anachem

Analytical Chemistry

Scope Solution Compositions Acids and Bases Gravimetric Analysis Titrimetric Methods Redox Reactions

Solutions

Solution Compositions

Units of Concentration

*Formality – identical to molarity that is used for solutions of ionic salts that do not exist as molecule in solid or in solution.

Units of Concentration

Units of Concentration Normality = f x Molarity f: for acids: no. of replaceable H+ for bases: no. of replaceable OHfor salts: net charge of an ion for redox reactants: no. of electrons lost or gain

Units of Concentration Mole fraction

Mole percent

Units of Concentration

Units of Concentration

1 ppm = 1 mg/L 1 ppb = 1 µg/L 1 ppt = 1 ng/L

Example 1 What is the molarity of water at 40C?

Ans. 55.5 M

Example 2 The concentration of glucose in normal spinal fluid is 75 mg/100 g. What is the molal concentration?

Ans. 4.2 x 10-3 M

Example 3 A 34.00%-by-mass solution of H3PO4 in water has a density of 1.209 g/cm3 at 200C. What is the molarity and molality of this solution?

Ans: 4.19 M, 5.26 m

Example 4 The hardness of water (hardness count) is usually expressed as parts per million (by mass of CaCO3), which is equivalent to milligrams of CaCO3 per liter of water. What is the molar concentration of Ca2+ ions in a water sample with hardness count of 175? Ans. 1.75 x 10-3 M.

Acids and Bases  Arrhenius Theory  Bronsted-Lowry Theory  Lewis Theory

Acids and Bases Arrhenius Theory of Acid and Bases:

“An acid produces H+ and a base produces OHin aqueous solutions.”

Acids and Bases Brønsted-Lowry Theory of Acid and Bases:

“An acid is a proton donor, and a base is a proton acceptor.”

Acids and Bases Brønsted-Lowry Theory of Acid and Bases: *For every acid, there is a conjugate base and for every base, there is a conjugate acid.

Acids and Bases Brønsted-Lowry Theory of Acid and Bases: Role of Solvent: - Makes to act the substance to behave as an acid or base.

Amphoteric – refer to a substance that can act as acid or base.

Acids and Bases Brønsted-Lowry Theory of Acid and Bases: Classification of Solvents:

1. 2. 3. 4.

Solvent Protophilic (proton seeking) Protegenic (proton generating) Amphiprotic (both 1 & 2) *Aprotic

Example H2O, NH3 H2O,HOAc H2O, EtOH C6H6, CCl4

*independent of proton seeking and generating.

Acids and Bases Lewis Theory of Acids and Bases “A Lewis acid is a species (an atom, ion or molecule) that is an electron-pair acceptor.

A Lewis base is a species that is an electronpair donor.

Acids and Bases Lewis Theory of Acids and Bases

Ligands – molecules or ions that behave as Lewis bases.

Acids and Bases Lewis Theory of Acids and Bases

Strengths of Acids and Bases STRONG

WEAK

Acids

Bases

Acids

Bases

HCl

LiOH

HF

NH3

HI

NaOH

HCN

Organic bases

HBr

KOH

H3PO4

Amines

HClO4

RbOH

H3BO3

HClO3

CsOH

H2CO3

HNO3

Mg(OH)2

H2SO3

HBrO3

Ca(OH)2

Carboxylic acids

*H2SO4

Ba(OH)2

Polyprotic acids

Strengths of Acids and Bases

*HClO4 , HBr, H2SO4, HCl and HNO3 are of same strengths in water; but in glacial acetic acid, their strengths are: HClO4 > HBr > H2SO4 > HCl > HNO3 This phenomenon is known as levelling effect of water.

p-Functions p-Functions – are a method of expressing concentrations, especially very large or very small values. pH – commonly known p-Function - defined as negative logarithm of hydronium ion concentration. pH = - log[H3O+]

* pH can be less than 0 or greater than 14.

pH Calculations

pH Calculations Other values of Kw...

pH Calculations For strong acids, wherein water contributes to H3O+ in solution;

I: Δ: F:

H2O + H2O  55.5 M 55.5 M -X -X 55.5-X 55.5-X

H3O+ + MSA +X MSA+ X

OH0 +X X

pH Calculations K w = [H3 O+ ] OH − = 1 x 10−14 −14 K w = [MSA+ X] X = 1 x 10 X 2 + MSA X − 1 x 10−14 = 0 Then, pOH = -log [X] pH = 14 + log [X]

pH Calculations For strong bases, wherein water contributes to OH- in solution;

I: Δ: F:

H2O + H2O  55.5 M 55.5 M -X -X 55.5-X 55.5-X

H3O+ 0 +X X

OHMSB +X MSB + X

+

pH Calculations

Example 5 What is the pH of 1 x 10 -8 HNO3 solution?

