cathode: 2Zn2+(aq) + 4e– → 2Zn(s) overall: H2O(l) + 2Zn2+(aq) → O2(g) + 2Zn(s) + 4H+(aq)

4

a

anode: 2Cl–(l) → Cl2(g) + 2e– cathode: Sr2+(l) + 2e– → Sr(s)

b

anode: Cu(s) → Cu2+(aq) + 2e– cathode: Cu2+(aq) + 2e– → Cu(s)

c

anode: 4OH–(aq → O2(g) + 2H2O(l) + 4e– cathode: 2H2O(l) + 2e– → H2(g) + 2OH–(aq)

Review exercise 17.2 1

If aqueous solutions of active metals are electrolysed, less energy is needed to reduce water rather than the metal. Hence molten compounds must be used in electrolysis.

2

Oxidation produces oxygen: 2H2O(l) → O2(g) + 4H+(aq) + 4e– Reduction produces hydrogen: 2H2O(l) + 2e– → H2(g) + 2OH–(aq)

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Solutions Manual: Module 4 Industrial chemistry

3

a

Water will be reduced before the sodium.

b

At the cathode, possible reactions are: Na+(aq) + e– → Na(s)

E° = –2.71 V

2H2O(l) + 2e– → H2(g) + 2OH–(aq)

E° = –0.41 V

Reduction of water has a more positive reduction potential, so hydrogen will be evolved. At the anode, possible reactions are: 2Cl–(aq) → Cl2(g) + 2e–

E° = –1.36 V

2H2O(l) → O2(g) + 4H+(aq) + 4e–

E° = –0.82 V

Since oxidation of water has a more positive E°, oxygen will be produced. 4 5

a

2NaCl(aq) + 2H2O(l) → 2NaOH(aq) + H2(g) + Cl2(g)

b

2Cl–(aq) + 2H2O(l) → 2OH– + H2(g) + Cl2(g)

Mercury process: Hg cathode absorbs Na to form an amalgam which is continually removed. Chlorine produced at the inert anode is thus prevented from mixing with other products. Diaphragm process: An asbestos diaphragm lines the steel-mesh cathode to prevent Cl2 mixing with other products. Membrane process: Anode and cathode compartments are separated by a waterimpermeable membrane constructed from synthetic polymers such as Teflon.

Chapter 17 Application and investigation 1

Metal: metal structure with sea of electrons allows flow of electrons as current. Molten ionic compound: ions are not confined to lattice, so the flow of ions conducts current. In aqueous solution of ionic compound, ions can move freely through the solution, conducting current.

2

Cathode:

2K+(l) + 2e– → 2K(l)

Anode:

2OH– → O2(g) + H2O(l) + 2e–

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3

a

b

Anode: 2H2O(l) → O2(g) + 4H+(aq) + 4e– 2Cl–(aq) → Cl2(g) + 2e– (with concentrated HgCl2) Cathode: Hg2+(aq) + 2e– → Hg(l)

4

a

anode:

2I–(aq)

→

colourless cathode:

I2(aq) + 2e– yellow/brown colour around anode

2H2O(l) + 2e–

→

H2(g) + 2OH–(aq) colourless gas evolved

b

anode:

Cu(s)

→

Cu2+(aq) + 2e–

→

Cu(s)

anode erodes cathode:

Cu2+(aq) + 2e–

copper deposited on cathode 2+

Equal loss and gain of Cu (aq) from solution maintains blue colour of solution. c

anode:

4OH–(aq)

→

O2(g) + 2H2O(l) + 4e– colourless gas evolved

cathode:

2H2O(l) + 2e

–

→

H2(g) + 2OH– colourless gas evolved

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Solutions Manual: Module 4 Industrial chemistry

d

anode:

2H2O(l)

O2(g) + 4H+(aq) + 4e–

→

colourless gas evolved cathode:

Cu2+(aq) + 2e–

→

Cu(s)

blue/green colour fades as pink/red copper deposited on cathode. 5

Electrolysis of molten sodium chloride involves: anode oxidation:

2Cl–(l)

→

Cl2(g) + 2e–

E° = –1.36V

cathode reduction:

Na+(l) + e–

→

Na(s)

E° = –2.71 V

With aqueous solutions, reactions at the electrodes will favour those with the most positive E° value, so with NaCl(aq): anode oxidation: 2H2O(l) → O2(g) + 4H+(aq) + 4e– E° = 0.82 V is favoured, unless a more concentrated NaCl(aq) is used, and then Cl2 may be evolved at the anode 2Cl–(aq) → Cl2(g) + 2e– E° = –1.36 V cathode reduction: 2H2O(l) + 2e– → H2(g) + 2OH–(aq) E° = –0.41 V 6

7

a

Water present in the aqueous solution would be reduced first at the cathode.

b

Unless separated, the chlorine product at the anode and sodium at the cathode would readily react to revert to sodium chloride.

c

Carbon is an inert electrode and would not lose electrons in an oxidation reaction, whereas iron would readily do so in preference to the chloride ions in solution.

Process

Mercury

Diaphragm

Membrane

a

Raw materials

Concentrated NaCl(aq)

Concentrated NaCl(aq)

Concentrated NaCl(aq)

b

Nature of electrodes

Cathode: flowing Hg(l)

Cathode: steel mesh Anode: carbon

Cathode: iron mesh Anode: titanium

2H2O(l) + 2e– → H2(g) + 2OH–(aq)

2H2O(l) + 2e– → H2(g) + 2OH–(aq)

2Cl–(aq) → Cl2(g) + 2e–

2Cl–(aq) → Cl2(g) + 2e–

Anode: inert (graphite) c

Cathode reaction

2Na+(aq) + 2e– + Hg(l) → 2Na(Hg)

Anode reaction 2Cl–(aq) → Cl2(g) + 2e–

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Solutions Manual: Module 4 Industrial chemistry

d

Advantages

High quality products formed; no chloride contamination

No Hg environmental problems

No contamination of NaOH product with NaCl or formation of NaOCl Inert synthetic diaphragm, only allows Na+ ions to pass. Can withstand long immersion in OH– solutions.

Disadvantages

Environmental issues with Hg vaporisation, leakage into waterways.

NaOH is contaminated with NaCl(aq) and needs to be separated for commercial use. Asbestos diaphragm is carcinogenic material.

CHAPTER 18: THE PRODUCTION OF SOAPS AND DETERGENTS Review exercise 18.1 1

Soap is a sodium or potassium salt of a fatty acid.

2

a

b

c

saponification

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Review exercise 18.2 1

2

3

a

Polar solvents tend to dissolve polar solutes and non-polar solvents tend to dissolve non-polar solutes.

b

A solute will dissolve in a solvent only if the intermolecular forces within the solute and those within the solvent are similar to those occurring between solute and solvent molecules.

a

CH3(CH2)14COO–Na+(s) → CH3(CH2)14COO–(aq) + Na+(aq)

b

CH3(CH2)4(CH2CH=CH)2(CH2)6COO–K+(s) → CH3(CH2)4(CH2CH=CH)2(CH)6COO–(aq) + K+(aq)

a

b

4

c

The oil would disperse within the soap solution to form an emulsion.

a

Hard water is water that contains significant concentrations of Ca2+, Mg2+, HCO3–, SO42–, Cl– ions.

b

The negative ion of soap combines with Ca2+, Mg2+ to form an insoluble, greasy precipitate known as scum.

c

2RCOO–Na+(s) + Ca2+(aq) → Ca(RCOO)2(s) + 2Na+(aq)

d

Hardness in water can be the result of the presence of Ca2+ ions. These can be removed by precipitation with CO32− ions.

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Solutions Manual: Module 4 Industrial chemistry

Review exercise 18.3 1 Soap

Detergent

Use

Cleaning agent, surfactant

Cleaning agent, surfactant

Nature

Has a long hydrocarbon ‘tail’ that readily dissolves in oily or waxy substances, and a polar or charged particle ‘head’ that easily dissolves in water.

Has a long hydrocarbon ‘tail’ that readily dissolves in oily or waxy substances, and a polar or charged particle ‘head’ that easily dissolves in water.

Reaction with Ca2+ and Mg2+

Precipitates

No precipitate

Types

One: soluble salts of fatty acids

Three : anionic, cationic, non-ionic

2

a

b

3

The positive ends of cationic detergents are attracted to the surface of glass, leaving their non-polar ends sticking out, forming a greasy layer.

Review exercise 18.4 1

Early detergents were non-biodegradable or degraded very slowly and caused excessive frothing in waterways. These early detergents had branched hydrocarbon chains. Current straight-chained hydrocarbon detergents have eliminated this problem. Phosphate compounds included in anionic detergents enhance their cleaning power by softening the water and increasing alkalinity. These phosphate compounds in waterways can lead to algal blooms, which can cause depletion in water oxygen levels and kill aquatic organisms. Measures to reduce the phosphate problem include the use of substitutes such as sodium citrate, as well as greater use of non-ionic detergents.

2

The hydrocarbon chain is branched, and therefore the detergent is non-biodegradable.

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Chapter 18 Application and investigation 1

a

b

It would be an oil as it contains only unsaturated fatty acids.

c

2

Investigation

3

a

b

4

a

The hydrophobic end of a soap or detergent is the non-polar hydrocarbon end, which will not dissolve in water. The hydrophilic end is the charged or polar end, which will dissolve in water.

b

The grease is surrounded by spherical aggregates of surfactant ions. The hydrophobic tails attach to the grease, while the hydrophilic heads dissolve in the water. The surfactant ion thus floats the grease spot from the cloth.

c

Non-polar dry cleaning fluids would dissolve the non-polar grease spot (solute).

d

Washing with detergent solution can remove both non-polar molecules and ionic compounds from the fabric into an aqueous solution.

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Solutions Manual: Module 4 Industrial chemistry

Dry-cleaning fluid contains no water and dissolves non-polar molecules such as grease into the fluid. 5

a

Formation of scum

b

2RCOO–(aq) + Ca2+(aq) → Ca(RCOO)2(s)

c

Detergents do not form insoluble precipitates with Ca2+ ions.

6

A scum ring indicates that the water is ‘hard’, that is, it contains Ca2+ and/or Mg2+ ions, which form insoluble precipitates with soap. The bubble bath contained detergent, which does not form precipitates with Ca2+ or Mg2+ ions.

7

a

‘Praise’ French salad dressing: vegetable oil, vinegar and water are immiscible. Emulsifier = soyabean lecithin

b

Many cationic surfactants are effective biocides—that is, they kill many organisms, specifically bacteria, when used in mouth washes.

c

Non-ionic detergents are used as substitutes for anionic detergents. Use of anionic detergents that contain phosphates is kept at a minimum because of the resulting environmental problem of eutrophication. Some non-ionic detergents also have greater cleaning power than anionic ones.

8

The negative, anionic detergent ends are attracted to the positive surface of the plastic. This leaves the long, non-polar hydrocarbon tails of the surfactant sticking out, forming a thin, greasy film on the plastic.

9

Investigation

10

Investigation

11

Investigation

CHAPTER 19: THE INDUSTRIAL PRODUCTION OF SODIUM CARBONATE Review exercise 19.1 1

Manufacture of glass, pharmaceuticals, soaps, detergents, dyes. To neutralise excess acid from industrial chemical processes. To remove ink from high-quality white waste paper in recycling process.

2 Formula

Systematic name

Common name

NaOH

sodium hydroxide

Caustic soda

Na2CO3

sodium carbonate

Soda ash

Na2CO3.10H2O

sodium carbonate decahydrate

Washing soda

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3

2HCl(aq) + Na2CO3(aq) → 2NaCl(aq) + H2O(l) + CO2(g)

Review exercise 19.2 1

CaCO3 is insoluble. (Reverse process occurs if product solutions are mixed).

2

a

CaCO3(s)

b

NH3(aq) + CO2(aq) + NaCl(aq) + H2O(l) → NaHCO3(aq) + NH4Cl(aq)

c

Na+(aq) + HCO3–(aq) → NaHCO3(s)

d

2NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(l)

e

CaO(s) + H2O(l) → Ca(OH)2(s)

f

Ca(OH)2(s) + 2NH4Cl(aq) → 2NH3(g) + CaCl2(aq) + 2H2O(l)

a

CO2 from 2a is used in 2b above, regenerated in 2d above for use in 2b again.

b

NH3 is used in 2b above, regenerated in 2f above for use in 2b again

3

heat

⎯⎯ ⎯→

CaO(s) + CO2(g)

Review exercise 19.3 1

Thermal pollution. Excess quantities of CaCl2 are produced and cannot be used in other applications. CaCl2 returned to the environment adversely affects aquatic ecosystems. Disposal of excess Ca(OH)2 which must be neutralised and stored before release into the environment.

2

Large quantities of heat from the process are absorbed by water from rivers or lakes. This resultant hot water, unless cooled before being returned to the environment, will lower O2 water concentrations and harm or destroy aquatic organisms—thermal pollution. To avoid thermal pollution, expensive heat diffusers are needed or alternative ways of dissipating heat must be devised.

3

Ca(OH)2(s) + 2HCl(aq) → CaCl2(aq) + 2H2O(l) 250 kg Ca(OH)2

=

250 × 103 g

=

250 × 10 3 mol 74.1

=

3373.82 mol

(Molar mass of Ca(OH)2 is 74.10 g mol–1) From equation, nHCl for neutralisation is 2nCa(OH)2 ∴ 6747.64 mol HCl needed

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4

c

=

n V

V

=

6747.64 L = 1124.6 L HCl 6

Availability and cost of land, raw materials, transport, energy and water supply. Supply of labour, accessibility to markets, social and safety issues as well as those of waste disposal and the environment. Solvay process plant: waste disposal plus thermal pollution, environmental issues and cost of raw materials.

Chapter 19 Application and investigation 1

a

Raw materials: limestone CaCO3(s), water, brine NaCl(aq), ammonia NH3(g).

b

End products: i CaCl2(aq) and ii Na2CO3(s). i

In the recovery process to obtain NH3(g), which is recycled, the by-product CaCl2(aq) is formed: Ca(OH)2(s) + 2NH4Cl(aq) → 2NH3(g) + CaCl2(aq) + 2H2O(l)

ii

Na2CO3(s) is formed by heating NaHCO3(s): 2NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(l)

c

Na2CO3(s) is produced by heating NaHCO3(s) (equation b ii) NaHCO3(s) is obtained from bubbling CO2(g) through a concentrated brine solution which has absorbed NH3(g): CO2(g) + NaCl(aq) + NH3(g) + H2O(l) → NaHCO3(aq) + NH4Cl(aq)

d

Recycling of CO2(g) and NH3(g) make the process efficient. CO2(g) is first generated by heating limestone: CaCO3(s)

heat

⎯⎯ ⎯→

CaO(s) + CO2(g)

and it is then used to make NaHCO3(aq) (equation c ii). CO2(g) is regenerated in the heating of NaHCO3(s) (equation b ii). NH3(g) is used to form NaHCO3(aq) (equation c ii) and is regenerated in the reaction with Ca(OH)2(s) and NH4Cl(aq) (equation b i) e

Main environmental problems associated with the Solvay process are: i

thermal pollution: can be minimised by use of heat diffusers to cool the heated water before it is returned to the environment. Heated water discharged into ocean where significant dilution occurs may avoid critical thermal pollution.

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ii

disposal of excess by-product CaCl2: excess CaCl2 may be used up in other processes. If not, rapid release into environment/waterways would cause ecosystem damage. Discharge of diluted CaCl2 into the ocean could also be less harmful, given the high [Ca2+] in the sea. Other environmental problems could include leaks of NH3 gas and problems associated with the mining and extraction of ores for limestone and sodium chloride.

2

3

heat

⎯⎯ ⎯→

a

CaCO3(s)

CaO(s) + CO2(g)

b

CaO(s) + H2O(l) → Ca(OH)2)(s)

c

2NH4Cl(aq) + Ca(OH)2(s) → 2NH3(g) + CaCl2(aq) + 2H2O(l)

From: NH3(aq) + CO2(aq) + NaCl(aq) + H2O(l) → NaHCO3(aq) + NH4Cl(aq) 1 mol NaCl produces 1 mol NaHCO3 Molar mass of NaCl

=

58.44 g mol–1

Molar mass of NaHCO3

=

84.01 g mol–1

1000 kg NaCl

=

1000 × 10 3 mol NaCl 58.44

=

1.711 × 104 mol NaCl

=

mol NaHCO3 produced

From: 2NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(l) 2 mol NaHCO3 produces 1 mol Na2CO3 thus 1.711 × 104 mol NaHCO3 produces 0.5 × (1.711 × 104) mol Na2CO3 Mass of Na2CO3 produced =

4

moles × molar mass

=

(0.5 × 1.711 × 104)(84.01) g

=

718 706 g = 718.7 kg

From overall chemical reactions in the Solvay process, the production mole ratio of Na2CO3(s) : CaCl2(aq) is 1 : 1. n(Na2CO3) in 1000 kg

=

mass molar mass

=

1000 × 10 3 mol Na 2 CO 3 105.99

=

9434.85 mol Na2CO3

=

n(CaCl2)

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∴ mass CaCl2 5

=

n × M = (9434.85 × 110.98) g

=

1047 kg

NH3(aq) + CO2(aq) + NaCl(aq) + H2O(l) → NaHCO3(aq) + NH4Cl(aq) 100.00 L of 6.00 mol L–1 NaCl(aq) c(NaCl)

=

n V

n(NaCl)

=

(6.00 × 100.0) mol = 600.0 mol NaCl

n(NH3)

=

600 mol

At 0°C and 101.3 kPa 1mole

=

22.41 L

∴volume of NH3 = (n × 22.41) L

=

13 446 L of NH3(g)

nNH3 : nNaCl is 1 : 1

6

2NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(l) 1000 kg NaHCO3 = =

7

m 1000 × 10 3 = mol NaHCO3 M 84.01 11903.34 mol

n(Na2CO3) produced =

0.5n(NaHCO3)

mass Na2CO3 produced =

n(Na2CO3) × M

=

(5951.67 × 105.99) g

=

631 kg of Na2CO3 produced

Investigation

Module 4 REVIEW 1

Natural material = rubber a

Original raw rubber from the bark of rubber trees was vulcanised to increase its elasticity, hardness and resistance to chemical attack. As the automotive industry increased, the demand for rubber used in tyres escalated. The source of natural rubber was restricted during World War II, which promoted the need to develop synthetic rubber.

b

Styrene–butadiene. Also vulcanised. Raw materials for production are plentiful as they are by-products of petroleum refining.

