Ace Ahead Chemistry (CD-Rom)1st(17.2.11)
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PRACTICAL GUIDE
CONTENTS SECTION 1 Volumetric analysis 1.1 Important formulae 1.2 Acid-base titration Colour change of acid-base indicators at end point Chemical equations for acid-base titrations 1.3 Potassium manganate(VII), KMnO4, titration Balancing half-equations Balancing ionic equations for redox reactions Important half-equations for KMnO4 titrations Important ionic equations for KMnO4 titrations Precautions in KMnO4 titration 1.4 Iodine-sodium thiosulphate (Na2S2O3) titration Important equations for iodine-sodium thiosulphate titrations Precautions in iodine-sodium thiosulphate titrations
SECTION 2 Physical chemistry Thermochemistry
SECTION 3 Qualitative analysis Table 1 Table 2 Table 3 Table 4 Table 5 Table 6 Table 7 Table 8 Table 9 Table 10
Reactions with dilute hydrochloric acid Reactions with aqueous sodium hydroxide Reactions with aqueous ammonia Reactions with aqueous iron(III) chloride Reactions with silver nitrate solution Reactions with potassium chromate(VI) solution Reactions with potassium iodide solution Reactions of sodium ethanoate solution Reactions of sodium carbonate solution Identification of functional groups in organic compounds
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SECTION 1 Volumetric analysis
1.1 Important formulae Concentration (in g dm–3) Concentration (in mol dm–3) = ——————————————— … (1.1) Relative molecular mass Concentration (in dm–3) = Concentration (in mol dm–3) Relative molecular mass … (1.2) If the reaction between X and Y to form P and Q is represented by the chemical equation: aX + bY → cP + dQ Then, (M V1)X a ——1—— – = — … (1.3) (M2V2)Y b Example Calculate the concentration in mol dm–3 of the following solutions: (a) F1 is hydrochloric acid of concentration 0.913 g dm–3. (b) F2 is a solution containing 3.4 g of OH– per dm3. (c) F3 is a solution containing 2.38 g MnO4– per dm3. (d) F4 is a solution containing 0.775 g KMnO4 per 250 cm3. (Relative atomic mass: H, 1; O, 16.0; Cl, 35.5; K, 39.1; Mn, 54.9) Solution (a) Relative molecular mass of HCl = 36.5 0.913 Concentration = ———– = 0.025 mol dm–3 36.5 (b) Relative formula mass of OH– = 17.0 3.4 Concentration = ———– = 0.20 mol dm–3 17.0 (c) Relative formula mass of MnO4– = 118.9 2.38 Concentration = ———– = 0.020 mol dm–3 118.9 (d) Relative molecular mass of KMnO4 = 158 0.775 Concentration = ———– = 0.0049 mol per 250 cm3 158 = 0.0049 4 = 0.0196 mol dm–3
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1.2 Acid-base titration Colour change of acid-base indicators at end point Solution in the burette
Solution in the conical flask
Indicator in the conical flask
Colour change
Acid
Base
Phenolphthalein
Pink to colourless
Base
Acid
Phenolphthalein
Colourless to pink
Acid
Base
Methyl orange
Yellow to orange
Base
Acid
Methyl orange
Red to orange
Chemical equations for acid-base titration a The ratio of — in the second column is obtained by considering b a = number of moles of the first reactant b = number of moles of the second reactant Equation
a Ratio of — b
H+ + OH– → H2O
1 — 1
NaOH + HCl → NaCl + H2O
1 — 1
HClO4 + NaOH → NaClO4 + H2O
1 — 1
Ba(OH)2 + 2HCl → BaCl2 + 2H2O
1 — 2
H2SO4 + 2NaOH + → Na2SO4 + 2H2O
1 — 2
H2C2O4 + 2NaOH + → Na2C2O4 + 2H2O
1 — 2
Example 1 To determine the identity of X in X(OH)2 F1 is hydrochloric acid of concentration 0.1 mol dm–3. F2 is a solution containing 6.85 g dm–3 of X(OH)2. 25.0 cm3 of F2 required 19.85 cm3 of F1 for complete neutralisation. (a) Write the equation for the reaction between X(OH)2 and hydrochloric acid. (b) Calculate the concentration of X(OH)2 in solution F2. (c) Hence calculate (i) the relative molecular mass of X(OH)2 and (ii) the relative atomic mass of X. (d) Suggest an identity for X. Solution (a) X(OH)2 + 2HCl → XCl2 + 2H2O (M1V1)X(OH) 1 2 (b) — — — — — — — — — — — = — 2 (M2V2)acid
1 0.1 19.85 — — — — — — — — — — = 0.0397 mol dm–3 Concentration of X(OH)2 = — — 2 25
(i) Concentration (in g dm–3) = Concentration (in mol dm–3) Relative molecular mass 6.85 Relative molecular mass = — — — — — — — = 172.5 0.0397 (ii) 172.5 = X + (2 17) X = 138.5 (d) X is barium.