Ans. 6.98

Example 6 What is the pH of pure water at 50 0C?

Ans. 6.63

Example 7 How much water is to be added to 12 M HCl in order to prepare 1600 mL solution of pH = 1.50?

Ans. 1595.784 L

Example 8 Calcium hydroxide solution has a concentration of 0.05 M. Calculate its pH.

Ans. 13

Example 9 A solution is prepared by adding 125.0 mL of 0.025 M HNO3 to 150.0 mL of 0.020 M HCl. Determine the concentration of pOH of the resulting mixture.

Ans. 12.35

pH Calculations For weak acids:

I: Δ: F:

MWB 55.5 M -X -X MWB – X 55.5-X

0 +X X

0 +X X

pH Calculations For weak acids:

pH Calculations For weak bases:

I: Δ: F:

MWB 55.5 M -X -X MWB - X 55.5-X

0 +X X

0 +X X

pH Calculations For weak bases:

pH Calculations For weak acids and acidic salt: X 2 + Kx − MK = 0 * K = Ka for Weak acid K = Kw/Kb for Acidic salt pH = -log [X]

For weak bases and basic salt: X 2 + Kx − MK = 0 * K = Ka for Weak base K = Kw/Kafor Basic salt

pH = 14 + log [X]

pH Calculations Percent Ionization equilibrium concentration of ionized acid initial concentration of acid

X 100

Example 10 Calculate the pH of a 0.010 M solution of iodic acid (HIO3, Ka = 0.17).

Ans. 2.02

Example 11 What mass of benzoic acid, HC7H5O2 is needed to dissolve in 350.0 mL of water to produce a solution having a pH of 2.85? Ka = 6.3 x 10-5.

Ans. 1.4 g

Example 12 Caproic acid, HC6H11O2, found in small amounts in coconut and palm oils, is used in making artificial flavors. A saturated aqueous solution of the acid contains 11 g/L and has pH = 2.94. Calculate the Ka for the acid. Ans. 1.4 x 10-5

Example 13 What is the percent ionization of propionic acid in a solution that is 0.45 M HC3H5O2? pKa = 4.89.

Ans. 0.53%

Example 14 What is the % ionization in 0.10 M NH3?

Ans. 1.33 %

Common Ion Effect Common Ion Effect • Shift in equilibrium when one or more ions that are part of the equilibrium are introduced from an outside source. • reduction in the dissociation of the weak electrolyte

Example 15 • Calculate the [H3O+] in a 0.0045 M benzoic acid (HC7H5O2) solution. Ka = 6.3 x 10-5. • Calculate the [H3O+] in a 0.0045 M benzoic acid a solution which contains 0.001 M NaC7H5O2.

Hydrolysis Reaction of Salts • Acidic Salt: NH4Cl NH4+ + H2O KH = KW /KNH3

• Basic Salt: NaCN CN- + H2O KH = KW /KHCN

H3O+ + NH3

HO- + HCN

pH of Salts pH of Salts Acidic Salt: pH = 7 – ½ log[Csalt/Kb] when Csalt/KH >>> 1000 Basic Salt: pH = 7 + ½ log[Csalt/Ka] when Csalt/KH >>> 1000

Example 16 What is the pH of an aqueous solution that is 0.089 M NaOCl? pKa = 7.54

Ans. 10.24

Example 17 What weight (in grams) of NH4Cl is needed to be dissolved in 200 mL of water to provide a solution having a pH of 4.50? Kb = 1.8 x 10-5

Ans. 19.2 g

Buffered Solutions “A buffered solution is one that resists a change in pH when either hydroxide ions or protons are added.”

Buffered Solutions Buffer Capacity - refers to the amount of acid or base that a buffer can neutralize before its pH changes.

Buffered Solutions

Buffered Solutions pH of a Buffered Solution

or

Example 18 A buffered solution contains 0.25 M NH3 and 0.40 M NH4Cl. Calculate the pH.

Ans. 9.05

Example 19 How many mL of pure formic acid (s.g = 1.22) must be mixed to 325 mL of 0.0664 M NaOH solution to obtain a buffer solution of pH 3.25? Ka = 1.7 x 10-4

Ans. 3.51 mL

Example 20 What is the pH of the resulting solution made by mixing 5 mL of 0.2178 M HCl and 15 mL of 0.1156 M NH3? Kb = 1.8 x 10-5

Ans. 9.02

Acid-Base Indicators Acid-Base Indicator – is a substance whose color depends on the pH of the solution to which it is added.