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2

a

2SO2(g) + O2(g) → 2SO3(g)

b

high pressure, low temperature

c

Conditions used: moderately high temperatures 400–600°C, V2O5 catalyst in excess O2 and atmospheric pressure.

d

Since the reaction is exothermic, the yield would be better at low temperatures. However, at low temperatures the reaction rate is very slow. High pressure equipment is very expensive. It has been found that an excellent yield is achieved at atmospheric pressures if V2O5 catalyst is used.

e

K=

f

Q=

[SO 3 ] 2 [SO 2 ] 2 [O 2 ] (2.00) 2 2

(0.250) (0.450)

= 1.42 × 10 2

This is less than the K value of 6.50 × 102, so the system is not at equilibrium. The system must shift to increase products, [SO3], in order to reach equilibrium. 3

a

b

i

Superheated steam pumped into underground deposits melts sulfur. Compressed air pumped into deposits transports sulfur to surface.

ii

Low melting point, insoluble in water, low density.

iii

Care must be taken to prevent possible oxidation or reduction of sulfur at the Earth’s surface. Low concentrations of SO2 and H2S are serious pollutants. Impurities dissolved in superheated water may also cause environmental damage on discharge to surrounds, plus thermal pollution. Possibility of earth subsidences over mining area where the underground caverns left after extracting sulfur are difficult to back-fill.

i

High temperature, large surface area of S to increase reaction rate.

ii

A mixture of SO2 and excess O2 is passed over several layers of V2O5 catalyst in pellet form. Temperature is at 600°C, then cooled slightly (400°C) between each pass over catalyst layer.

iii

SO3(g) + H2SO4(l) → H2S2O7(l) H2S2O7(l) + H2O(l) → 2H2SO4(l)

c

Manufacture of fertilisers such as superphosphate and ammonium sulfate. Cleaning iron before galvanising or electroplating.

d

i

Cu(s) + 2H2SO4(aq) → CuSO4(aq) + 2H2O(l) + SO2(g)

ii

2 4 C2H5OH(l) ⎯ ⎯ ⎯ ⎯→ C2H4(g) + H2O(l)

iii

Ba(NO3)2(aq) + H2SO4(aq) → BaSO4(s) + 2HNO3(aq)

H SO (l )

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4

a

2Na(Hg) + 2H2O(l) → 2NaOH(aq) + H2(g) + 2Hg(l)

b

2Cl–(aq) → Cl2(g) + 2e–

c

Overall equation for mercury cell is 2NaCl(aq) + 2H2O(l) → 2NaOH(aq) + Cl2(g) + H2(g) 1 mole NaCl produces 1 mole NaOH 1.00 × 103 kg of NaCl = n(NaCl) =

m mol of NaCl M 1 × 10 3 × 10 3 mol NaCl 58.44

= 1.711 × 104 mol NaCl = 1.711 × 104 mol NaOH produced mass (NaOH) = n × molar mass = (1.711 × 104) (40.00) = 6.84 × 105 g = 6.84 × 102 kg d

e

i

Main hazard is Hg pollution. Hg can be lost from the process in vapour form or in contaminated brine. The Hg in brine released to the environment (usually the ocean) is taken up by organisms and passed along the food chain, becoming more concentrated at each stage. Mercury can damage the human nervous system/brain.

ii

Using the membrane process, a concentrated NaCl solution is electrolysed to produce NaOH and H2 at the cathode and Cl2 at the anode. The anode and cathode compartments are separated by a water-impermeable membrane made of synthetic polymers. This membrane only allows Na+ ions to pass through it and so prevents contamination of the NaOH with NaCl and the formation of other products such as NaOCl.

Uses of NaOH: •

manufacture of soaps and detergents

•

manufacture of rayon and plastics

•

extraction of alumina from bauxite.

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5

a

b

The hydrophobic tail of soap attaches to oily particles and the other charged end dissolves in water. Soap ‘floats off’ the oil particles from the plate.

6

c

In hard water, the action of soap is much less effective because it reacts with Ca2+ and Mg2+ ions to form insoluble precipitates (scum).

d

Early detergents were anionic branched-chain alkylbenzene sulfonates. They were not easily biodegradable and caused foam build-up in waterways. Phosphates were added to detergents to enhance their cleaning power by softening the water and increasing alkalinity. In waterways these phosphates caused algal blooms and eutrophication.

a

Manufacture of glass, pharmaceuticals, soaps, detergents, dyes. To neutralise excess acid from industrial chemical processes. To remove ink from high-quality white waste paper in recycling process.

b

X = brine (NaCl(aq)) Y = limestone (Na2CO3)

c

CaCO3(s) + 2NaCl(aq) → Na2CO3(s) + CaCl2(aq)

d

The statement ‘involves the conversion of two cheap abundant raw materials into two useful products’ infers a simple reaction, whereas five chemical reactions take place in the process. NH3(g) is not mentioned, although it is essential in the process. Recycling of NH3 reduces the cost, but it is an expensive raw material need for the initial process. There is a possibility of pollution with any NH3(g) loss to the atmosphere.

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The other recycled substance, CO2, is used continuously and still needs additional CO2 supplied from heating limestone. Geographic location dominates access to raw materials which, even if present in abundance, may still have to be mined, treated for undesirable side products, and purified. Calcium chloride may not be such a ‘useful’ product. Often its production far exceeds its demand. Although the total outcome of the process produces thermal pollution, considerable energy input is still necessary to make the whole process work—pumping, bubbling, heating limestone, NaHCO3 and so on.

7

e

Thermal pollution; disposal of excess Ca(OH)2 which must be neutralised and stored before release into the environment.

f

Availability and cost of land, raw materials, transport, energy and water supply need to be determined; supply of labour, accessibility to markets, social and safety issues as well as those of waste disposal and the environment..

a

b

anode: 2Cl–(aq) → Cl2(g) + 2e–

oxidation

cathode: 2H2O(l) + 2e– → H2(g) + 2OH–(aq) 8

reduction

Production of SO3(g) during the industrial manufacture of H2SO4(aq): a

In a closed system a reaction that is in dynamic equilibrium is one in which the rate of the forward reaction equals the rate of the reverse reaction. The macroscopic (observable) properties of a dynamic system in equilibrium are constant, but both forward and reverse reactions are occurring at equal rates at the molecular level.

b

The production of SO3(g) is an exothermic reaction: 2SO2(g) + O2(g) ↔ 2SO3(g) + energy

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Lowering the temperature would cause the system to shift to the right-hand side of the equation (products) in order to counteract the drop in temperature, since the formation of products releases energy. However, the drop in temperature would also lead to a very low reaction rate. Thus a compromise must be reached between yield and rate. Similarly, an increase in pressure would cause the reaction to move to the products side in order to minimise the increase in pressure. Increasing pressure would increase both reaction rate and yield, but the necessary equipment is very expensive to use on an industrial scale.

9

c

Industrially, a very high yield is obtained by using moderately high temperatures and atmospheric pressure together with a very effective catalyst, V2O5, in the presence of excess O2.

a

K=

[H 2 O] 2 [Cl 2 ] 2 [HCl] 4 [O 2 ] 4HCl

b

[initial]

2

+

O2 0.5

↔

2H2O 0

+

2Cl2 0

–1

Let x mol L of O2 react in order to achieve equilibrium. Then 4x mol L–1 of HCl will react with x mol L–1 O2 and 2x mol L–1 are formed of H2O and Cl2. Then at equilibrium: [HCl] = (2 – 4x) mol L–1 [O2] = (0.5 – x) mol L–1 [H2O] = 2x mol L–1 [Cl2] = 2x mol L–1 = K (2x) 2 (2x) 2 ( 2 − 4 x ) 4 ( 0 .5 − x )

x, and thus equilibrium concentrations, can only be expressed as a function of K, which is unknown. Insufficient information. c

Since the reaction is exothermic, if temperature is decreased, the system will shift to the product side of the equation in order to counteract the change in temperature until equilibrium is re-established. Thus the K value would increase.

d

If pressure is decreased, the system moves to the side of the equation which has more gas particles in order to counteract the loss of pressure imposed on it. The reactants side has 5 mol of reactants compared to 4 mol of products. So the system shifts to the left and the K value decreases.

e

High pressure, low temperature. Low temperature would decrease the reaction rate so that it became an unviable commercial proposition, so probably a catalyst to increase reaction rate would have to be introduced. Alternatively, one of the

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products could be removed from the system on formation, thus producing a small but constant flow of products. 10

a

Extraction of a metal from its sulfide ores; and extracted from natural gas and petroleum crudes in which it is a pollutant.

b

S(s) + O2(g) → SO2(g)

c

SO3(g) + H2O(l) → H2SO4(aq)

d

•

SO3(g) is absorbed in concentrated H2SO4 to form oleum, H2S2O7 SO3(g) + H2SO4(l) → H2S2O7(l)

•

Water is added to produce sulfuric acid H2S2O7(l) + H2O(l) → 2H2SO4(l)

e

The exothermic reaction forms a fog of H2SO4 droplets, which make working conditions difficult.

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Solutions Manual: Module 5 Shipwrecks, corrosion and conservation

CHAPTER 20: THE OCEAN ENVIRONMENT Review exercise 20.1 1

The bonding and intermolecular forces within the solute. The forces between the solute and water. The temperature of the water.

2

High intermolecular forces between gas molecules and water molecules, resulting in ionisation reactions.

Review exercise 20.2 1

As temperature increases, solubility of gas decreases. As pressure of the gas above water increases, solubility of gas increases.

2

CO2(aq) ↔ CO2(g) so when the container is opened, the pressure of CO2(g) above the liquid rapidly decreases. The equation shifts to the right-hand side in order to counteract the fall in pressure by producing more CO2(g) molecules from the solution (Le Chatelier’s principle).

3

[O2] is greatest near the surface due to atmospheric O2 and O2 produced by photosynthetic aquatic organisms. [O2] decreases as depth of ocean increases due to use of O2 in respiration of organisms and their decomposition. At great depths, [O2] may increase due to presence of dense, cold water containing high O2 levels—‘global conveyor belt’. [CO2] increases with ocean depth. At the surface, CO2 is used for photosynthesis. With increasing depth and less light penetration, less photosynthesis occurs and more CO2 is produced due to respiration and decomposition of organisms.

Review exercise 20.3 1

The leaching effects of rain water passing through terrestrial environments and hydrothermal vents in mid-ocean ridges.

2

Aquatic electric current transfer requires mobile ions and sea water has these.

3

No outlet to large sea mass. High evaporation, low precipitation due to regional weather interferes with water cycle.

4

Salts of chromium, copper, gold, manganese and nickel are generally insoluble. Superheated water from volcanic mid-ocean ridges readily dissolves these minerals, which become dispersed and form underwater mineral deposits.

Chapter 20 Application and investigation 1

KCl—soluble. Ions surrounded by polar water molecules. C6H12O6—soluble. Polar molecule forms hydrogen bonds with water molecules. © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2006. This page from Chemistry Contexts 2 Teacher’s Resource second edition may be reproduced for classroom use.

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CO2—slightly soluble. Reacts with water to form carbonic acid which then partially ionises. C2H6—insoluble. Non-polar solute in polar solvent. Will not react with water. 2

NH3, SO2, O2 decreasing order of solubility. Decreasing order of reaction with water. NH3 readily reacts with water forming ammonium and hydroxide ions. SO2 reacts with water, forming H2SO3 which partially ionises. O2 molecules have only dispersion forces with water molecules.

3

a

Weak dispersion forces between water and O2 decrease as temperature or average kinetic energy of particles increases. Hence, as temperature increases, solubility of O2 decreases. Water which has been boiled is depleted of dissolved O2. After cooling, a goldfish would not survive in this water.

b

Pressure increases the number of gas molecules per unit volume. This causes an increase in the rate at which the gas enters the solution and so the concentration of dissolved gas increases. Carbonated drinks are bottled at high pressures of carbon dioxide to ensure high concentration of dissolved CO2 in the liquid.

4

Investigation

5

Investigation

6

a

Weigh a sample of sea water in a sealed container (W1).

b

Remove top from container and warm.

c

To ensure all the gas has escaped from the seawater, replace top securely, shake conatiner and contents. When liquid has settled, remove top. If any further gas appears to be released, repeat b.

d

Replace top, re-weigh bottle (W2).

e

Mass of dissolved gases = (W1 – W2) g.

7

Investigation

8

Investigation

9

Investigation

CHAPTER 21: OXIDATION–REDUCTION REACTIONS Review exercise 21.1 1

Oxidation is loss of electrons. Reduction is gain of electrons. For electron transfer to occur, the two processes must occur simultaneously, e.g. zinc in a solution of copper(II) sulfate will lose electrons (oxidation) to copper(II) ions, which are reduced. Zn(s) → Zn2+(aq) + 2e– © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2006. This page from Chemistry Contexts 2 Teacher’s Resource second edition may be reproduced for classroom use.

Solutions Manual: Module 5 Shipwrecks, corrosion and conservation

Cu2+(aq) + 2e– → Cu(s) Overall: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) 2

→ Fe 2 + (aq) + 2e −

Fe(s) +

E° = 0.44 V

−

→ 2Ag(s) E° = 0.80 V 2Ag (aq) + 2e + 2+ Fe(s) + 2Ag (aq) → Fe (aq) + 2Ag(s) E° = 1.24 V

3

Cu(s) |Cu2+(aq)| |Ag+(aq)| Ag(s)

4

Galvanic: Chemical energy is converted to electrical energy. The reaction is spontaneous with a positive voltage. Electrolytic: Electrical energy is converted to chemical energy. The reaction is not spontaneous and energy must be supplied.

Review exercise 21.2 1

a

Galvani: The muscle contracted due to the passage of an electric current. The muscle had generated the electricity. Volta: The different metals in a solution generated the electricity, not the muscle.

b 2

Volta made the first galvanic cell to produce an electric current.

Na, K, Mg and Ca are found in nature in compounds which contain their stable ions. Electrolytic techniques enabled electrons to be added to these metal ions and convert them to elements.

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Solutions Manual: Module 5 Shipwrecks, corrosion and conservation

Review exercise 21.3 1

To obtain a reactive metal by electrolysis, the metal ion must be reduced at the cathode. If any water is present, it will be reduced preferentially at the cathode, since the energy required to do so is less than that needed to reduce the metal ion. Thus the metal ion must be in a liquid, not aqueous, state.

2

Oxidation: oxygen

2H2O(l)

→

O2(g) + 4H+(aq) + 4e–

Reduction: hydrogen

2H2O(l) + 2e–

→

H2(g) + 2OH–(aq)

3

Possible cathode reactions: Cu2+(aq) + 2e– 2H2O(l) + 2e

–

→ →

Cu(s)

E° = 0.34 V –

H2(g) + 2OH (aq)

E° = –0.83 V

Deposition of Cu(s) is favoured as it has a more positive E° value. This remains unaltered as [CuCl2] increases. Possible anode reactions: 2Cl–(aq) 2H2O(l)

→

Cl2(g) + 2e–

→

+

E° = –1.36 V –

O2(g) + 4H (aq) + 4e

E° = –1.23 V

Under standard conditions, the oxidation of water is favoured as it has a more positive E° value. However, as [Cl–] increases, the discharge of chlorine gas becomes more favourable. 4

a

Cathode: silver 2Ag+(l) + 2e– → 2Ag(l) Anode: iodine 2I–(l) → I2(l) + 2e−

b

Cathode: silver 2Ag+(aq) + 2e– → 2Ag(s) Anode: oxygen 2H2O(l) → O2(g) + 4H+(aq) + 2e−

c

Cathode: hydrogen 2H2O(l) + 2e− → H2(g) + 2OH−(aq) Anode: oxygen 2H2O(l) → O2(g) + 4H+(aq) + 2e−

Chapter 21 — Application and investigation 1

a

Fe(s) + Cu2+(aq) → Fe2+(aq) + Cu(s) brown

b

Copper ions are reduced to copper atoms by the addition of electrons. These electrons come from iron atoms which are oxidised. The transfer of electrons from one species to another constitutes a redox reaction involving simultaneous gain and loss of electrons.

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Solutions Manual: Module 5 Shipwrecks, corrosion and conservation

2

a, b

3

Investigation

4

Investigation

5

Electrons are gained in reduction and lost in oxidation. The number of moles of electrons lost in oxidation equals the number of electrons gained in reduction. The charge on ions is equal in magnitude to exactly one, two, three or four times that of one electron. This corresponds to the transfer of electrons during the formation of ions.

Q , it is possible to work out the 9.65 × 10 4 number of moles of electrons which flow through a circuit when a given current flows for a fixed time.

By using the formula Q = It and n(e − ) =

Volumes and masses of products can be calculated, knowing the numbers of electrons involved at each electrode half equation. 6

CrO42–(aq) + 8H+(aq) + 6e– → Cr(s) + 4H2O(l)

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Solutions Manual: Module 5 Shipwrecks, corrosion and conservation

7

8

Silver anode would dissolve as the silver metal is oxidised. Ag(s) → Ag+(aq) + e–

E° = –0.80 V

which has a higher positive value of E° and therefore occurs in preference to 2H2O(l) → O2(g) + 4H+(aq) + 4e– 9

E° = –1.23 V

In electrolysis, the most positive E° values of possible half reactions at the electrodes will nominate which reaction occurs, e.g. reduction at cathode: Na+(l) + e– → Na(s) or 2H2O(l) + 2e– → H2(g) + 2OH–(aq)

E° = –2.71 V E° = –0.83 V

Reduction of water is preferred. Other factors affecting products at anode and cathode: i

nature of electrodes—refer to text page 378

ii

molten or aqueous electrolyte—refer to text page 377

iii

concentration of electrolyte—refer to text page 378

CHAPTER 22: CORROSION Review exercise 22.1 1

Corrosion involves the oxidation of a metal to form positively charged ions by the loss of electrons. The electrons are generally transferred to atmospheric O2 or other substances acting as oxidising agents.

2

A very thin coating of metal oxide on the surface of the metal can form a protective layer, preventing any further reaction between the metal and the oxidising agent. © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2006. This page from Chemistry Contexts 2 Teacher’s Resource second edition may be reproduced for classroom use.

Solutions Manual: Module 5 Shipwrecks, corrosion and conservation

3

Chromium, zinc, aluminium or manganese.

Review exercise 22.2 1

Ready availability. Economical extraction and refining methods.

2

Iron is an active metal susceptible to corrosion. Moist, salty sea conditions accelerate the rusting process. Rusting requires moisture and oxygen—conditions to which ships are always exposed.

3

An iron bar on the ground can trap moisture between the ground and the bottom layer. The underside becomes an anodic site where the iron oxidises. The surfaces of the iron bar exposed to atmospheric oxygen become cathodic sites of reduction. The moisture between the ground and the iron bar would contain certain dissolved salts which act as an electrolyte and would accelerate the rusting process. Iron nails stored in a sealed bottle would have atmospheric oxygen surrounding them but only water vapour in the air, so the rate of corrosion is much less.

4

a

The differential aeration principle applies. Between the tightly woven central strands, water may be trapped and oxygen is excluded. These surfaces become anodic sites. The external strands are exposed to air and become cathodic sites.

b

Central strands, anodic sites: Fe(s) → Fe2+(aq) + 2e– Outer strands, cathodic sites: O2(g) + 2H2O(l) + 4e– → 4OH–(aq)

5

Copper is a less reactive metal than iron, and galvanic action occurs. The iron becomes anodic and corrodes: Fe(s) → Fe2+(aq) + 2e– The copper becomes cathodic where oxygen from the air is reduced: O2(g) + 2H2O(l) + 4e– → 4OH–(aq)

Review exercise 22.3 1

The magnesium is a sacrificial anode. It corrodes preferentially to the steel tank.

2

Pipelines may be painted, put at a negative potential at intervals, or have attached sacrificial anode blocks at intervals. They may also be mounted above ground to reduce corrosion by the differential aeration principle.

3

The standard reduction potential: Cu2+(aq) + 2e– → Cu(s)

E° = 0.34 V

This is more likely to occur than the oxidation of copper atoms or the Cu+ ion: © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2006. This page from Chemistry Contexts 2 Teacher’s Resource second edition may be reproduced for classroom use.

Solutions Manual: Module 5 Shipwrecks, corrosion and conservation

Cu(s) → Cu2+(aq) + 2e–

E° = –0.34 V

Cu+(aq) → Cu2+(aq) + e– E° = –0.16 V The higher E° value of the reduction half equation for Cu2+ ions protects the metal from corrosion. Chapter 22 Application and investigation 1

In coastal areas such as Kiama and Byron Bay, sea spray produces the necessary conditions for rusting—water, oxygen and electrolytic salts. These conditions promote accelerated rusting of steel structures compared to inland areas such as Dubbo where there is less water in the atmosphere and negligible electrolytic salts present in this atmospheric water.

2

Aluminium is a passivating metal which forms a very thin, but strongly adhering, layer of aluminium oxide on its surface. This oxide layer protects the metal from further oxidation and is not soluble in water.

3

Investigation

4

a

i

Yes. No reaction between the nickel and iron(III) salt would occur.

ii

No. The nickel can react with the iron(III) salt: Ni(s) + 2Fe3+(aq) → Ni2+(aq) + 2Fe2+(aq)

b

No. The steel tank would corrode: Fe(s) + Cu2+(aq) → Fe2+(aq) + Cu(s)

5

E°(cell) = +1.03 V

E°(cell) = +0.78 V

Magnesium is a more active metal than iron. It would form its own protective oxide coating and protect the underlying steel. It would also act as a sacrificial anode if the magnesium layer were cracked. However, magnesium is readily attacked by mild acids. Gold, copper and nickel are less reactive metals than iron. Copper and nickel will form their own protective oxide coating and gold will remain untarnished. All will protect the iron. However, any tiny gaps in the surface layer will lead to galvanic action in which the iron is preferentially oxidised. In addition, these are expensive metals.

6

Tin is a less reactive metal than iron. Once the tinplate of a can is broken, the iron will preferentially corrode by galvanic action. Galvanised iron has a protective zinc coating. If the protective coating is broken, the zinc will preferentially corrode before iron because zinc is a more active metal than iron.

7

Aluminium forms a tenacious, impervious oxide layer on its surface. Incorporating it with zinc in Zincalume® will enhance the protective oxide layer that zinc would form by itself.

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Solutions Manual: Module 5 Shipwrecks, corrosion and conservation

8

Just below the surface the steel becomes the anodic site for oxidation. The sea water, rich in electrolytes, transfers the electrons to the steel surface exposed to atmospheric oxygen above the water, which then acts as the cathodic site. Iron just under water dissolves away, while rust forming on the cathodic area flakes away faster due to the action of the sea.

9

a b

10

Magnesium bar:

Mg(s) → Mg2+(aq) + 2e–

Steel pipe:

2H2O(l) + 2e– → H2(g) + 2OH–(aq)

Electrons flow from the magnesium bar to the pipe.

Investigation

CHAPTER 23: CAN CHEMISTRY SAVE THE TITANIC? Review exercise 23.1 1

Sunlight would not affect the temperature of water in deep, dark depths. There are fewer significant convection currents that would mix water from different depths.

2 Variable

Effect

Temperature

Increase temperature, increase reaction rate of corrosion.

Light

Increase light, increase photosynthetic aquatic organisms and their consumers, which could increase corrosion below water line.

Dissolved O2

Corrosion is greatest where [O2] is greatest. O2 is reduced at the cathode, as Fe is oxidised at anode site.

Dissolved CO2 As [CO2] increases, acidity increases as carbonic acid provides H+ ions to promote reduction of O2 and so increase corrosion rate. Salinity

As salinity increases, the number of ions increases. The electrolyte becomes more effective and so accelerates corrosion.

Review exercise 23.2 1

The lower temperatures would decrease the rate of the corrosion reaction.

2

Anaerobic bacteria are prevalent on ships since the hull provides a more ready supply of iron than sea water. The bacteria reduce sulfates to hydrogen sulfide, which in turn increases the acidity of the sea water. This promotes the reduction of O2 as well as aiding in dissolving rust. Hydrogen sulfide also reacts with other metals to form insoluble metal sulfides.

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Solutions Manual: Module 5 Shipwrecks, corrosion and conservation

3

Less active metals in sea water close to the steel hull accelerate the corrosion of iron. An electrochemical cell is set up with, for example, the less active copper acting as a cathode to the iron anode of the hull.

Review exercise 23.3 1

In oak artefacts, the cellular structure of the wood is destroyed by crystallising sulfate and chloride ions. Salt crystals in fine pores or cracks can distort the shape of an artefact, break it up or react chemically with it.

2

Oxidation of any exposed silver would take place, removing more silver from the coin.

3

Preservative of wood and it restores supple structure to leather.

Chapter 23 Application and investigation 1 Factor

a

b

Surface Deep sea water sea water

2

Difference from a → b in rate of Fe rusting

Temperature

higher

low

Rate decreases with decrease in temperature.

Light intensity

high

low

Damaging sea life decreases with decrease in photosynthetic organisms together with associated consumers.

Dissolved O2

high

low

Oxidation of Fe decreases with reduction of oxygen levels.

Dissolved CO2

low

higher

As pH decreases with dissolving CO2, the oxidation of Fe would be promoted.

Salinity

high

higher

Improved electrolytic action with more ions present would increase corrosion rate.

a

In deep waters it was expected that the lack of damaging sea life, salinity levels, freezing temperatures and low oxygen content would leave the Titanic well preserved.

b

Galvanic action was set up between the steel and the presence on board of a number of other less active metals. Differential aeration caused by sediment or calcareous deposits from marine organisms resulted in high levels of localised corrosion. Oxygen levels were higher than anticipated. This increased the corrosion rate.

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Solutions Manual: Module 5 Shipwrecks, corrosion and conservation

Slight lowering of pH levels due to the presence of anaerobic bacteria accelerated the reduction of O2 and hence the oxidation of iron. Hydrogen ions may aid in the dissolving of rust. 3 4

a

SO42–(aq) + 10H+(aq) + 8e– → H2S(aq) + 4H2O(l)

b

2Ag(s) + H2S(aq) → Ag2S(s) + H2(g)

a

In oak artefacts, the cellular structure of the wood is destroyed by crystallising sulfate and chloride ions. Salt crystals in fine pores or cracks can distort the shape of an artefact, break it up or react chemically with it.

b

Treatment with weak acids such methanoic acid or ammonium citrate. Mechanical cleaning.

c

Chloride and sulfate ions are allowed to diffuse out of the wood by replacing the ionic solution with polyethylene glycol as a preservative. In leather, the salts are leached out by placing it in distilled water and then in a 30% PEG solution for approximately 2 weeks. The leather absorbs PEG to keep it supple even after drying.

5

a

The formation of silver sulfide can be reversed by making the silver coin the cathode in an electrolyte cell: cathode:

Ag2S(s) + 2e– → 2Ag(s) + S2–(aq)

also

2H2O(l) + 2e– → H2(g) + 2OH–(aq)

The resulting H2(g) helps to loosen any corroded material on the surface of the coin. b

Oxides and chlorides of iron may be reduced to form elementary iron if the affected object is placed in an alkaline solution and made into the cathode in an electrolysis reaction. Fe2O3(s) + 3H2(g) → 2Fe(s) + 3H2O(l) FeCl2(s) + H2(g) → Fe(s) + 2HCl(g)

6–8 Investigation

Module 5 REVIEW 1

Galvanic action was set up between the steel and the presence on board of a number of other less active metals. Differential aeration caused by sediment or calcareous deposits from marine organisms resulted in high levels of localised corrosion. Oxygen levels were higher than anticipated. This increased the corrosion rate. © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2006. This page from Chemistry Contexts 2 Teacher’s Resource second edition may be reproduced for classroom use.

Solutions Manual: Module 5 Shipwrecks, corrosion and conservation

Slight lowering of pH levels due to the presence of anaerobic bacteria accelerated the reduction of O2 and hence the oxidation of iron. Hydrogen ions may aid in the dissolving of rust. 2

3

a

Paint prevents oxygen and water coming into contact with the iron.

b

Cathodic protection of steel—use of a sacrificial anode such as Mg or Zn which oxidises preferentially to the iron, preventing the formation of Fe2+ ions. A sacrificial anode is attached to the hull at regular intervals and must be covered by sea water so that ions can migrate between anode and cathode.

c

Iron alloyed with elements such as chromium, carbon, molybdenum and manganese form steels which are resistant to corrosion. Pure stainless steel effectively resists rust but is prohibitively expensive. Hulls of new ships now have surface stainless steel alloys which protect the bulk of the steel underneath from corrosion.

d

Corrosion of steel requires the presence of O2, water and electrolytes in the water. The ocean environment has an abundance of sea water and sea spray containing many ions. O2 is in the air and is dissolved in upper layers of the ocean. Unless steel is protected from the water and oxygen, the corrosion rate is very high.

Iron corrodes to form Fe3+ ions: 4Fe(s) + 3O2(g) + 2H2O(l) → 2Fe2O3.H2O(s) The resulting oxide is soft and porous, allowing O2 and H2O to pass through it to another Fe layer, which corrodes again. Aluminium oxidises to form a very thin coat of aluminium oxide, which strongly adheres to the metal layer, is insoluble in water and thus forms a protective barrier against more oxidation occurring. 4Al(s) + 3O2(g) → 2Al2O3(s)

4

a

For a reaction to occur spontaneously, the sum of the two half reactions in the redox reaction must have a positive E° value. Anode: Fe(s) → Fe2+(aq) + 2e– –

E° = 0.44 V –

Cathode: O2(g) + 2H2O(l) + 4e → 4OH (aq)

E° = 0.40 V

The sum of these two reactions is E° = +0.84 V Hence, providing an electrolyte is present to allow migration of ions between electrodes, the reaction is spontaneous. b

Acidic conditions could change the cathode reaction to: O2(g) + 4H+(aq) + 4e– → 2H2O(l)

E° = 0.82 V

which would increase the overall value of the redox reaction and hence accelerate the rusting process. © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2006. This page from Chemistry Contexts 2 Teacher’s Resource second edition may be reproduced for classroom use.

Solutions Manual: Module 5 Shipwrecks, corrosion and conservation

5

Soft drink is carbonated: CO2(aq) ↔ CO2(g) Releasing the pressure above the drink when the can is opened causes more CO2(g) to come out of solution and the solution to go ‘flat’. As temperature increases, the reaction rate would increase as well. As temperature increases outside the refrigerator, the solubility of CO2 in solution decreases and more CO2 is released to the environment.

6

a

Ca2+ ions are present in sea water. Decomposing organisms which act on paper and leather produce CO2(g) which in turn reacts with water to form HCO3–(aq). CO2(g) + H2O(l) ↔ H2CO3(aq) H2CO3(aq) ↔ H+(aq) + HCO3–(aq) Thus the two ions Ca2+(aq) and HCO3–(aq) can react to form calcium carbonate.

7

b

Careful mechanical cleaning and chemical cleaning with relatively weak acids such as methanoic acid or ammonium citrate.

a

When twisted, the surface oxide layer will develop fine cracks and expose the metal. The orderly crystal structure of iron is distorted, making it easier for individual Fe atoms to break away from the crystal as Fe2+ ions from anodic oxidation.

b

Greater moisture in the air which contains sea salts acts as an effective electrolyte, which accelerates corrosion.

c

Zinc is a passivating metal which forms a coating to protect steel nails. If the coating is damaged, exposing the steel beneath, zinc will act as a sacrificial anode since it is oxidised more readily than iron. A tin coating also excludes oxygen and water from the steel nail surface, but if the steel is exposed, the iron has a lower ionisation energy than tin and will oxidise first.

8

Faraday established the basic laws of electrolysis that govern the quantitative relationships in electrolytic processes. Faraday’s 1st law—refer to text page 375 Faraday’s 2nd law—refer to text page 375

9

a

It is a sacrificial anode.

b

Zn(s) → Zn2+(aq) + 2e–

E° = 0.76 V

As the zinc is oxidised, it forms Zn2+(aq), so must be replaced regularly. c

The iron in ocean oil and gas rigs can corrode or oxidise according to the halfreaction

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Solutions Manual: Module 5 Shipwrecks, corrosion and conservation

Fe(s) → Fe2+(aq) + 2e–

E° = 0.44 V

Any metal which has a more positive E° value than Fe(s) for the oxidation halfreaction will oxidise first and so can act as a sacrificial anode, provided it does not react with water itself. Instead of Zn, you could use Cr, Mn, Al, Mg. 10

a

In a galvanic cell, a spontaneous chemical reaction takes place in which stored chemical energy is converted to electrical energy.

b

c

anode:

Zn(s) → Zn2+(aq) + 2e–

cathode:

Ag+(aq) + e– → Ag(s)

For every mole of Zn (65.39 g) lost at the anode, 2 mol of Ag (2 × 107.9 g) are deposited at the cathode.

11

a

Mass Zn lost

:

mass Ag deposited

65.39 g

:

215.8 g

1

:

3.3

Copper is resistant to corrosion and has a higher ionisation energy value than Mg. This enables Mg to act as a sacrificial anode. The oxidation half reaction for

Cu(s) has E°

=

–0.34 V

Fe(s) has E°

=

+0.44 V

Mg(s) has E°

=

+2.37 V

The highest E° value will occur first. © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2006. This page from Chemistry Contexts 2 Teacher’s Resource second edition may be reproduced for classroom use.

Solutions Manual: Module 5 Shipwrecks, corrosion and conservation

12

b

From 11a, magnesium will have the greatest tendency to oxidise. As well as reducing O2, the electrons from magnesium can reduce any Fe2+ ions that are formed and so inhibit corrosion.

c

Underground pipes are exposed to O2 and particularly water containing dissolved salts. The presence of an efficient electrolyte accelerates corrosion.

a

Chemical weathering from terrestrial environments. Hydrothermal vents in mid-ocean ridges.

b

Cl–, SO42–

c

Desalination: Leaching takes time. Chloride ions are difficult to remove and further electrolysis is required. Materials often have to be treated with preservatives. Electrolysis: An example is the reduction of Ag2S on silver objects with an inert anode in an alkaline solution. It is very effective as this method allows restoration of Ag markings on objects. Hydrogen gas produced by water reduction at the cathode on the surface of the object can help to loosen any flakes of corroded material.

13

Galvani joined two different metallic wires together and placed the unjoined ends into a freshly extracted frog muscle. The electric current that was generated contracted the muscle. Galvani thought the muscle had generated electricity, not the wires in the solution. Galvani is credited with the first generation of an electric current. Alessandro Volta showed that Galvani’s frog muscle contraction was the consequence of ‘metallic electricity’ resulting from contact between the two different metals when he made the first battery with two different metals coming into moist contact with each other. He went on to develop the Voltaic Pile, the first direct current battery. Davy developed improved versions of Volta’s pile, which he recognised as producing a source of electrons from a chemical reaction. Electric current could be used in electrolysis reactions to decompose water and prepare samples of the very reactive elements of potassium and sodium from their molten salts. Thus, Galvani unknowingly introduced the notion of electricity as being associated with differences between two different metals, Volta knowingly developed this idea and generated electricity in a prototype battery with the idea that something was able to flow between the metals, and Davy extended the concepts to chemical compounds.