Two forms of acid-base indicators: 1. Weak acid (represented by HIn. 2. Conjugate base (represented by In-.

HIn + acid color

H2O

H3O+ +

Inbase color

Acid-Base Indicators HIn + acid color

H2O

H3O+ +

Inbase color

Acid-Base Indicators Name

pH range

pKa

Color change

Thymolphthalein

1.70

1.2-2.8

R-Y

Methyl Orange

3.46

3.1-4.4

R-Y

Bromocresol Green

4.66

3.8-5.4

Y-B

Methyl Red

5.00

4.2-6.3

R-O

Bromothymol Blue

7.10

6.2-7.6

Y-B

M-cresol Purple

8.32

7.6-9.2

Y-Purple

Thymol Blue

8.96

8.0-9.6

Y-B

Phenolphthalein

9.00

8.3-10.0

C-Pink

Thymolphthalein

10.0

9.4-10.6

C-B

Example 21 A particular indicator has a color red and a color blue in its acid and base form respectively. If this indicator has a Ka = 3 x 10-5 , by how much must the pH change in order to change the indicator from 75% red to 75% blue? Ans. 0.95

Gravimetry

Gravimetry Gravimetric Methods: - are methods that depend upon measuring the mass (i.e., gravity).

Gravimetry Three types of gravimetric methods: 1) Volatilization Methods 2) Extraction Methods 3) Precipitation Methods

Gravimetry Three types of gravimetric methods: Volatilization Method The analyte are volatilized at suitable temperature. The volatilized species are collected. The collected samples are weighed directly or weighed by difference.

Gravimetry Three types of gravimetric methods: Extraction Method The analyte to be determined is extracted w/ appropriate solvent.

The mass of the purified extract is related to the amount of the analyte.

Gravimetry Three types of gravimetric methods: Precipitation Method A sample is dissolved in an appropriate solvent. The analyte is precipitated by a reagent that yields a sparingly soluble product. The precipitate is converted to a product of known composition by heat treatment.

Gravimetric Methods Basic calculations: Percentage of analyte in a sample is calculated using a Gravimetric factor GF.

Gravimetric Methods Basic calculations:

Example 22 A sample containing NaBr and KBr only weighs 312.54 grams. The sample was dissolved in water and treated with excess AgNO3. The precipitate formed was found to weigh 532.55 grams. Calculate % KBr in the sample. Ans. 49%

Example 23 • A 0.1005 gram sample of an ionic compound containing chloride ions and an unknown metal is dissolved in water and treated with an excess of AgNO3. If 0.0445 g of AgCl precipitate forms, what is the % by mass of Cl- in the original compound? Ans. 10.95%

Example 24 What weight of Mn ore should be taken so that the percentage of MnO2 in the ore would be twice the mass of Mn3O4 precipitate obtained in milligram? Ans. 57.0 mg

Example 25 A 0.8715 gram sample containing chloride and iodide ions gave a silver halide precipitate which weighs 1.8561 grams. This was heated with Cl2 gas to convert silver iodide precipitate to AgCl. The resulting precipitate weighs 1.7223 gram. Calculate the percentage of chloride in the sample. Ans. 43%

Titrimetric Methods

Titrimetric Methods Titrimetric methods are analytical procedures in which the amount of an analyte is determined from the amount of a standard reagent required to react completley with the analyte.

Titrimetric Methods Types: 1. 2. 3. 4.

Precipitimetry Acidimetry/Alkalimetry Compleximetry Reductimetry/Oxidimetry

Titrimetric Methods Standard Solution - a reagent used to titrate the analyte. It must: 1. Have a precisely known concentration. 2. Generally is added from a buret.

Titrimetric Methods Standard Solution

Titrimetric Methods Primary Standard

-

a highly purified compound that serves as a reference materials.

Titrimetric Methods Primary Standard It must have the ff. characteristics: a. High purity b. Stable in air c. Absence of hydrated water molecules d. Moderate cost and easy availability e. Solubility in the titration solutions f. Large formula weight (molecular weight) *Compounds that do not meet all these criteria are called secondary standards.