14

a

zinc, magnesium

b

Zn(s) → Zn2+(aq) + 2e–

E° = +0.76 V

Fe2+(aq) + 2e– → Fe(s)

E° = –0.44 V

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Solutions Manual: Module 5 Shipwrecks, corrosion and conservation

The E° value for the overall reaction is + 0.32 V, which indicates that the reaction can occur spontaneously. Electrons are lost at the zinc anode (oxidation) and transferred to the iron cathode.

15

c

Cathode. Reduction of any Fe2+(aq) occurs through the electrons being lost by the sacrificial anode.

a

2H2O(l) + 2e– → H2(g) + 2OH–(aq)

b

1 mole of electrons has a charge of 9.649 × 104 coulombs. For one mole of H2(g) to be produced,

E° = –0.83 V

2 × (9.649 × 104) C are needed. For 7 × 108 mol of H2 gas, 2 × (9.649 × 104) × (7 × 108) C are needed = 1.351 × 1014 C c

7.0 × 108 mol of H2 gas is approximately 1.71 × 1010 L at 25°C and 101.3 kPa. Vast quantities of H2 bubbling to the surface would displace the air above the H2 source, leading to a lack of O2 for the crew. H2 gas is highly flammable. With air it forms an explosive mixture.

16

a

Reduced as silver ions gain electrons: Ag+(aq) + e– → Ag(s)

b

Aluminium metal acts as the anode in the galvanic cell that is set up, donating electrons to the silver ions. Anodic oxidation: Al(s) → Al3+(aq) + 3e–

c 17

An abrasive cleaner removes the silver content in the tarnish layer on the metal surface. Reduction of silver ions restores the silver to the object.

Crystallisation of Cl– and SO42– salts can cause considerable damage to the structure of materials such as wood and leather. Non-metal porous artefacts must be leached of salts, and protected by preservatives and fungicides before they are brought to the surface and dried. Artefacts must also be cleaned of calcium carbonate encrustations and sediments.

18

Shipwreck A:

Much greater corrosion rate of hull if it is made of iron. Plentiful supply of O2. Does not dry out. Temperature high, good for reaction rate due to location. Hull in greater light, more organisms to aid in breakdown of materials. Because of shallow waters, artefacts of different metals have probably been removed.

Shipwreck B:

Limited supply of sea O2. Cold water. Lower number of destructive sea organisms. Anaerobic bacteria play larger role in corrosion. Interacting different metal artefacts could accelerate corrosion.

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Solutions Manual: Module 5 Shipwrecks, corrosion and conservation

19

a

Low O2 content of water; very low temperatures; lack of damaging sea life.

b

Higher O2 content in the deep water currents than expected, accelerated rate of corrosion of steel structure as oxygen reduced cathodically. Presence of different metal artefacts set up galvanic corrosion in localised areas of shipwreck.

20

a

In an electrolytic cell, electrical energy is converted to chemical energy. The reaction is not spontaneous and energy must be supplied.

b

c

Cu2+(aq) + 2e– → Cu(s)

d

Concentration of ions in electrolyte. Quantity of electrical charge passing through the circuit.

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Solutions Manual: Module 6 Forensic chemistry

CHAPTER 24: THE ROLE OF FORENSIC CHEMISTRY Review exercise 24.1 1

Theft, fraud, counterfeiting, drink driving, arson

2

Forensic chemists may test samples associated with a crime and be asked to testify in court. They do not solve a crime individually; rather, they provide specific pieces of evidence used in constructing a case.

3

All scientific tests, including chemical ones, have errors associated with them. It is possible (although sometimes highly unlikely) to get a false positive (or false negative) result. A forensic chemist has an ethical responsibility to explain how likely or unlikely this is to happen for a particular test. This can make it difficult to answer questions in court that require a ‘yes’ or ‘no’ response, such as ‘Are you 100% sure that the sample you tested came from the suspect?’ The answer might be ‘no’ even when the test is positive and 99.9% accurate! One way to solve this ethical dilemma would be for the forensic chemist to try to give a more complex answer that conveys the accuracy of the test and meets the requirements of the court. A forensic chemist must be impartial. This can raise ethical difficulties if the forensic chemist is employed by the prosecution or the defence. Ideally, this should not be the case.

Review exercise 24.2 1

The possibility of sample contamination can be used to discredit chemical forensic evidence. For example, this happened in the 1997 trial of O.J. Simpson, in which DNA evidence to be used in the case against Simpson was successfully discredited. Forensic laboratories need to ensure that such contamination cannot occur.

2

A DNA sample can be contaminated by a flaking skin cell from another person. An unsealed or unsecured sample of blood or urine from an athlete could be contaminated by the addition of a banned substance.

3

‘Chain of custody’ refers to the full tracking of samples from athlete to final result. It includes observation of athletes while samples are being provided, labelling and packing of samples in the athlete’s presence and transportation of samples under security.

4

Napoleon died in 1821 after a long period of ill health. Doctors performing the autopsy concluded that a perforated, cancerous stomach ulcer was the cause of death. In the 1950s forensic scientists found that samples of Napoleon’s hair contained very high levels of arsenic and suggested that Napoleon had been poisoned with arsenic oxide or a similar compound. By the 1950s it was known that arsenic bonds strongly to sulfur atoms, including those found in the hair protein keratin. It remains bound to keratin as hair grows, thus providing evidence of arsenic intake. This knowledge was unavailable in 1821. © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2006. This page from Chemistry Contexts 2 Teacher’s Resource second edition may be reproduced for classroom use.

Solutions Manual: Module 6 Forensic chemistry

Another forensic chemist showed that the wallpaper in Napoleon’s residence was a source of arsenic that could have adversely affected Napoleon’s health. Wallpaper made in the 1800s commonly used a green pigment called Scheele’s green, containing copper(II) arsenite. This chemical breaks down, releasing a vapour containing arsenic when the wallpaper became damp or mouldy. The scientist suggested that Napoleon probably died of stomach cancer as proposed in the original diagnosis but that arsenic in the wallpaper caused a decline in his health. The chemical contents of the green pigment and factors relating to its breakdown were unknown in 1821. All of the evidence relating to Napoleon’s ill health and/or death due to arsenic poisoning was gathered by analysing trace quantities of arsenic from hair and wallpaper samples. Such analyses require sophisticated extraction methods and the use of instrumentation such as spectrometers, as well as knowledge unavailable at the time of Napoleon’s death. Chapter 24 — Application and investigation 1

Investigation

2

Investigation

3

Forensic pharmacists study medicinal drugs and forensic toxicologists study poisons.

4

Investigation

5

Analytical chemists seek to identify or work out the concentrations of chemicals often in trace quantities. Sophisticated separation and analytical techniques have been developed by analytical chemists to aid this process. Such techniques are invaluable in forensic chemistry where samples to be used as evidence only contain trace quantities of the chemicals to be identified and/or matched with the victim, suspect or scene of the crime.

6

If samples may have been contaminated, the evidence associated with them can be discredited. This was illustrated in the 1997 trial of O.J. Simpson in the US, in which Simpson’s lawyers used the possibility of sample mishandling or tampering to discredit DNA evidence presented against Simpson. A chain of custody in sample testing helps to overcome this problem. In this process (which is often used with athletes), all samples are tracked throughout the testing process from suspect to final result.

7

Duplicate samples are collected so that a second independent test can be performed if the first test identifies the presence of a banned substance. Athletes personally select and seal their sample containers to prevent the possibility of someone else tampering with samples. Athletes have the right to a second analysis if a positive test result is obtained in case the test result is a false positive, i.e. a positive result has appeared for some reason other than the presence of a banned substance in the athlete’s blood or urine.

8

Investigation

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Solutions Manual: Module 6 Forensic chemistry

CHAPTER 25: ANALYSIS OF MATERIAL Review exercise 25.1 1

Compounds can be classified according to (i) their original source; (ii) their solubility in water; (iii) the types of particles found in the substance.

2

Millions of compounds exist on Earth. The process of identifying a compound utilises the classification process. For example, if a compound undergoes combustion, producing carbon dioxide or carbon monoxide and/or soot under reduced oxygen conditions, then it is organic. Organic compounds can be further classified into groups including hydrocarbons, alkanols and alkanoic acids. Having established that a compound is organic, further tests can be performed to determine to which group of organic compounds it belongs. This process can be continued until the compound is specifically identified. Forensic chemists need to be able to classify compounds because the process of classifying a compound using chemical testing enables identification of the compound.

3

Soil particle size, microorganism content and chemical composition (such as percentages of quartz and calcium carbonate) can be used to provide specific location details. The presence of foraminifers indicates that a soil is of marine origin and may sometimes provide specific location details due to the wide variety of foraminifers geographically and over geological time. Glass fragments can be used to match cars with particular car accidents and criminals with crime scenes. This is possible because glasses vary in composition and physical properties. Density and refractive index measurements are used to identify glass samples. Metals and metallic ions occur in soils, paints, glass and metal tools associated with crimes. Metal ion compositions can be used to match a sample with its source. Techniques utilised include mass spectra, atomic absorption spectra and scanning tunnelling microscopy. It is possible to use ‘fingerprinting’ to determine the sources of gold samples because two different samples mined from the same source should have the same distribution of impurities.

4

5

a

The forensic chemist would be involved in the detection and identification of substances such as accelerants as well as substances that might be matched with the suspect, such as trace fibres, paint and glass.

b

Matching paint fragments from victim’s clothing with the car, matching blood from the victim with residues found on the car, matching the rubber of the tyres, and oil from the car with samples taken from the scene of the accident.

Gold contains trace quantities of specific impurities relating to the mineralisation event in which it was formed. Forensic chemists use mass spectrometry to match gold associated with crimes with samples mined in the same location. Both samples should

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Solutions Manual: Module 6 Forensic chemistry

have the same distribution of impurities (‘fingerprint’) if they come from the same location. 6

The fibres do not have the same composition. In a gas chromatograph, different components are identified by different peak locations and the area under each peak provides a measure of mole per cent concentration. The two samples have some significantly sized peaks in quite different locations. For example, sample 1 has a large peak at 35 minutes but sample 2 has no peak in this position. Sample 2 has a moderately sized peak at 21 minutes but sample 1 only has a small peak in this position. Sample 1 has a large peak at 13 minutes but sample 2 has no peak in this position. This implies that sample 1 has some components in significant proportions which are not in sample 2 and vice versa.

Review exercise 25.2 1

Decolourisation of bromine water. Ethene reacts with Br2(aq), forming 1,2-dibromoethane. The red-brown colour of the Br2 disappears. Ethane does not react with Br2(aq).

2

Primary alkanols are oxidised most readily by acidified potassium permanganate. Secondary alkanols are oxidised more slowly. Tertiary alkanols are resistant to oxidation by acidified permanganate because a large amount of energy is necessary to break a C–C bond in order to form an oxidised product.

3

a

Orange to green as the chromium is reduced from an oxidation state of +7 in the dichromate ion to an oxidation state of +3

3CH3CH2OH(aq) + Cr2O72–(aq) + 8H+(aq)

→

3CH3CHO(aq) + 2Cr3+(aq) + 7H2O(l)

(orange) b

(green)

The driver would blow into the breathalyser bag. If there was a high concentration of alcohol in the driver’s blood, there would also be considerable alcohol in his or her breath. This alcohol (ethanol) would dissolve in the dichromate solution, changing it from orange to green.

Review exercise 25.3 1

a Carbohydrate class

Carbohydrate

Origins in nature

Monosaccharide

glucose

fruits, honeys

Monosaccharide

fructose

flowers, fruits, honeys

Disaccharide

sucrose

cane sugar, sugar beet

Disaccharide

lactose

milk

Disaccharide

maltose

grains—hydrolysis of starch

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Solutions Manual: Module 6 Forensic chemistry

b

Polysaccharide

starch

plants—energy source

Polysaccharide

glycogen

animals—energy source

Polysaccharide

cellulose

plants—structural material

cellulose

2

In the case of maltose, which is a reducing sugar, a red precipitate of copper(I) oxide should be formed. Benedict’s solution is an alkaline solution containing copper(II) ions. The copper(II) ions will be reduced, forming red copper(I) oxide. The carbonyl group in the maltose will be oxidised to a carboxylic acid group. Sucrose is not a reducing sugar because the normally reactive carbonyl group within the glucose and fructose units making up sucrose is involved in the glycosidic bond linking the two molecules. Sucrose therefore cannot be readily oxidised, so no reaction should take place.

3

Starch, glycogen and cellulose are all polymers consisting of linked glucose units. They have the general formula (C6H10O5)n. Starch and glycogen consist of long chains of glucose units but the chains are longer and more highly branched in glycogen than in starch. Cellulose consists of long chains of glucose units but the glucose molecules are oriented differently to give a more linear molecule. This permits the polymer molecules to pack more closely together in bundles linked together by hydrogen bonds. (Both starch and glycogen are made up of α-glucose monomers; cellulose is made up of βglucose monomers with every second glucose monomer inverted.)

4

Substances containing starch change colour to blue-black in the presence of an iodine/potassium iodide solution. Starch

+

iodine solution

→

(brown)

starch–iodine complex (blue-black)

This test gives a negative result with cellulose, due to the strong hydrogen bonds between cellulose polymer chains that prevent an interaction with iodine. 5

a

The two identifying functional groups of amino acids are the amine (–NH2) and carboxylic acid (–COOH) groups.

b

Amino acids can link together by undergoing condensation reactions to form peptide bonds. This process can be repeated many times to form polypeptide chains or proteins which consist of a series of amino acids linked through peptide bonds.

Chapter 25 — Application and investigation 1

Inorganic evidence: soils, glass and paint fragments, metals. Organic evidence: carbohydrates, lipids, proteins, nucleic acids.

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Solutions Manual: Module 6 Forensic chemistry

2

3

4

a

Add silver nitrate solution. The formation of a white precipitate that is insoluble in nitric acid and the absence of a precipitate when barium chloride is added to a fresh sample of water indicates the presence of chloride ions.

b

Calcium carbonate should react with hydrochloric acid, forming carbon dioxide gas. Carbon dioxide turns a drop of barium hydroxide in an eyedropper milky. Calcium sulfate should not react with hydrochloric acid.

a

2-hexene decolourises bromine solution (i.e. the red-brown bromine solution changes to colourless). Hexane does not react with bromine solution.

b

1-pentanol is oxidised readily by acidified permanganate solution, i.e. the purple colour of the permanganate ion changes to colourless. 2-methyl-2-propanol resists oxidation by acidified permanganate.

c

Glucose, a reducing sugar, can be oxidised by Benedict’s solution, forming a red precipitate of copper(I) oxide. Sucrose cannot be oxidised by Benedict’s solution.

d

Starch turns blue-black in the presence of an iodine/potassium iodide solution, i.e. the solution changes from brown (the colour of iodine) to blue-black. Glucose does not react with iodine in this way.

a

CH3CH=CHCH2CH3(l) + Br2(aq) → CH3CHBrCHBrCH2CH3(l)

b

3CH3CHOHCH2CH3(l) + Cr2O72–(aq) + 8H+(aq) → 3CH3COCH2CH3(l) + 2Cr3+(aq) + 7H2O(l)

c

2Cu2+(aq) + 2H2O(l) + C6H12O6(aq) → Cu2O(s) + 4H+(aq) + C5H9O5COOH(aq)

5

The refractive index of glass is a measure of the change in direction of light as it enters or leaves glass. Glass will be invisible if it is placed in a medium with the same refractive index. This makes it possible to quickly determine whether a sample consists of a particular type of glass and therefore to ascertain its possible uses.

6

a

Test for the presence and concentration of ethanol (an alkanol) in the driver’s blood; mass spectrometry.

b

Identify the accelerant (a fuel) and try to determine its source; mass spectrometry.

c

‘Fingerprint’ test the gold, comparing it against gold mined at the minesite. Compare the mass spectra of the two samples.

d

Match the paint sample with the suspect’s car using gas chromatography.

7 Type of compound

Elements present

Monomer units

Carbohydrate

carbon, hydrogen, oxygen

C6H10O5 (for polysaccharides)

Lipids

carbon, hydrogen, oxygen

Proteins

carbon, hydrogen,

amino acids

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Solutions Manual: Module 6 Forensic chemistry

oxygen, nitrogen Nucleic acids 8

carbon, hydrogen, oxygen, nitrogen

nucleotides

a

Starch, glycogen and cellulose are all polymers consisting of linked glucose units. They have the general formula (C6H10O5)n. Starch and glycogen consist of long chains of glucose units but the chains are longer and more highly branched in glycogen than in starch. Cellulose consists of long chains of glucose units but the glucose molecules are oriented differently to give a more linear molecule. This permits the polymer molecules to pack more closely together in bundles linked together by hydrogen bonds.

b

Cellulose is insoluble in water because the hydrogen bonds that could be formed between water molecules and cellulose are not strong enough to disrupt the regular arrangement of hydrogen bonds among the closely arranged polymer molecules.

9

Reducing sugars contain a reactive carbonyl (—C=O) group and are therefore easily oxidised by weak oxidising agents such as copper(II) or silver cations. Two examples of reducing sugars are glucose and lactose.

10

Investigation

11

a

Investigation

b

Proteins consist of amino acids linked together by peptide bonds. The 20 amino acids found in the human body can link together in a myriad of different combinations, giving rise to the vast number of proteins found in humans.

12

Investigation

13

DNA is characteristic of a particular organism and is independent of the organism’s age and the type of tissue sampled. This means that it can be used to identify a species or an individual member of a species. Comparisons can be made between different types of biological samples because every cell contains DNA. DNA can remain intact for thousands of years, making it possible to preserve evidence. DNA analyses have high analytical sensitivity. Only tiny samples (about 100 molecules) are needed for an analysis.

CHAPTER 26: TOOLS OF A FORENSIC CHEMIST Review exercise 26.1 1

Chromatography involves the separation of a mixture (often in solution) by passing it over an inert substance like alumina, silica or special paper. The mixture is called the mobile phase and the inert substance is called the stationary phase. The components of the mixture are separated because they are adsorbed onto the surface of the stationary phase with different strengths depending on their particular properties. © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2006. This page from Chemistry Contexts 2 Teacher’s Resource second edition may be reproduced for classroom use.

Solutions Manual: Module 6 Forensic chemistry

2

Gas chromatography: used to measure the blood alcohol (ethanol) level of drivers. Paper chromatography: used to provide evidence about inks and pigments used in forgeries. Ion exchange chromatography: used to separate different proteins from each other. Column chromatography: used for the separation of phenols.

3

Some peaks in the athlete’s sample coincide with peaks in the standard. It is likely that the weight lifter was using steroids.

4

Stepwise elution involves passing solutions containing different salt concentrations and/or different pHs through the column. Proteins with the lowest affinities are eluted through the column first, followed by those with higher affinities.

5

a

Electrophoresis involves separation of a mixture using the property of different rates of ion migration towards electrodes. For proteins, the rate of migration varies with molecular mass and total charge of the molecules. The charge on proteins changes with the pH of the solution. At higher pH values, the charge on proteins is more negative, owing to the neutralisation of any acidic side chains. After separation, the proteins are stained and identified. A reference sample is often used to aid in identification.

b

The proteins should be negatively charged because of the neutralisation of any acidic side chains by the hydroxide ions present in an alkaline solution of pH 9.

Review exercise 26.2 1 Relative mass

Fragment

27 29

CH3CH2

41 57

CH3CH2CH2CH2 and CH3CH2CO

72

2

85

COCH2CH2CH2CH3

114

molecular ion

a

infra-red, ultraviolet, visible, gamma

b

An atom is in its ground state when it has the lowest possible energy, i.e. when the electron in a hydrogen atom is in the first energy level, the hydrogen atom is in its ground state.

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Solutions Manual: Module 6 Forensic chemistry

c

An ‘excited state’ is produced by supplying the atom with sufficient energy for one or more electrons to move to a higher energy level than it would normally be in if it were in the ground state.

3

Strontium produces a scarlet colour in a flame test and barium a yellow-green colour. These colours are characteristic of the predominant wavelength of light emitted when electrons in each type of atom jump from higher energy levels to lower energy levels. Different atoms have different electron arrangements and energy level differences and hence emit different colours.

4

a

The emission spectrum of an element is the characteristic pattern of wavelengths emitted by atoms of that element when they are in an excited state.

b

The emission spectrum of an element consists of a series of lines because electrons in the atoms of the element cannot have an infinite number of possible energy values greater than zero. They can only have energy characteristic of the energy level they occupy. When electrons are transferred from a higher energy level to a lower one, radiation of characteristic energy (equivalent to the difference between the energy levels) is emitted. This radiation has a characteristic wavelength and colour and gives rise to a particular line in the emission spectrum of that element.

c

Each element has a signature line emission spectrum because it has a unique pattern of energy level differences between the shells (and subshells) that electrons can occupy in excited and ground states. This means that the emission spectrum of an element can be used to identify the element. For example, the elements in stars can be identified using this technique.

5

Forensic chemistry often involves the analysis of minute samples. For example, one drop of blood may be the only organic material available for analysis. Ideally, the sample should be preserved throughout the investigation, i.e. the techniques used should be non-destructive. Atomic force spectroscopy and scanning-tunnelling microscopy are two less destructive techniques.

Review exercise 26.3 1

The protein cytochrome c is found is almost all living things. Different species have slightly different sequences of amino acids in their cytochrome c. The degree of similarity of the amino acid sequences among different species is an indicator of the degree of genetic similarity among these species.

2

a

A DNA fingerprint consists of single strands of DNA, which have been treated with DNA probes to become visible. A DNA fingerprint is unique to a particular individual and can be compared with DNA from suspects and victims to locate matches.

b

DNA is extracted from a sample, then cut into fragments using a restriction enzyme. DNA fragments are then separated into bands according to size using

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Solutions Manual: Module 6 Forensic chemistry

electrophoresis. DNA probes are bound to the bands. The DNA band pattern is then transferred to a nylon membrane, which is developed using X-ray film or other photographic methods. c

DNA fingerprinting is a tremendously powerful tool for forensic chemists because a DNA fingerprint is unique to an individual.

d

If the sample is very small, e.g. a drop of blood, the DNA is amplified by a polymerase chain reaction (PCR). In this process, the DNA is gently heated to separate the two strands of the double helix. An enzyme is added which causes each strand to produce its complementary ‘other half’. This process can be repeated several times until sufficient DNA is available for further analysis.

3

If F and M are both the parents of the child, then each band in the child’s DNA fingerprint should correspond with a band from at least one parent. The child’s DNA fingerprint has some bands corresponding with those of the mother and others that do not correspond with either parent. This suggests that M is the child’s mother but F is not the child’s father.

4

a

b

Advantages:

Many criminals reoffend. A DNA database would make it easier to identify and convict criminals and reduce the chances of convicting someone who is innocent.

Disadvantages:

It would interfere with the freedom of criminals because they would be forced to contribute to the database.

Advantages:

It would be possible to conduct extensive medical research using the material in the database together with other information gathered from medical records (if this was permitted).

Disadvantages:

Once information is readily available in a DNA database, it has the potential to be misused. For example, many people are concerned that insurance companies and some employers might use such information in making decisions about who to insure, employ, promote etc.

Chapter 26 — Application and investigation 1

Many samples used in chemical analysis are very small quantities of mixtures. Chromatography is particularly useful for separating such samples into the elements and compounds that make them up.

2

Ion exchange chromatography is used to separate different proteins. A solution containing the protein mixture is passed through an ion-exchange column. Different proteins bind to the ion-exchanger with different affinities. Proteins with low affinities for the ion exchanger move quickly down the column. Those with higher affinities move more slowly. The affinity of proteins for the ion-exchanger depends on the nature of their amino acid side chains and on the pH. © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2006. This page from Chemistry Contexts 2 Teacher’s Resource second edition may be reproduced for classroom use.

Solutions Manual: Module 6 Forensic chemistry

3

In electrophoresis, the ions in a mixture migrate to either the positive or the negative electrode at characteristic rates. For proteins, the rates of migration depend on molecular mass and total charge of the molecules. Different animal species including different species of fish have characteristic protein profiles. This means that electrophoresis can be used to distinguish between them.

4

Flame tests: sodium chloride gives a bright yellow colour (characteristic of sodium) when a solution is sprayed into a flame. Potassium chloride solution gives a lilac colour characteristic of potassium.

5

The presence of characteristic absorption and emission spectra for particular elements suggests that electrons in an atom have only certain allowed energy levels. The energy of the radiation emitted in an emission spectrum or absorbed in an absorption spectrum represents the energy difference between the level the electron occupied and the level to which it transferred.

6

Ultraviolet radiation damages the hereditary material in the skin cells (DNA).

7

Destructive analysis occurs when a sample is destroyed during the analysis. This is a problem in forensic chemistry, because often only very small quantities of samples are available to be used as evidence.

8

Peaks in the athlete’s sample match the standard peaks for amphetamine and ephedrine.

9

a

The mass spectrum has peaks corresponding to the ions of copper. The height of each peak indicates the relative abundance of each ion.

b

In this case, 63Cu has an abundance of approximately 70% and 65Cu has an abundance of approximately 30%. These relative abundances would give rise to an average atomic mass of 63 × 0.7 + 65 × 0.3 = 63.6, which is very close to the atomic mass of copper (63.55).

10

Blood, skin, hair and semen.

11

Investigation

12

Investigation

13

a

The molecular ion peak corresponds to the relative mass of the molecular ion. The fragmentation pattern provides information about molecular structure, i.e. the relative mass of the groups present in the molecule. This information can be used to determine some of the groups present in the molecule.

b

i

Top spectrum = 2-methylhexanal Bottom spectrum = benzaldehyde

ii

Top spectrum: 29 = CHO, 43 = CH3CH2CH2, 100 = CH3CH2CH2CH(CH3)CHO+. Bottom spectrum: 29 = CHO, 77 = C6H5, 106 = C6H5CHO+

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Solutions Manual: Module 6 Forensic chemistry

14

a

The two semen DNA fingerprints (left and centre) match exactly. If one was taken from a suspect and the other from the victim then it is highly likely that this suspect was responsible for the crime.

b

The child’s entire DNA fingerprint features match with either the mother or possible father. It is highly likely that the possible father is the father of the child.

Module 6 REVIEW 1

2

3

a

Glucose is a reducing sugar and will give a brick-red precipitate when warmed with Benedict’s solution. The two remaining solutions will not. If acidified potassium permanganate is added to the remaining two compounds, glycerol will be oxidised (i.e. the purple colour of the permanganate ion will change to colourless due to the formation of Mn2+). Ethanoic acid will not react with acidified potassium permanganate, i.e. the solution will remain purple.

b

Forensic chemists could distinguish between the three substances using a mass spectrometer. Each would give a different relative mass characteristic of the molecular ion and a different pattern of fragments characteristic of the groups in each compound.

a

1-propanol can be oxidised by acidified potassium permanganate more rapidly than 2-propanol can. In each case, the rate of reaction can be estimated from the rate of disappearance of the purple colour of permanganate ion from the solution.

b

These two alcohols could be distinguished used mass spectroscopy. The spectrum of 1-propanol should have a peak at 43 corresponding to the fragment CH3CH2CH2. This peak should be absent from the spectrum of 2-propanol.

a

Pure: metal (elements pure) Impure: concrete, cloth, metals (alloys impure), paint and glass. Any of these samples could also contain organic material such as drops of blood or other fragments of skin.

4

b

Chromatography: this separation process is based on the differing adsorbances of the pure substances in the mixture for the material in the column.

c

Scanning-tunnelling microscope: this type of microscope scans the surface of substances using the current flowing between the surface and the probe to map the sample surface.

Electrophoresis could be used to identify the types of vegetables in the vegetable debris. Electrophoresis uses the rate of migration of ions to identify the biological molecules in a mixture. Each species of vegetable would have a characteristic protein profile that could be matched with a reference sample. DNA fingerprinting could be used to identify © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2006. This page from Chemistry Contexts 2 Teacher’s Resource second edition may be reproduced for classroom use.

Solutions Manual: Module 6 Forensic chemistry

the species of rodent and even matched with hairs from the actual rodent or rodents. In this process, DNA is extracted from a sample, cut into fragments using a restriction enzyme and the fragments are separated into bands using electrophoresis. A radioactive probe is then used to bind the bands onto a membrane and the bands are detected using X-ray film. The developed film can be matched with reference samples. 5

a

Starch turns blue-black in the presence of an iodine/potassium iodide solution, i.e. the solution changes from brown (the colour of iodine) to blue-black. Glucose does not react with iodine in this way.

b

Glucose, a reducing sugar, can be oxidised by Benedict’s solution, forming a red precipitate of copper(I) oxide. Sucrose cannot be oxidised by Benedict’s solution.

6 Polysaccharide Monomer units

7

Shape of polymer

Function

Starch

α–glucose

branched

energy storage in plants

Glycogen

α–glucose

highly branched

energy storage in animals

Cellulose

β–glucose

straight–every second monomer is inverted

structural, in plants

a

amino acids

b

c

The two compounds have different molecular masses, so their ions migrate towards the electrodes at different speeds when a voltage is applied.

d

At high pHs the acidic side chains on these molecules will be neutralised, forming negatively charged ions, which will be attracted to the positive electrode.

8

If samples become contaminated with other DNA-containing material, the results of the analysis will be invalid. Some of the bands on the DNA fingerprint may be caused by the DNA of the contaminating substance rather than by the DNA being analysed. Sources of potential contamination include drops of blood, flakes of skin and hair.

9

a

The emission spectrum of an element is the characteristic pattern of wavelengths emitted by atoms of that element when they are in an excited state.

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Solutions Manual: Module 6 Forensic chemistry

10

b

Each element has an individual emission spectrum because it has a unique pattern of energy level differences between the shells (and subshells) that electrons can occupy in ‘excited’ and ground states. This means that the emission spectrum of an element can be used to identify the element.

c

The flame colours are the colours produced by combining the brightest lines in the emission spectrum of the metal ion of the salt.

a

Non-destructive analysis includes analytical methods in which the sample is not destroyed during the process of the analysis.

b

Often only minute quantities of samples are available for analysis, e.g. one drop of blood. It is desirable to carry out as many tests as possible on these samples and to preserve them in case further testing is required.

c

Atomic force microscopy: in this process, a microscope scans a very sharp tip with a flexible lever over the surface of the substance. The tip moves up and down as the forces between it and the sample surface vary. A computer uses the variation in the forces to produce a topographic image of the surface of the sample.

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Solutions Manual: Module 7 The biochemistry of movement

CHAPTER 27: THE FOODS WE EAT Review exercise 27.1 1

carbon, hydrogen and oxygen

2

Glucose is a six-membered ring compound with the formula C6H12O6. It is the compound from which all livings derive chemical energy during respiration. It is also a building block for many other carbohydrate structures. Glycogen is a condensation polymer of glucose found in animals as a storage compound for glucose.

3

Glycogen is a polysaccharide made up of individual glucose monosaccharides. When two of these monosaccharides ‘link up’, a molecule of water is eliminated. This is a condensation reaction.

Review exercise 27.2 1

One of the major functions of fats in the diet of humans is to provide energy.

2

a Palmitic acid is saturated. The ester is a fat.

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Solutions Manual: Module 7 The biochemistry of movement

b

Two saturated fatty acids and one unsaturated. The ester is a fat.

c

Both are unsaturated fatty acids. The ester is an oil. glycerol + 1 oleic acid + 2 linoleic acid → 1-oleoyl-2-linoleoyl-3-linoleoylglycerol

3

a

As the number of carbon atoms increases, their melting points rise.

b

As the number of carbon atoms increases, the molar mass increases and the intermolecular attractions increase. This will elevate the melting points.

Review exercise 27.3 1

Fibrous proteins tend to be long, tough molecules that provide structure for many body components including bone, hair, skin, tendon and cartilage. Globular proteins tend to be compact, roughly spherical molecules that can move freely in blood and other tissues. They transport materials, fight invading organisms and catalyse many biochemical reactions.

2

a

Amino acids contain an amine group (–NH2) and an alkanoic (carboxylic) acid group (–COOH).

b

Essential amino acids, of which there are eight (nine for children), must be taken in with the diet as they cannot be sythesised in the body.

c

When amino acids link, between the carboxyl group of one molecule and the amine group of the other molecule, a peptide bond is formed; water is eliminated during this condensation reaction.

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Solutions Manual: Module 7 The biochemistry of movement

3

When biochemists describe protein structure they do so using the following sequence: The primary structure of proteins refers to the sequence of amino acids contained within the molecule. The secondary structure refers to the spatial arrangement of the polypeptide chain. The tertiary structure refers to the overall way the protein folds. The quaternary structure refers to the way several polypeptide chains form an aggregated whole.

4

If the complex three-dimensional shape of a protein is destroyed by such agents as heating, a change in pH or reaction with other chemicals, it no longer functions as it should and is described as having been denatured. Denaturing destroys the secondary, tertiary and any quaternary structure of the protein.

Review exercise 27.4 1

Enzymes: •

are specific to particular reactions

•

provide a pathway of lower activation energy

•

speed up biochemical reactions

•

remain unchanged at the completion of the chemical reaction

•

do not alter the equilibrium of the reaction

•

function within restricted pH and temperature ranges.

2

Chymotrypsin is one of the pancreatic enzymes that breaks down protein into polypeptides. Catalase breaks down toxic hydrogen peroxide into oxygen and water.

3

a

In order to work, enzymes must come in contact with the material they react with. This material is the substrate.

b

Where substrate and enzyme meet is known as the active site.

c

If an enzyme is to function it must ‘fit’ the substrate site—hence the specificity of that enzyme.

d

Some enzymes have cofactors that are required for them to work effectively.

Chapter 27 — Application and investigation 1

Glucose, fructose and galactose all share the same molecular formula but each has a different structural formula and therefore each will behave chemically differently. Glucose and galactose are 6-membered rings but fructose is a 5-membered ring. Glucose and galactose differ only in the orientation of the OH group on one of the carbon atoms. © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2006. This page from Chemistry Contexts 2 Teacher’s Resource second edition may be reproduced for classroom use.

Solutions Manual: Module 7 The biochemistry of movement

2

Sucrose is a disaccharide made up of the monosaccharides glucose and fructose. When two monosaccharide molecules combine, a condensation reaction occurs and a water molecule is produced as a by-product. When disaccharides are broken down into their two separate monosaccharides, a water molecule is required. This reaction is the reverse of a condensation reaction and is known as hydrolysis. When sucrose is hydrolysed, glucose and fructose are produced.

3

Plants store surplus glucose as starch within the cells. Animals use another form of polymerised glucose known as glycogen to store their glucose. Glycogen is more highly branched than starch. The glucose molecules are connected through different carbon atoms in the two different compounds.

4

Approximately 1499 (243 000/162.08) glucose units make up this starch molecule. Each time a glucose unit is added to the starch molecule in the condensation reaction, one molecule of water is eliminated. This means that the repeating unit is C6H10O5.