Titrimetric Methods Some commonly used Primary Standards: For Bases:

• • • •

Benzoic Acid, C6H5COOH (f = 1) Oxalic Acid, H2C2O4∙2H2O (f = 2) Potassium Biiodate, KH(IO3)2 (f =1) Potassium Hydrogen Phthalate (KHP), C6H4(COOH)(COOK) (f =1) • Sulfamic Acid (HSO3NH2) (f =1)

Titrimetric Methods Some commonly used Primary Standards: For Acids:

• • • •

Calcium Carbonate, CaCO3 (f = 2) Mercuric oxide, HgO (f = 2) Sodium Carbonate, Na2CO3 (f = 2) Tris-hydroxymethylaminomethane (THAM), (CH2OH)3CNH2 (f =1)

Example 26 How many grams of KHP are needed to neutralize 167.33 mL of 0.99955 M NaOH?

Ans. 34.157 grams

Example 27 In standardizing a solution of NaOH against 1.431 gram of KHP, the analyst uses 35.50 mL of the alkali and has to run back with 6.12 ml of acid (1mL = 4.1 mg NaOH). What is the molarity of the NaOH solution? Ans. 0.2151 M

Applications of Acid-Base Titration 1. Determination of Organic NitrogenKjeldahl Method 2. Double indicator method for mixture of bases – Warder Titration 3. Acid Number 4. Saponification Number

Kjeldahl Method (Determination of Organic Nitrogen) Step 1. Digestion • The sample is oxidized in hot, concentrated sulfuric acid, H2SO4 and turns black. . Step 2. Distillation • The oxidized solution is cooled and then treated with NaOH to liberate ammonia gas: NH4+ + OHNH3(g) + H2O

Kjeldahl Method (Determination of Organic Nitrogen) • Step 3. Titration Using an excess amount of HCl. . . NH3 + HCl NH4Cl The excess HCl is determined using a standard NaOH solution HCl + NaOH NaCl + H2O

Kjeldahl Method (Determination of Organic Nitrogen)

Ammonia distilled is collected in a boric acid solution. . . NH3 + H3BO3 NH4+ + H2BO3-1 Titrate the H3BO3-NH3 solution with standard acid. . . H2BO3-1 + H3O+ H3BO3 + H2O

Kjeldahl Method (Determination of Organic Nitrogen)

• Percentage Protein in the sample %protein =%N * f f = 5.70 (cereals) = 6.25 (meat products) = 6.38 (dairy products)

Example 28 A 758-mg sample of full cream milk was analyzed by the Kjeldahl method; 38.61 mL of 0.1078 M HCl were required to titrate the liberated ammonia. Calculate the % N in the sample.

a.12.04% b. 7.69% c. 15.59% d. 10.93%

Example 29 A 7.443-gram sample beef was analyzed for its N content and the liberated NH3 was collected in a 43.25 mL of 0.4330 M HCl and a 15.00 mL back titration with 0.0250 M NaOH was required. Determine the % protein in the sample using 6.25 as factor for meat products. a.12.44% b. 21.57% c. 32.54% d. 10.98%

Example 30 A 3000 mg sample of flour was taken through a Kjeldahl method. Upon digestion, the ammonia liberated was collected into 200 mL of 0.995 M H3BO3 solution. If this solution required 25.25 mL of 0.3315 M HCl for titration to methyl red end point, what is the percentage of protein in flour? Use 5.70 for cereal products. Ans. 22.27%

Example 31 A 500 mg sample of each mixture was analyzed for its alkaline content using 0.1025 M HCl via double indicator method. Mixture

V0->Ph

V0->MR

1

4.27

10.18

2

0.01

6.19

3

5.12

10.24

4

6.37

6.38

5

5.63

9.04

Determine the weight compositions of each mixture.

Acid Number • Acid number - mass (mg) of KOH that will neutralize the acid produced fromwater degradative reaction of one gram of fat or oil Acid no. = (VxM)KOH X MW KOH gram fat or oil

Saponification no. Saponification no./Koettstorfer no. = mass of KOH req’d to saponify 1 gram fat or oil

Sap. no. = (Vblank, mL – Vsample, mL ) (MHCl)(56.10) gram fat or oil Molar mass of Fat or Oil. = 168,300 Sap. Value

Example 32 • The saponification no. of triglycerides is 200. The average MW of the triglycerides is:

A. 200

B. 280

C. 600

D. 840

Precipitation Titrations Common Technique: Argentometric Titrations: 1. Mohr method 2. Volhard method 3. Fajans method

Method

Mohr

Titrant

Indicator

AgNO3

K2CrO4

Rxn : Ag  ( aq )  Cl   AgCl ( s ) 2 Ag  ( aq )  CrO 4 2 ( aq )  Ag 2CrO 4( s )

KSCN Backtitration : Volhard

Ag  ( aq )  Cl   AgCl( s )

ferric alum, NH4Fe(SO4)212H2O Fe 3 ( aq )  SCN 1 ( aq )  Fe( SCN ) 2 ( aq )