5

a

The structural formula of glycerol is:

The structural formula of stearic acid is:

The structural formula of glyceryl tristearate is:

b

Glycerol: The glycerol molecule has three hydroxy groups which are hydrophilic. The remaining three-carbon hydrocarbon chain is hydrophobic. Stearic acid: The carboxyl group forms hydrogen bonds and is hydrophilic. The 17-carbon chain is non-polar and hydrophobic. Glycerol tristearate has three carboxyl groups which will form hydrogen bonds. This end of the molecule is hydrophilic. However, the three 17-carbon atom hydrocarbon chains are hydrophobic.

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Solutions Manual: Module 7 The biochemistry of movement

c

Glycerol is soluble in water. It has a large hydrophilic and a small hydrophobic component. Stearic acid has a small hydrophilic component but a large hydrophobic component and will be insoluble in water. Glycerol tristearate has a small hydrophilic component but a very large hydrophobic component and will be even less soluble in water.

6

Saturated fatty acids belong to a homologous series with the general formula CnH2n+1COOH or CH3(CH2)nCOOH. They contain no double bonds. Stearic acid is an example. Monounsaturated fatty acids contain one C=C double bond in the hydrocarbon chain. Oleic acid is an example. Polyunsaturated fatty acids contain two or more C=C bonds in the hydrocarbon chain. Linoleic acid is an example.

7

Investigation

8

a

CH3(CH2)4(CH=CHCH2)2(CH2)6COOH

b

Linoleic acid has two double bonds. Each molecule of linoleic acid requires four hydrogen atoms for saturation. One mole of linoleic acid reacts with 2 moles of H2.

9

Take as a typical fat, say, glycerol monostearate C21H42O4. Compare this with a carbohydrate with the same number of carbon atoms, C21H42O21. Fat

Carbohydrate

358.336

630.336

% Carbon

70.3

40.0

% Hydrogen

11.8

6.7

% Oxygen

17.9

53.3

Molar mass

The molecular formula of fat shows it to have much less oxygen per mole than carbohydrate. The percentage composition shows fats to be higher in carbon and hydrogen and lower in oxygen. The combustion of fats and carbohydrates yields carbon dioxide and water in both cases. Molecules with more oxygen/less hydrogen release less energy on combustion than molecules with less oxygen. This makes sense, as combustion is oxidation and a molecule with oxygen in it is already partly oxidised. Thus, fats release more energy upon metabolism than carbohydrates because they contain more hydrogen and less oxygen per carbon than carbohydrate.

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Solutions Manual: Module 7 The biochemistry of movement

10

Weight lifters and body builders are attracted to protein supplements because the nature of their sport requires the development and maintenance of larger-than-usual muscles. A pure supply of protein therefore is seen as a quick route to the deposition of bulk tissue.

11

a

When two amino acids link together, a condensation reaction occurs, with the elimination of a water molecule. This link is between the anion (COO–) of one amino acid and the cation (NH3+) of another. This link is known as a peptide bond.

b

Glycine plus alanine gives the dipeptide Gly-Ala or Ala-Gly. i

ii

12

c

6 (= 3 × 2 × 1)

d

6.7 × 1011 (= 20 × 19 × 18 × 17 × 16 × 15 × 14 × 13 × 12 × 11)

a

Fibrous proteins are long, tough molecules that provide structure for many body components including bone, hair, skin, tendon and cartilage. Globular proteins are more compact, roughly spherical molecules that perform a vast array of functions including the transport of oxygen, fighting disease as antibodies, and in catalysing biochemical reactions.

b

When biochemists describe protein structure, they do so using the following sequence: The primary structure of proteins refers to the sequence of amino acids contained within the molecule. The secondary structure refers to the spatial arrangement of the polypeptide chain. The tertiary structure refers to the overall way the protein folds. The quaternary structure refers to the way several polypeptide chains form an aggregated whole.

c

The secondary structure is held in place by hydrogen bonding, while the tertiary and quaternary structures are held in place by electrostatic forces, hydrogen bonding, hydrophobic forces and covalent disulphide bonds. Of course, the primary structure is held in place by peptide bonds.

d

Disruption of the shape of the protein molecule renders it biologically inactive. This is known as denaturation. It involves disrupting the forces and bonds which determine secondary, tertiary and, if relevant, quaternary structure of the protein, while leaving the chain of amino acids intact. Extremes of temperature, pH and a variety of chemicals can denature proteins.

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Solutions Manual: Module 7 The biochemistry of movement

13

Investigation

14

Investigation

15

Blood contains protein. Warm water is not recommended for the removal of blood stains from garments because the proteins may denature and unravel and then coagulate into solid lumps, making the stain harder to remove.

16

a

Enzymes are responsible for catalysing almost all of the chemical reactions that occur in our bodies. Clearly, without enzymes, living organisms including ourselves could not survive.

b

Because of the need for a precise ‘fit’ between the enzyme and substrate molecules, the mechanisms of enzyme-catalysed reactions are often discussed in terms of a ‘lock and key’ mechanism. If the active site of the enzyme has a specific shape (similar to a lock) then only the right-shaped substrate (key) will fit that site.

17

Investigation

18

a

i

Each hydroxyl group on the glycerol bonds with the corresponding hydroxyl of the stearic acid. The linkage is an ester linkage.

ii

The hydrogen of the hydroxyl group of the first molecule reacts with the –OH of the hydroxyl group of the second molecule. The linkage is a glycosidic linkage.

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Solutions Manual: Module 7 The biochemistry of movement

b

c

iii

Amino acids link together with dipeptide bonds from the amino group of one and the hydroxyl group of the other.

i

Esterification–condensation reaction

ii

Condensation polymerisation

iii

Condensation polymerisation

Proteins and polysaccharides are biopolymers.

CHAPTER 28: CELLULAR RESPIRATION Review exercise 28.1 1

ATP is adenosine triphosphate. ADP is adenosine diphosphate.

2

When glucose in the cell is oxidised to water and carbon dioxide, this energy is stored in the form of ATP. This ATP can be used to provide energy for various endothermic biochemical reactions.

Review exercise 28.2 1

a

The reactants of glycolysis are glucose, nicotinamide adenine dinucleotide (NAD+), ADP and free phosphate. The products of glycolysis are pyruvate ion, reduced NAD+ (NADH), ATP, water and hydrogen ions.

b

38 molecules of ATP

c

The lactate ion

d

Glycolysis, a non-aerobic process, occurs within the cytoplasm of the cell and is the preliminary stage to aerobic respiration in humans and most other species.

2

pyruvate ion + coenzyme A + nicotinamide adenine dinucleotide → acetyl-coenzyme A + carbon dioxide + reduced NAD+

3

a

Carbon dioxide and coenzyme A

b

For each molecule of acetyl-CoA that is oxidised, one molecule of ATP, three molecules of NADH and one molecule of FADH2 are produced.

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Solutions Manual: Module 7 The biochemistry of movement

4

c

The TCA cycle occurs within the mitochondrial matrix.

a

The oxidation of NADH and FADH2 is coupled with the phosphorylation of ADP to produce ATP. This oxidative phosphorylation is a multi-stage pathway, where a total of 10 redox reactions occur.

b

30 molecules of ATP are generated from the oxidation of one pyruvate ion.

c

O2 + 4e– + 4H+ → 2H2O

5

One molecule of CO2 is produced during the conversion of glucose to pyruvate and two molecules are produced in the TCA cycle. Chapter 28 — Application and investigation 1

2

a

Cellular respiration involves the conversion of glucose and oxygen to carbon dioxide, water and energy. The energy evolved is stored in the form of ATP and enables the cell to perform other biochemical reactions. The reactants are glucose and oxygen. The products are water, carbon dioxide and energy.

b

Combustion of carbon compounds involves the oxidation of the compound with oxygen to produce carbon dioxide, water and energy. In cellular respiration the energy produced is stored as ATP and only a small amount of heat energy is evolved. Simple combustion requires an initial and considerable amount of heat before it occurs and releases its energy as heat. Such heat would destroy living tissue. Cellular respiration occurs as a many-stepped pathway requiring enzymes to proceed. Simple combustion is straightforward and requires no enzymes.

C6H12O6 + 2NAD+ + 2ADP + 2Pi → 2CH3COCOO– + 2NADH + 2ATP + H2O + 4H+

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Solutions Manual: Module 7 The biochemistry of movement

3

Cellular respiration in the absence of oxygen is known as anaerobic respiration. The products of anaerobic respiration are the pyruvate ion and water. In plant and fungi tissue, anaerobic respiration produces ethanol and carbon dioxide. In animal tissue, anaerobic respiration produces lactic acid.

4

pyruvate ion + coenzyme A + nicotinamide adenine dinucleotide → acetyl-coenzyme A + carbon dioxide + reduced NAD+

5

The following equation shows the conversion of acetyl-CoA to its final product, carbon dioxide. Note the conversion of ADP to ATP, the reduction of NAD+ to NADH and the reduction of FAD to FADH2. Restored coenzyme A becomes a product. acetyl-CoA + ADP + 3NAD+ + FAD + Pi → 2CO2 + ATP + 3NADH + FADH2 + CoA

6

–

OOC-CH=CH-COO–

7

Investigation

8

a

NADH → NAD and ADP + Pi → ATP

b

FADH2 → FAD and ADP + Pi → ATP

9

Investigation

10

Investigation

11

The radioactive carbon-14 would emerge in carbon dioxide at steps 3 and 4 of the TCA cycle.

12

i

(a) glycolysis; (b) tricarboxylic acid cycle or Kreb’s cycle; (c) respiratory chain

ii

X is 2pyruvate; Y 2CO2; Z ATP

13

Investigation

14

Investigation

15

This amount of heat energy in one burst would cause excess heat within the cells and destroy living tissue.

CHAPTER 29: MUSCLE ACTION AND EXERCISE Review exercise 29.1 1

Skeletal muscle: voluntary movement of muscles. Cardiac muscle: involuntary muscles of the heart. Smooth muscle: involuntary muscles of the digestive and circulatory systems.

2

Myosin, actin, tropomysin and troponin.

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Solutions Manual: Module 7 The biochemistry of movement

3

Thick filaments are made up of the protein myosin. This molecule is unusual in having at one end a fibrous structure and at the other, a globular end. Several hundred of these myosin molecules make up a filament. Thin filaments are made up from the protein actin, which has a fibrous structure. Two other proteins, tropomysin and troponin, combine with actin to form thin filaments. Tropomysin and troponin together prevent actin and myosin interacting when the muscles are in a relaxed state. Both thin and thick filaments are found in skeletal (striated) muscles.

Review exercise 29.2 1

Exothermic. When ATP is hydrolysed to ADP and phospate ion, energy in the form of heat is given out.

2

During muscle contraction, an interaction between the proteins tropomysin and troponin and calcium ions occurs. A site on the actin becomes available, allowing the myosin head to bond. This process allows the myosin to catalyse the hydrolysis of ATP. The energy released allows the head of the myosin molecule to bend, putting it into a higher energy state. Sufficient energy of the myosin allows it to bind with another myosin molecule further along the filament, which liberates phosphate and ADP.

Review exercise 29.3 1 Type 1 cells

Type 2 cells

Slow twitching (contraction)

Fast twitching (contraction)

Rich supply of blood cells

Reduced supply of blood vessels

Adequate oxygen supply

Limited amount of oxygen available

Many mitochondria

Fewer mitochondria

ATP is derived from the process of oxidative phosphorylation (aerobic respiration)

Use muscle-stored glycogen as initial fuel (anaerobic respiration)

Contain fewer contractile filaments than do type 2 cells Can use glucose, fatty acids and amino acids as fuels 2

Anaerobic glycolysis; the pyruvate produced is reduced to make lactate. This lactate is harmful to the muscle cells and if it accumulates will lead to cramp. It occurs in skeletal muscle cells. (The blood takes the lactate away from the muscles to the liver where it is oxidised back into pyruvate when oxygen is again available.)

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Solutions Manual: Module 7 The biochemistry of movement

Review exercise 29.4 1

Fats, carbohydrates, proteins.

2 Aerobic respiration

Anaerobic respiration

Energy is derived from carbohydrates, fats and proteins.

It is carbohydrate dependent, in particular, glucose.

Initially it involves glycolysis, which leads into the tricarboxylic acid cycle, part of which involves oxidative phosphorylation.

Stored glycogen (muscles and liver) is converted to glucose.

Acetyl CoA provides the initial fuel for the TCA cycle.

The breakdown of glycogen followed by glycolysis ensures a rapid and immediate supply of ATP, which can be supplied in a short period of time but only in short bursts.

3

Athletes who compete in short but rapid events such as the 100 m sprint or the 100 m freestyle benefit from the anaerobic respiration of glycolysis because it provides high levels of ATP at a rapid rate.

4

The metabolism of both glucose and fatty acids requires the presence of the carrier molecule, coenzyme A, but there is a limited amount of this within the cell at any time. If high levels of fatty acids enter the cell, the CoA is consumed, leaving insufficient quantities for glucose metabolism.

5

In long-distance running, most of the energy required is derived from the metabolism of carbohydrates, fats and proteins by aerobic respiration. At rest, approximately one-third of the energy required comes from carbohydrate metabolism; the remaining two-thirds comes from the metabolism of fat. Once into the run, carbohydrates become the primary fuel. The muscles require more oxygen for aerobic respiration as the demand for oxygen rises. During this period, anaerobic respiration supplies the extra ATP needed. With oxygen levels restored, metabolism is fuelled by fats.

Chapter 29 — Application and investigation 1

It is composed of fibre-like cells and has a banded appearance.

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Solutions Manual: Module 7 The biochemistry of movement

2

a

Thick filaments are made up of the protein myosin. This molecule is unusual in having at one end a fibrous structure and at the other, a globular end. Several hundred of these myosin molecules make up a filament. Thin filaments are made up from the protein actin, which has a fibrous structure. Two other proteins, tropomysin and troponin, combine with actin to form thin filaments.

b

Thick filaments appear as dark bands in striated muscle. They alternate with thin filaments, which appear light in colour.

3

Skeletal muscle consists of thin actin filaments which overlap thick myosin filaments. During muscle contraction, these filaments slide over each other but do not individually alter in length. Calcium ions are liberated in the region of the sarcomeres, interacting with the proteins tropomysin and troponin. A site on the actin becomes available, allowing the myosin head to bond. This process allows the myosin to catalyse the hydrolysis of ATP. The energy released allows the head of the myosin molecule to bend, putting it into a higher energy state. Sufficient energy of the myosin allows it to bind with another myosin molecule further along the filament, which liberates phosphate and ADP.

4

a

Slow twitch muscles contract slowly and contain mostly type 1 muscle cells. They are richly supplied with blood vessels, making them suited to aerobic respiration because oxygen is well supplied. They have abundant supplies of mitochondria and ATP is generated from them through the process of oxidative phosphorylation. Slow twitch muscles are able to provide the long-distance runner or the migratory bird with a steady and sufficient supply of energy from fuel reserves of fats, proteins and carbohydrates. Fast twitch muscles contract relatively quickly and contain mostly type 2 muscle cells; they contain fewer mitochondria and are poorly supplied with blood vessels compared with slow twitch muscle. Respiration is mostly by anaerobic means. Glucose available in the blood and glycogen stores provide the primary fuel. Fast twitch cells supply small amounts of high energy, making them suitable for highintensity athletic events such as sprints.

b

5

Cheetahs hunt their food by making short but rapid attacks on their prey. Their skeletal muscles would be mostly made up of the type 2 fast twitch muscles cells. Elephants, being herbivores, do not ambush their prey, but they do need muscles that will sustain their large bulk. Elephants have muscles made up primarily of type 1 slow twitch muscles.

Investigation

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Solutions Manual: Module 7 The biochemistry of movement

6

7

Investigation

8

The ‘fuels’ of food compounds include glucose, amino acids, fatty acids, lactate and ‘ketone bodies’. All of them can be converted into acetyl-CoA by a variety of biochemical pathways. Acetyl-CoA is the starting compound for the TCA cycle. It is the common intermediate product formed by the breakdown of the major primary ‘fuels’. During the TCA cycle, the acetyl group of acetyl-CoA is oxidised to give carbon dioxide and energy, which can be used to generate ATP. ATP generated can thus be used as a raw fuel in other biochemical processes in the body.

9

Anaerobic (without air) respiration or glycolysis is the preliminary stage to aerobic (with air) respiration in many living organisms. Anaerobic respiration does not use oxygen as a reactant. It produces carbon dioxide and alcohol and relatively small amounts of ATP. Yeasts used in fermentation employ this kind of metabolism. Aerobic respiration uses copious amounts of oxygen in the cellular metabolism of food. It produces energy in the form of ATP, carbon dioxide and water as products. Muscles in higher organisms obtain most of their energy (ATP) from aerobic respiration.

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Solutions Manual: Module 7 The biochemistry of movement

10

For much of their time on the competition circuit, basketball players and triathletes rely on sudden bursts of muscular activity. Fast twitch muscles that provide these sudden exertions need to be supplied with a steady supply of glucose. A consequence of this activity is an accumulation of acid in the muscles and a drop in pH. This leads to fatigue and cramp. Such athletes train to utilise as much oxygen as possible and eat carbohydrates (the major source of glucose) prior to such events.

11–13

Investigation

Module 7 REVIEW 1

2

3

4

a

Glucose

b

The pyruvate ion, ATP, water, hydrogen ions and reduced NAD.

c

The pyruvate ion becomes acetyl-CoA, which enters the TCA cycle to be finally oxidised into water and carbon dioxide.

a

X is water and Y is carbon dioxide.

b

There are some 140 intermediate compounds before the waste products are produced.

a

(i) amino acid, (ii) fatty acid, (iii) carbohydrate

b

glycosidic linkage forms:

a

carbohydrates

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Solutions Manual: Module 7 The biochemistry of movement

b

5

a

peptide bond

b

6

7

a

The sudden release of the energy derived from the breakdown of glucose to water and carbon dioxide would generate too much heat at one time; the energy could not all be converted to useful ATP molecules.

b

Glycolysis, tricarboxylic acid cycle, oxidative phosphorylation. TCA cycle and oxidative phosphorylation release most energy.

Food is essential for the healthy functioning of the human body. A balance of carbohydrates, fats, proteins, vitamins, minerals and water is required. As a fuel, food supplies the necessary energy for body functions. Food provides the raw materials from which the body is made and repairs damaged tissue. Some essential nutrients play a role in the biochemical processes that take place within the body. Carbohydrates provide the bulk of energy and should generally make up about 60% of the total calorie intake. Most elite athletes rely on sudden bursts of muscular activity, so they must be supplied with a steady supply of glucose. The major source is carbohydrates. Fats serve as energy sources, particularly long-term reserves. Proteins provide structure for many body parts including bone, hair, skin, tendon and cartilage. © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2006. This page from Chemistry Contexts 2 Teacher’s Resource second edition may be reproduced for classroom use.

Solutions Manual: Module 7 The biochemistry of movement

Others aid in the transport of oxygen to the tissues, fight disease or catalyse biochemical reactions. 8

The primary structure of proteins refers to the sequence of amino acids contained within the molecule which are linked by covalent peptide bonds. The secondary structure refers to the spatial arrangement of the polypeptide chain (made up from amino acids). It is held in place by H-bonding between the CO and nearby NH groups. The tertiary structure refers to the overall way the protein folds and it is held in place by bonds and forces from the side chains of the amino acids. These include electrostatic forces caused by ionic interactions between amino acids, hydrogen bonds, hydrophobic forces derived from the side chains of amino acids and some cross-linking of parts of proteins caused by disulphide bonds. Any quaternary structure is determined by the same type of bonds and forces that determine tertiary structure.

9

a

Enzymes function as catalysts for many biochemical reactions. Enzymes are specific to particular biochemical reactions speeding up reaction rates. They are more specific than most chemical reactions, and act upon specific substrates.

b

Enzymes are proteins that catalyse biochemical reactions.

c

The enzyme is A. An enzyme remains unchanged at the end of the reaction.

10

The two functional groups are amine (–NH2) and carboxylic acid (–COOH).

11

a

Glucose

b

Glycogen (from storage) and glucose (in blood).

c

Glycolysis is the series of biochemical reactions in which glucose is broken down to pyruvate with the release of usable energy in the form of ATP. The net energy yield is two ATP molecules per glucose molecule.

d

A side-effect is the production of lactate and hydrogen ions. The hydrogen ion surplus lowers the pH in the muscle, causing cramps and fatigue.

a

Glycerol and alkanoic or fatty acids.

12

b

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Solutions Manual: Module 7 The biochemistry of movement

13

c

Ester; the ester linkage occurs between the glycerol and each of the fatty acids.

d

Fuel

a

Glycogen

b

There are two types of muscle cells: Type one cells contain slow twitch fibres and have a rich blood supply and many mitochondria. Type two cells contain fast twitch fibres, have fewer mitochondria and a reduced supply of blood.

c

14

15

Skeletal muscle is striated in appearance. It is composed of fibre-like cells containing many nuclei. The banded appearance is due to the presence of alternating bands of thick and thin filaments. Thin filaments appear lighter in colour than the thick filaments.

a Monosaccharides

Disaccharides

Polysaccharides

glucose

sucrose

starch

fructose

lactose

glycogen

b

Glucose

c

At rest, about one-third of the body’s energy comes from the metabolism of carbohydrates, with about two-thirds being supplied by the metabolism of fats. As exercise gets under way, the total energy consumption increases and the fuel balance changes, with the metabolism of carbohydrates becoming dominant. Athletes should consume more carbohydrates in the 24 hours before competing.

a

The enzyme fumarase acts on the compound fumarate. Also citrate synthase (acetyl-CoA), aconitase (citrate), isocitrate dehydrogenase (isocitrate), malate dehydrogenase (malate), succinate dehydrogenase (succinate), succinyl-CoA synthetase (succinyl-CoA), α-ketoglutarate dehydrogenase (α-ketoglutarate).

b

The coenzymes NADH/NAD+ and FADH2/FAD. Reversible reactions between the two forms facilitate the flow of electrons through the cycle.

c

ATP is adenosine triphosphate. It is an ‘energy-rich’ compound used to drive many biochemical reactions. In the TCA cycle, ATP is generated as a product and stores potential chemical energy.

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Solutions Manual: Module 8 The chemistry of art

CHAPTER 30: PIGMENTS IN PAINTING Review exercise 30.1 1

Early artists relied upon the pigments available to them from the Earth. The only technique available for producing them at the time was grinding of local materials. These materials were red, brown and yellow ochres (containing iron), white clays, chalk and gypsum and black charcoal, graphite and manganese dioxide.

2

All the ochres contain the metal iron, either as anhydrous or hydrated iron(III) oxide or as limonite, FeO(OH).

3

All paints are made up of a pigment (the colouring matter) and a medium (the liquid that carries the pigment).

4

The pigment needs to be insoluble so that as the colourless medium dries, the colour remains.

Review exercise 30.2 1

Egyptian ores: Ore used as pigment

Main metal in ore

Colour

Malachite

Copper

Green

Azurite

Copper

Blue

Cinnabar

Mercury

Red

Orpiment

Arsenic

Yellow

Realgar

Arsenic

Orange-scarlet

Egyptian blue

Copper

Blue

Stibnite

Antimony

Black

2

A pigment is a coloured insoluble solid suspended in a medium; the pigment remains after the medium dries, e.g. malachite, orpiment, realgar. A dye is a soluble colouring agent that colours the medium, usually water, e.g. Madder lake, indigo.

3

‘Lakes’ are produced from dyes. The soluble colouring agent in the dye is fixed onto an insoluble white powder, such as chalk or gypsum, which can then be extracted and dried.

4

CuCO3.Cu(OH)2(g) → 2CuO(s) + CO2(g) + H2O(g)

Review exercise 30.3 1

Some cosmetics, such as cinnabar, white lead, red lead, stibnite and manganese dioxide as kohl, contain heavy metals that act as cumulative poisons in the body. Romans would © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2006. This page from Chemistry Contexts 2 Teacher’s Resource second edition may be reproduced for classroom use.

Solutions Manual: Module 8 The chemistry of art

not have been aware of the connection between the long-term use of these cosmetics and any illnesses they may have developed. Orpiment, containing arsenic, may have produced more immediate effects but the Romans may have seen no obvious connection. 2

mercury

+

sulfur

→

mercury(II) sulfide

Hg(s)

+

S(s)

→

HgS(s)

(vermilion)

Review exercise 30.4 1

When hydrated iron oxides such as siennas and umbers are heated, the water of crystallisation is driven off, so the oxides become anhydrous. This changes the colour of the compounds.

2

Current price of gold (September 2005) is AUS$577 an ounce. One ducat weighs 0.12 of an ounce, so eight ducats will be worth 8 × 0.12 × $577 = $553.92 .

Review exercise 30.5 1

Chromite, FeO.Cr2O3, was discovered in commercial quantities in America in 1820. Chromite is the source of the pigments chrome yellow, chrome red and chrome green. These pigments give colours of great depth and covering ability.

2

Chrome yellow is manufactured by the precipitation of lead(II) chromate from a solution of a lead salt, such as lead(II) nitrate, and chromate compound, such as sodium chromate.

3

4

lead(II) nitrate

+ sodium chromate

→ lead(II) chromate

+ sodium nitrate

Pb(NO3)2(aq)

+ Na2CrO4(aq)

→ PbCrO4(s)

+ 2NaNO3(aq)

cadmium(II) chloride

+

hydrogen sulphide

→ cadmium sulphide

+ hydrochloric acid

CdCl2(aq)

+

H2S(aq)

→ CdS(s)

+ 2HCl(aq)

titanium(IV) chloride +

oxygen

→

titanium dioxide

+

chlorine

TiCl4(l)

O2(g)

→

TiO2(s)

+

2Cl2(g)

+

Review exercise 30.6 1

Most paintings consist of the following layers: a support (such as wood panel or canvas), a ground (such as gesso), paint and then varnish.

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Solutions Manual: Module 8 The chemistry of art

2

Gesso was a common ground for paintings. It was usually a mixture of chalk or gypsum with animal glue. It was applied as a warm, thick liquid and allowed to set. It dried to a creamy colour and was then scraped and rubbed smooth.

3

Varnishing a painting protects the paint layers, enhances clarity and depth of saturation. Some varnishes were chosen for the golden glow they gave to paintings.

4

Colour can be produced in glass by: • grinding metal oxide pigments to a fine powder and adding them to the glass mixture before melting • ‘flashing’, which involves the coating of clear (‘white’) glass with a thin layer of coloured glass at the molten stage • ‘staining’, which involves painting silver nitrate onto the glass and then firing it in the kiln • mixing pigments (from powdered copper or iron oxides or ground enamel pigments) with a medium such as wine or urine, then painting onto the glass and firing it again.

Review exercise 30.7 1

Tempera is used as a paint medium made from egg yolk mixed with water. It was mixed with pigments and used immediately.

2 Pigment

Metal component

Red ochre

Iron

Minium (red lead)

Lead

Ultramarine

Sodium, aluminium

Crimson lake (cochineal)

(none)

White lead

Lead

Lead tin yellow

Lead, tin

Vermilion

Mercury

Green earth

Iron

Gilding

Gold

Review exercise 30.8 1

a

The common elements of pigments are metals, especially transition metals.

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Solutions Manual: Module 8 The chemistry of art

b

Most metals are found in the d-block; and the heavier p-block metals

Chapter 30 — Application and investigation 1–6 Investigation 7

Early Aboriginal people, with their limited technology, derived their pigments directly from their immediate environment or from occasional trade with other groups. Therefore their pigment palette was limited to easily extracted earth products and some plant and animal products.

8

Early Aboriginal people attached pigments, such as red ochre, to surfaces such as rock faces, using media such as saliva, water and birds’ eggs to carry the pigment and adhere to the surface. Early Egyptians used saliva to carry and adhere kohl to their eyes.

CHAPTER 31: THE STRUCTURE OF THE ATOM AND COLOURS Review exercise 31.1 1

Many metallic elements emit light of a distinctive colour when their compounds are heated in a flame. If it is known that the sample of compound tested is pure and a particular colour is given out in a flame test, then a comparison can be made with known flame colours and a possible identification of the metal made.

2

If a flame test is made of both potassium and sodium chlorides, the sodium salt will give a yellow flame and the potassium salt will give a violet flame. (The chloride does not give a colour.)

Review exercise 31.2 1

Electromagnetic radiation includes gamma rays, X-rays, ultraviolet light, visible light, infra-red radiation, microwaves, TV and radio waves. As the wavelength gets shorter (from radio waves to gamma rays), the electromagnetic radiation becomes more energetic and penetrating.

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2

A line emission spectrum is the spectrum of visible light that results from energetically excited atoms; it is seen as a series of discrete lines of colour separated by blank areas. A line absorption spectrum results when light is passed through a substance in the vapour phase and the resulting spectrum observed; it is seen as a pattern of dark lines across a continuous spectrum. For a given element, its line emission spectrum is the reverse of its line absorption spectrum.

3

a b

i

Sample A is pure and is cadmium (Cd).

ii

Sample B is impure and contains sodium (Na) and lithium (Li).

Sample A contains seven spectral lines and they correspond exactly to the seven spectral lines of cadmium below 600 nm; they do not match any other elements. Sample B contains the three lithium lines that can be found between 300 and 600 nm and the two visible sodium lines; these lines do not exactly match any other elements.

Review exercise 31.3 1

Niels Bohr (1885–1962) was studying the line spectrum of hydrogen and proposed that in a hydrogen atom the electron moves around the nucleus without radiating energy. Only orbits of certain energy levels are allowed, and electrons can ‘jump’ from one orbital to another by absorbing or emitting energy of a particular quantised amount. The allowed orbits are designated a principal quantum number, n, with values of n = 1, n = 2, etc.

2

An atom is in its ground state when its electrons are in their lowest possible energy levels. An atom is in an excited state when any of its electrons are in higher energy levels than their ground state. An atom can become excited when its electrons absorb photons of energy and can move to a higher energy level.

3

Bohr postulated that every line in the emission spectrum of hydrogen represents a transition of electrons from one energy level to a lower one. When an electron jumps from one energy level to a lower energy level, a photon of radiation is emitted. For some of these emissions, the energy released is in the visible spectrum range and so can be seen as the line spectrum.

Review exercise 31.4 1

Bohr’s model of the atom successfully explained the line spectrum of hydrogen and introduced the idea of quantised energy states for electrons in atoms. This feature is incorporated into our current model of the atom.

2

Bohr’s model was not as successful at explaining the atomic spectra of atoms other than hydrogen. Nor could it explain the different intensities of spectral lines, the fact that some spectral lines actually consist of a number of very fine close lines, or the splitting of such spectral lines in magnetic fields.

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3

The double yellow lines that appear in the sodium line spectrum at 589 and 590 nm are the result of emission from excited electrons as they fall back from two different p subshell orbitals to the s sub-shell in the n = 3 energy level.

Review exercise 31.5 1

Both absorption and emission spectra are used to identify different components of pigments, media and varnishes by comparing the spectrum produced to known spectra. This allows conservators to decide on the products and techniques they will use to restore or conserve different paintings.

2

A reflectance spectrum is produced by shining white light on the surface of a painting and analysing the reflected light with a spectrophotometer. The reflectance spectrum obtained can then be compared to known spectra gathered using the same technique for identification of the pigments used in the painting.

3

Laser is an acronym for light amplification by stimulated emission of radiation. Laser microspectral analysis can be used to identify pigments by either destructively vaporising a small sample and subjecting the vapour to analysis or by non-destructively reflecting the laser light from the surface of the pigment to be analysed. Either way, the resulting spectrum is compared to known spectra gathered using the same technique and a match and identification made.

4

5

Radiation

Zinc oxide pigment

Copper-containing pigments

Infra-red

Change from white to lemonyellow, change back again when IR light removed

Permanent chemical change to black copper(II) oxide

Ultraviolet

Fluoresces a pale yellow

Malachite fluoresces a ‘dirty’ mauve

Laser microspectral analysis is a useful analytical tool for pigment identification in the art world. As the equipment becomes more portable and less expensive, it can be used by more conservators to aid in the conservation and restoration of paintings, and by those needing to establish the autheniticity of some paintings.

Chapter 32 — Application and investigation 1

Investigation

2

In a flame test, potassium ions give a violet colour and strontium gives a red colour. You could demonstrate that it is the metal ion that produces the colour by using potassium chloride and strontium chloride for testing, or both potassium nitrate and strontium nitrate. In each case, the same flame colours of violet and red will result.

3

Laser light with a wavelength of 780 nm will appear in the red light region of the electromagnetic spectrum. © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2006. This page from Chemistry Contexts 2 Teacher’s Resource second edition may be reproduced for classroom use.

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4

Different elements can produce different colours when, in the vapour state, they are subjected to electric discharge. Neon gas, with its distinctive red colour, was one of the first elements used in this way for lighting. Since then other elements have been used to produce lights of different colours.

5

The complex will appear green. The complex absorbs most light from the blue (400 nm) and red (650–750 nm) regions of the visible light spectrum and so will transmit mostly in the green region.

6–8 Investigation 9

He+ has two protons in its nucleus, unlike hydrogen, which has only one proton. This increased nuclear attraction changes the energy levels of the helium, so the He+ ion gives spectral lines with higher energy/frequency.

10

760 ppm

CHAPTER 32: ELECTRONS IN THE ATOM Review exercise 32.1 1

Bohr’s model of the hydrogen atom postulated allowed energy levels for electrons. The quantum mechanical model suggests that the electrons should be viewed as waves and that the orbits do not represent the paths of the electrons but rather the amplitudes of the electron waves at different points in space. It is possible to determine only the probability of locating an electron at a certain point, and thus we speak of the electrons existing in an electron cloud.

2

An orbit is a region in space with a fixed radius in which Bohr postulated that an electron could be found. An orbital is a region of space around a nucleus through which an electron with a given energy may move.

3

a

Shell n = 2: two sub-shells, an s and a p; one orbital in the s sub-shell and three orbitals in the p sub-shell; maximum of eight electrons can be accommodated in this shell.

b

Shell n = 4: four sub-shells, an s, a p, a d and an f; one orbital in the s sub-shell, three orbitals in the p orbital, five orbitals in the d sub-shell and seven orbitals in the f sub-shell; a maximum of 32 electrons can be accommodated in this shell.

a

16

b

3

c

0

d

1

e

5

f

7

4

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Review exercise 32.2 1

2

3

4

a

helium

b

beryllium

c

nitrogen

d

sodium

e

sulfur

a

1s2 2s1

b

1s2 2s2 2p6 3s2 3p4

c

1s2 2s2 2p4

d

1s2 2s2 2p6 3s2 3p5

e

1s2 2s2 2p1

f

1s2 2s2 2p6 3s2 3p6 4s2

a

Ground state atoms: i, iii and iv. Excited atoms: ii and v.

b

i oxygen, 8; ii neon, 10; iii aluminium, 13; iv calcium, 20; v zinc, 30

a b c d e

Review exercise 32.3 1

a

s2

b

p4

2

fluorine: 1s2 2s2 2p5

3

a

i

Group 2

ii

3

iii

12

iv

Mg

b

i

Group 17

ii

3

iii

17

iv

Cl

c

i

Group 15

ii

4

iii

33

iv

As

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Review exercise 32.4 1

The negative electrons are held in place by their electrostatic force of attraction for the positively charged nucleus. It takes energy to overcome this force.

2

a b c

B(g) F(g) K(g)

→

B+(g) +

→

F (g)

→

+

K (g)

+

e–

–∆H

+

e

–

–∆H

e

–

–∆H

+

Review exercise 32.5 1

The first ionisation energy for the successive elements boron, carbon and nitrogen increases. As you move across the period, the outer electrons (which are in the same shell) are held by progressively increasing nuclear charge and so it takes progressively more energy to remove them from this electrostatic hold.