Ag  ( aq )  SCN  ( aq )  AgSCN ( s )

Fajans

Ag



( aq )

AgNO  3

 Cl  AgCl ( s )

Fluorescein Dicholro-fluorescein End point: greenish-yellow to pink

Example 33 A 1.500-gram sample of impure aluminum chloride was dissolved in water and treated with 45.32 mL of 0.1000 M AgNO3 using K2CrO4 as indicator. Express the analysis in % AlCl3

Ans. 13.43 %

Titration with EDTA EDTA – Ethylenediaminetetraacetic acid • combines w/ any metal ion in ratio of 1:1 Indicators: 1. Eriochrome Black T or Solochrome 2. Calmagite

Example 34 An EDTA solution was prepared by dissolving the disodium salt in 1 L of water. It was standardized using 0.5063 gram of primary standard CaCO3 and consumed 28.50 mL of the solution. The standard solution was used to determine the hardness of a 2 L sample of mineral water, which required 35.57 mL of the EDTA solution. The concentration (ppm) in terms of CaCO3 is… Ans. 315.95 ppm

Example 35 The 300 mg sample of impure Na2SO4 (142.04) was dissolved in sufficient water and the sulfate was precipitated by the addition of 35.00 mL of 0.1022 M BaCl2. The precipitate was removed by filtration and the remaining BaCl2 consumed 6.79 mL of 0.2467 M EDTA for titration to the Calmagite endpoint. Calculate the purity of the sample.

Redox Reactions

Example 36 Balance the reaction:





NO 2  MnO4  NO3  MnO2 

RULES: • • • • • • • •

• •

An atom in its free or elemental form has oxidation equal to zero For monoatomic ions, the oxidation number is equal to its charge Metals have positive oxidation number such as alkali metals (+1), alkaline earth metals (+2), aluminum (+3), zinc (+2) and silver (+1) Nonmetals usually have negative oxidation numbers: Oxygen is usually –2, except in peroxides (–2) and superoxides (–1) Hydrogen is usually +1, except in hydrides (–1) Fluorine has –1 oxidation state; other halogens are usually in the – 1 oxidation state, except when combined with oxygen, they are positive; when different halogens are bound to each other, –1 is assigned to the more electronegative halogen The sum of oxidation number of elements in a compound is zero. The sum of oxidation number of elements in a polyatomic ion is the charge of the ion

Example 37 What is the molarity of a KMnO4 solution standardized against 1.356 gram Na2C2O4 (134 g/mol) requiring 25.1 mL of the solution in acidic medium?

a. 0.161 M b. 0.403 M c. 1.008 M d. 0.856 M

Example 38 A sample of pyrolusite weighing 0.2400 gram was treated with excess KI. The iodine liberated required 46.24 mL of 0.1105 M Na2S2O3 solution. Calculate % MnO2 (86.94) in the sample.

a. 46.27% b. 30.85% c. 92.54% d. 76.12%

Example 39 A sample of iron ore weighing 385.6 mg was dissolved in acid and passed through a Jones reductor. The resulting solution 52.36 mL of 0.01436 M K2Cr2O7 for titration to the diphenylamine sulfonic acid endpoint. Calculate % Fe3O4 (231.55 g/mol) in the ore sample. a. 15.05% b. 45.15% c. 90.30% d. 67.98%

Example 40 A 10.00 gram sample of cooked-ham was puréed with 200 mL of water, filtered and the resulting solution containing dissolved potassium nitrite was acidified. This solution was treated with 25.00 mL of 0.00514 M KMnO4 was back titrated with 14.97 mL of 0.01678 M FeSO4. Calculate the amount of nitrite (46.01) in ppm. Ans. 90 ppm

Masking Example 41: A 0.8521 gram sample of an alloy was found to contain Cu (63.55) and Zn (65.41) with small amounts of Pb (207.2) and Hg (200.59). The sample was dissolved in nitric acid and diluted to 500 mL. A 10 mL aliquot was treated with KI to mask the Hg and the resulting solution required 7.06 mL of 0.0348 M EDTA solution. A second 25 mL aliquot was treated with ascorbic acid and the pH was adjusted to 2.00 to reduce Hg+2 and the metallic Hg was removed from the solution. To this solution, thiourea was then added to mask the Cu and the resulting solution required 8.58 mL for titration. The lead ion was titrated in a 250 mL in the presence of NaCN to mask Cu, Zn and Hg and required 3.11 mL for titration. Calculate the percentage of Cu and Hg in the sample of alloy. Ans. 47.08% Cu, 3.48% Hg