2

The first ionisation energy for the elements Mg, Ca and Sr decreases. As you move down the group to which they all belong, the outer electrons are held in progressively further out energy levels. They are increasingly further away from the electrostatic attraction of the nucleus and are increasingly shielded from this same pulling attraction, so it takes increasingly less energy to remove an outer shell electron.

3

Carbon will have a higher first ionisation energy than boron, as its outer electrons are held in place by a higher nuclear charge. Aluminium will have a lower first ionisation energy than boron and carbon, as its outer electrons are further away from the pull of the nucleus and shielded by an extra shell. Carbon will have the largest first ionisation energy (B, I1 807 kJ mol–1; C, 1093 kJ mol–1; Al, 584 kJ mol–1).

Review exercise 32.6 1

2

a

B(g)

→

B+(g)

+

e–

B+(g)

→

B2+(g)

+

e–

B2+(g)

→

B3+(g)

+

e–

B3+(g)

→

B4+(g)

+

e–

b

Removing the first three electrons from boron removes the three that occupy the outer n = 2 shell. The fourth electron to be removed occupies the n = 1 energy level, which is closer to the positive nucleus and so this requires much more energy.

c

Three valence electrons.

The first three electrons removed from aluminium occupy the n = 3 energy level. The fourth and fifth electrons occupy the n = 2 energy level, which is closer to the nucleus and so require much more energy to remove than those in the further out n = 3 shell. © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2006. This page from Chemistry Contexts 2 Teacher’s Resource second edition may be reproduced for classroom use.

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Review exercise 32.7 1

Electronegativity is a measure of the ability of an atom in a molecule to attract electrons to itself in a chemical bond. Atoms with strong ‘electron pulling power’ in covalent bonds are said to be highly electronegative. Electronegativity can be used to determine the polarity of a covalent bond: the greater the difference in electronegativity, the more polar the bond.

2

The most electronegative element, fluorine, has been arbitrarily assigned a value of 4.0 and all other values are relative to this. The least electronegative element is caesium, with a value of 0.7. The values for all other elements lie between these two extremes.

3

As you move from left to right across a period, electronegativity steadily increases. This reflects the increasing nuclear charge across the period.

Chapter 32 — Application and investigation 1

2

a

2

b

6

c

2

d

10

e

14

a

Period 3, Group 1, sodium

b

Period 2, Group 16, oxygen

c

Period 3, Group 14, silicon

d

Period 3, Group 18, argon

3

Ionisation energy is influenced by position across a period (group number—increasing number of protons), position down a group (period number—distance from nucleus and number of shells shielding outer electrons).

4

a

Ne: 1s2 2s2 2p6; F: 1s2 2s2 2p5; C: 1s2 2s2 2p2; O: 1s2 2s2 2p4; Cl: 1s2 2s2 2p6 3s2 3p5; He: 1s2

b

Ne: p block; F: p block; C: p block; O: p block; Cl: p block; He: s block

c

Ne: period 2, group 18; F: period 2, group 17; O: period 2, group16; Cl: period 3, group 17; He: period 1, group 18; C: period 2, group 14

d

Number of valence electrons: Ne: 0; F: 7; C: 4, O: 6; Cl: 7; He: 0

e

Increasing first ionisation energies: C< Cl < O < F < Ne < He.

5

The first ionisation energy of sodium is less than that of magnesium because the effective nuclear charge of sodium is 11+ and that of magnesium is 12+, yet their valence electrons are in the same energy level (at the same distance). Thus sodium’s © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2006. This page from Chemistry Contexts 2 Teacher’s Resource second edition may be reproduced for classroom use.

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electrons are not as strongly held as magnesium’s. The second ionisation energy of sodium is higher than that of magnesium because the second electron to be removed from sodium must come from a closer energy level to the nucleus than that of magnesium’s. 6

a

1s2 2s2 2p6 3s2 3p3

b

1s2 2s2 2p6 3s2 3p6

7

Investigation

8

Electronegativity values are determined by giving a high value to the known most electronegative element, fluorine, and then comparing the other elements to this. Investigation for the other scales possible.

9

Investigation, however detail provided in Chapter 33.

CHAPTER 33: THE TRANSITION METALS Review exercise 33.1 1

The transition metals have electron configurations that include the gradual filling of the d-block orbitals. The other elements in the main groups of the periodic table have their d-block orbitals either completely empty or completely filled.

2

a

V: 1s2 2s2 2p6 3s2 3p6 3d3 4s2

b

Ni: 1s2 2s2 2p6 3s2 3p6 3d8 4s2

c

Cr: 1s2 2s2 2p6 3s2 3p6 3d5 4s1

d

Pd: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d8 5s2

3

The d-block elements have similar properties because their electron configurations (which give rise to their properties) have their outer shells as s1 or s2. The p-block elements have greater differences in their outer shells, ranging from p1 to p6, and hence a greater difference in their properties.

Review exercise 33.2 1

Chromium, [Ar]4s1 3d5, exhibits several oxidation states in its compounds as a result of being able to lose or bond its 3d electrons as well as its 4s electron. Aluminium, [Ne]3s2 3p1, has only the three valence electrons, which are all lost or shared during bonding, resulting in only the +3 oxidation state.

2

a

The transition metals commonly have s2 electrons as their outer electrons, making +2 a common oxidation state for them.

b

Transition metals can also lose or combine their many d orbital electrons, which can give rise to many oxidation states.

3

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Cr(s) + 2H+(aq) → Cr2+(aq) + H2(g) 4

Vanadium, [Ar]3d3 4s2 and 23 protons, forms VCl2 by releasing its two 4s2 electrons. Scandium, [Ar]3d1 4s2 and 21 protons, does not readily form ScCl2; its 3d and 4s electrons are not as strongly held by the attraction from the nucleus, so it more readily releases these three electrons rather than just two of them.

Review exercise 33.3 1

2

a

Ni2+: [Ar]3d8; Cu2+: [Ar]3d9; Zn2+: [Ar]3d10

b

Both Ni2+ and Cu2+ will form coloured compounds as they both have unfilled d-orbitals which can be occupied by excited electrons as they absorb light. Zn2+ is colourless because the d-orbitals are completely filled and no electrons can be promoted, so no light is absorbed.

c

Because Cu2+ ions are typically blue they must be absorbing light of the complementary colours, that is, in the red-orange range.

Sc3+: [Ar] and Ti4+: [Ar] do not have coloured compounds because they have no electrons occupying d-orbitals that can be promoted to higher energy level orbitals. However, Ti3+: [Ar]3d1 and V3+: [Ar]3d2 both have partially filled d-orbitals with electrons that can be promoted into higher energy d-orbitals when they absorb photons of light.

Review exercise 33.4 1

a

Cr3+ acts as a Lewis acid (electron pair acceptor) and the water molecules act as Lewis bases.

b

Monodentate

2

The ethylenediamine (en) has two sites with lone pairs of electrons (see Figure 33.9 in text), which it is able to donate and hence act as a Lewis base. Three en molecules donate both their lone pairs to Ni2+, which thus acts as a Lewis acid and forms [Ni(en)3]2+.

3

a

[Ir(NH3)3Cl3]

b

[Cr(H2O)2(C2O4)2]–

c

[Co(CN)6]3–

Review exercise 33.5 1

The colour depends on the type of ligand present; the water molecule produces the violet colour and the cyanide ion produces the yellow colour.

2

Cu2+: [Ar]3d9 and Zn2+: [Ar]3d10; the copper ion will produce coloured complexes because it has an unfilled d-orbital which can be occupied by electrons as they absorb

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light. However, the zinc ion will produce colourless complexes, as there are no unfilled d-orbitals. 3

a

b

c

4

i

The metal ion is Ni2+ and its oxidation state is +2.

ii

Each water molecule is bonded to the metal ion through one of the lone pairs of electrons on the oxygen atom.

i

The metal ion is Ni2+ and its oxidation state is +2.

ii

Each ammonia molecule is bonded to the metal ion through the lone pair of electrons on the nitrogen atom.

i

The metal ion is Ni2+ and its oxidation state is +2.

ii

Each en molecule is bonded to the metal ion through the lone pair of electrons on each on the two nitrogen atoms.

Because it is impossible to actually see the bonding arrangement with complexes containing chelated ligands, such as iron(III)hexahydrate, we use models such as diagrams, modelling kits and computer images to show what we think the arrangement is like. We can illustrate the way we believe the six water molecules are arranged around the iron(III) ion and can use models to represent the coordinate covalent bonding in the Lewis acid–base interaction. However, we cannot show how this happens, the positions and movements of electrons, the energy changes involved or the rate at which this happens. Overall, modelling is the most useful method we have at the moment to depict our understanding.

Chapter 33 — Application and investigation 1

Investigation

2

a

MnO4–(aq) (purple) + 8H+(aq) + 5e– ↔ Mn2+(aq) (colourless) + 4H2O(l)

b

2CrO42–(aq) (yellow) + 2H+(aq) ↔ Cr2O72–(aq) (orange) + H2O(l)

c

Fe3+(aq) (brown) + (SCN)–(aq) ↔ [Fe(SCN)]2+(aq) (deep red)

d

3Cu(s) (colourless)+ 2NO3–(aq) + 8H+(aq) ↔ 3Cu2+(aq) (blue) + 2NO(g) + 4H2O(l)

e

Cr2O72–(aq) (orange) + 2OH–(aq) ↔ 2CrO42–(aq) (yellow) + H2O(l)

3

+2

4–6 Investigation

Module 8 REVIEW 1

a

Red ochre, Fe2O3

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2

3

b

The colour of red ochre comes from the presence of the Fe3+ ion. Iron, [Ar]3d6 4s2, is a transition metal with unfilled d-orbitals. The electrons in the 3d subshell are able to absorb quanta of energy and move to the available unfilled orbitals. The energy absorbed is from the green range of the visible spectrum and so the compound appears as the reflected complementary colour red. The Australian Aboriginal people commonly applied their pigments to rock surfaces by mixing them with saliva, fats, waxes, water or blood, and allowing these media to dry.

a

Colour can be added to the bulk of a material such as glass by finely powdering metal oxide pigments and adding them to the glass mixture before melting the mixture. Other methods, such as staining and flashing, involve coating the previously prepared glass and then refiring it.

b

Cobalt oxide, manganese oxide, iron oxide, copper(II) oxide and silver nitrate can be used to add colour to glass.

In the mediaeval painting of about 1365, St John the Baptist with St John the Evangelist (?) and St James, by Nardo di Cione, the picture is on wood and the ground is gesso. This gesso is a mixture of gypsum, CaSO4.2H2O, and animal glue. The figures were drawn onto the ground and then painted using an egg tempera medium. The paints involved in this painting include the following: Pigment

Chemical composition

Method of preparation

Red ochre

Fe2O3

Grinding of the natural ore.

Minium (red lead)

PbO2.2PbO

Oxidation of molten lead.

Ultramarine

3Na2O.3Al2O3.6SiO2.2Na2S

Grinding, kneading in lye solution with a special dough. Insoluble colour in suspension is then allowed to settle, is washed and dried.

Crimson lake (cochineal)

C11H12O17

Colour from solution is fixed onto an insoluble white powder such as gypsum, which is then extracted and dried.

White lead

2PbCO3.Pb(OH)2

Stack process—lead in clay pots with vinegar and manure and left to ferment.

Lead tin yellow

PbSnO4

Roasting of lead and tin mixture.

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Vermilion

HgS

Heating mercury and sulfur together and HgS condenses from the vapours.

Green earth

Various Fe, Mg, Al and K compounds

Grinding of the natural ores.

Gilding

Au

Found in native condition.

This painting was not varnished; the egg tempera of the paint layer provided a selfprotective finish. 4

a

An emission spectrum is produced when the visible light emitted from excited atoms of an element is passed through a narrow slit and then a prism. An absorption spectrum is the result of light first passing through a substance in the vapour phase and then producing a pattern of dark lines across a continuous spectrum. A reflectance spectrum results when white light is shone onto a surface and the reflected light is analysed.

b

A reflectance spectrum can be obtained from a painting by shining white light on to the surface of the chosen pigment and analysing the resulting light reflected from the surface of the pigment. The reflectance spectrum can then be compared to those of known compounds and an identification made of the pigment. A 5% tint of cobalt blue shows over 80% reflectance of light with a wavelength of 480– 490 nm (blue at the green end) and 90% reflectance of light with a wavelength of 700 nm (deep red). A 5% tint of Prussian blue has over 60% reflectance of light in the 440–490 nm range (violet blue). Laser microspectral analysis can use laser reflectance to identify pigments very accurately.

5

The Bohr model of the atom described the orbit of electrons around the atom as being at fixed radii and energy; the electrons could absorb energy in fixed quanta and move to a higher energy level and then emit these same quanta of energy when they fell back to a lower energy level. This model was very useful in explaining the line spectrum of hydrogen, and it supplied the notion of quanta of energy able to be absorbed and emitted, but it was not as useful for explaining the atomic spectra of atoms other than hydrogen. Nor could it explain the different intensities of spectral lines, the fact that some spectral lines actually consist of a number of very fine close lines, or the splitting of such spectral lines in magnetic fields.

6

A piece of platinum wire was thoroughly cleaned by immersion in hydrochloric acid and heating in a non-luminous Bunsen burner flame. Then the wire was dipped in a small sample of sodium chloride and placed in the non-luminous Bunsen flame and the colour of the flame observed. Sodium ions produced a strong yellow flame. This colour is produced because the electrons of the sodium absorb quanta of energy and move to © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2006. This page from Chemistry Contexts 2 Teacher’s Resource second edition may be reproduced for classroom use.

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higher energy levels. This absorbed energy is emitted as the excited electrons jump back to their ground states. The emitted energy is the yellow range of visible light. 7

a b

The 27 electrons of the cobalt atom fill in the shell order 1s, 2s, 2p, 3s, 3p, 4s, then 3d. In filling the orbitals, two restrictions apply: the Pauli exclusion principle states that an orbital can only hold a maximum of two electrons (and that these must have opposite spins); and Hund’s rule states that if two or more orbitals are available (as in the 3d orbitals for cobalt), then one electron goes into each orbital until all the orbitals are half-full, and then a second electron can go into the orbitals.

8

The transition metals have valence electrons in both the 3d and 4s sub-shells. Because there is very little difference in energy between these sub-shells and their orbitals, the transition metals can lose these electrons with almost equal ease, resulting in a number of oxidation states for most of these metals. The unfilled orbitals in the sub-shells of some of the transition metals also have good electron acceptor properties in forming coordination compounds and so make good oxidising agents because they are themselves so easily reduced. Metals from the s and p blocks do not have the same d-orbital configurations and so have little or no variation in their oxidation states.

9

Colour in aqueous ions of transition metals results from the ability of these metals to absorb quanta of energy and promote electrons to unfilled d-orbitals. The absorbed energy is in the range of visible light and the compounds appear the complementary colour of the energy absorbed: the [Cr(H2O)6]3+ ions appear violet because they have absorbed energy in the yellow-green range and reflect or transmit in the violet range. Aqueous ions of metals other than transition metals do not have unfilled d-orbitals and so have no opportunity to produce colours.

10

a

The ligand EDTA forms a coordinate covalent bond with cadmium ions in complex formation.

b

The ligand EDTA has six sets of lone pair electrons through which it bonds to the Cd2+ ion. As the EDTA is donating the pairs of electrons, it is acting as a Lewis base and the metal ion is acting as a Lewis acid. Hence, the sequestering of cadmium into an EDTA complex is a Lewis acid–Lewis base interaction.

11

A strong oxidising agent is itself readily reduced. Potassium permanganate solution, KMnO4(aq), is a deep purple colour as a consequence of the +7 oxidation state of manganese in the MnO4–(aq) ion. If this ion is reduced, its colour should quickly change. Steps: 1

Make up a 0.1 M solution of acidified (with sulfuric acid) potassium permanganate, KMnO4(aq).

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2

Add it dropwise to a solution of violet vanadium(II) ions formed from oxidising dioxovanadium(V) ions using granulated or powdered zinc metal added to acidified (with about 9 M sulfuric acid) 0.1M ammonium metavanadate solution.

3

If you don’t agitate the mixture too much you will see the colours as the vanadium(II) is oxidised progressively to the vanadium(III) ion, to the oxovanadium(IV) ion, to the dioxovanadium(V) ion. The colour goes from violet to green to blue to yellow and the purple permanganate decolourises.

4

As the vanadate ion (converted to dioxovanadium(V) ion by the sulfuric acid) is also a strong oxidising agent, permanganate must be a very strong oxidising agent.

The net ionic equation for the permanganate reduction is: MnO4–(aq) (purple) + 8H+(aq) + 5e– → Mn2+(aq) + 4H2O(l) (colourless) The oxidation state of the Mn2+ ion is +2. The ionic equations for the vanadium(II) oxidation are: V2+(aq) → V3+(aq) + e− V3+(aq) + H2O(l) → VO2+(aq) + 2H+(aq) + e– VO2+(aq) + H2O(l) → VO2+(aq) + 2H+(aq) + e– Because acidification of the metavanadate is with concentrated acid, safety goggles, gloves and protective clothing must be worn. Also, the metavanadate is toxic and a possible mutagen; gloves, goggles and protective clothing must be worn. 12

A ligand is a species (acting as a Lewis base) with a lone pair of electrons able to be donated to a metal cation (acting as a Lewis acid) with orbitals available to receive them. Thus, a coordinate covalent bond forms in this Lewis acid–base interaction as both electrons in the bond originate from the one atom. Chelates are ligands able to form at least two coordinate bonds to the one cation. For example, ethylene diamine is a ligand that has two nitrogen atoms that can each bond to a cation such as nickel(II).

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