(c)
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Example 2 To determine the identity of X in HXO4 F3 is 0.10 mol dm–3 sodium hydroxide solution. F4 is an acid with molecular formula, HXO4. 25.0 cm3 of F4 required 21.00 cm3 of F3 for complete neutralisation. (a) Write the equation for the reaction between HXO4 and NaOH. (b) Calculate the concentration of HXO4 in solution F4. (c) Hence calculate (i) the relative molecular mass of X(OH)2 and (ii) the relative atomic mass of X. (The concentration of HXO4 is 8.44 g dm–3) (d) Suggest an identity for X. Solution (a) HXO4 + NaOH → NaXO4 + H2O (M1V1)HXO 1 4 (b) — — — — — — — — — — — = — 1 (M2V2)NaOH
0.1 21.00 — — — — — — — — — — = 0.084 mol dm–3 Concentration of HXO4 = — 25
(c)
(i) Concentration (in g dm–3) = Concentration (in mol dm–3) Relative molecular mass 8.44 Relative molecular mass = — — — — — — = 100.5 0.084 (ii) 100.5 = 1 + X + (4 16) X = 35.5
(d) X is chlorine. Take Note
Example 3 NaOH solution is not used to standardise You are asked to determine the accurate concentration of a monoprotic acid because it is a acid, HX in F5 solution from the following experiment. The rough deliquescent solid. Thus, concentration of HX is about 1.0 mol dm–3. it is difficult to find the By means of a pipette, 50.0 cm3 of F5 is transferred into a 250 cm3 accurate mass of NaOH as it absorbs the moisture volumetric flask. Distilled water is then added and make up to the mark from the air (deliquescent) on the volumetric flask. The solution is labelled as F6 solution. during weighing. F7 is prepared by dissolving 2.65 g of anhydrous sodium carbonate in 250 cm3 solution. 25.0 cm3 of F7 required 20.95 cm3 of F6 for complete neutralisation using methyl orange as indicator. (a) Calculate the concentration of sodium carbonate in F7 solution. (b) Write the chemical equation for the reaction between sodium carbonate and HX. (c) Calculate the concentration of F6. (d) Hence, calculate the accurate concentration of HX in F5. Solution (a) Relative molecular mass of Na2CO3 = 106 2.65 4 — — — — — — — = 0.1 mol dm–3 Concentration of Na2CO3 = — 106 (b) 2HX + Na2CO3 → 2NaX + CO2 + H2O (M1V1)HX 2 (c) — — — — — — — — — — — = — 1 (M2V2)Na CO 2
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2 0.1 25.0 Concentration of HX in F6 = — — — — — — — — — — — — = 0.239 mol dm–3 20.95 250 (d) Concentration of HX in F5 = 0.239 — — — = 1.20 mol dm–3 50 4
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1.3 Potassium manganate(VII), KMnO4, titration Balancing half-equations 1. A half-equation is a chemical equation containing electrons. Fe2+ → Fe3+ + e– Cl2 + 2e– → 2Cl–
… oxidation reaction … reduction reaction
2. The following steps are used to balance half-equations for redox reactions. Step 1 : Determine the oxidation number of the atom that undergoes oxidation or reduction. Step 2 : Balance the half-equation in terms of the total charge and the number of atoms on both sides of the equation. Step 3 : Calculate the number of oxygen atoms on both sides of the equation and add water (if necessary) on the side of equation that has insufficient number of oxygen atoms. Example 1 Balance the equation : MnO4– + H+ → Mn2+ Step 1 The oxidation number of Mn decreases from +7 to +2. Hence, 5e– must be added to the left-hand side of the equation. MnO4– + H+ + 5e– → Mn2+ … not balanced Step 2 Balance the total charge and number of atoms on both sides of the equation MnO4– + 8H+ + 5e– → Mn2+ + 4H2O … balanced Total charge on LHS of equation = –1 + 8 (+1) + 5 (–1) = +2 Total charge on RHS of equation = +2 Total number of atoms on LHS of equation = 1Mn + 4O + 8H Total number of atoms on RHS of equation = 1Mn + 4O + 8H Hence, the half-equation is a balanced equation. Example 2 Balance the equation : XO2– → XO3– Step 1 The oxidation number of X increases from +3 to +5. Hence, 2e– must be added to the right-hand side of the equation. XO2– → XO3– + 2e–
… not balanced
Steps 2 and 3 XO2– + H2O → XO3– + 2e– + 2H+ … balanced Total charge on LHS of equation = –1 Total charge on RHS of equation = (–1) + 2 (–1) + 2 (+1) = –1 Total number of atoms on LHS of equation = 1X + 2H + 3O Total number of atoms on RHS of equation = 1X + 3O + 2H Hence, the half-equation is a balanced equation.
Balancing ionic equations for redox reactions Step 1 : Write the half-equations for the oxidation and reduction reactions. Step 2 : Combine the half-equations to get an ionic equation that does not contain electrons. Example 3 Balance the equation for the reaction between HCOO– and MnO4–. Step 1 Write the half-equations for HCOO– and MnO4– HCOO– → H+ + CO2 + 2e– … oxidation (1) MnO4– + 8H+ + 5e– → Mn2+ + 4H2O … reduction (2)
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Step 2 Combine the half-equations into the ionic equation 5 equation (1) gives 5HCOO– → 5H+ + 5CO2 + 10e– … (3) 2 equation (2) gives 2MnO4– + 16H+ + 10e– → 2Mn2+ + 8H2O … (4) Adding equations (3) and (4) gives 5HCOO– + 2MnO4– + 11H+ → 5CO2 + 2Mn2+ + 8H2O
Important half-equations for KMnO4 titrations MnO4– + 8H+ + 5e– → Mn2+ + 4H2O Fe2+ → Fe3+ + e– C2O42– → 2CO2 + 2e– H2O2 → O2 + 2H+ + 2e– NO2– + H2O → NO3– + 2H+ + 2e–
Important ionic equations for KMnO4 titrations Note: a = number of moles for the first reactant, b = number of moles for the second reactant Ionic equation
a Ratio of — b
5Fe2+ + MnO4– + 8H+ → 5Fe3+ + Mn2+ + 4H2O
5 — 1
5C2O42– + 2MnO4– + 16H+ → 2Mn2+ + 10CO2 + 8H2O
5 — 2
5H2O2 + 2MnO4– + 6H+ → 5O2 + 2Mn2+ + 8H2O
5 — 2
5NO2– + 2MnO4– + 6H+ → 5NO3– + 2Mn2+ + 3H2O
5 — 2
5FeC2O4 + 3MnO4– + 24H+ → 5Fe3+ + 3Mn2+ + 12H2O + 10CO2
5 — 3
Precautions in KMnO4 titration 1. KMnO4 titration is used to determine the concentration of iron(II) salt, ethanedioate (also called oxalate, C2O42–), hydrogen peroxide and nitrites. 2. Solutions used for KMnO4 titrations must be acidified. In neutral or alkaline solution, brown MnO2 is precipitated and detection of end point is difficult. MnO4– + 2H2O + 3e– → MnO2 + 4OH– 3. Sulphuric acid is always used for acidification. Hydrochloric acid and nitric acid must not be used for acidification. 4. KMnO4 titrations involving Fe2+, H2O2 and NO2– should be carried out at room temperature. 5. KMnO4 titration involving C2O42– should be carried out at temperatures above 70 °C. If the titration is carried out at lower temperatures, the redox reaction is incomplete, precipitation of MnO2 may occur and the result obtained is inaccurate. 6. KMnO4 solution should be added from the burette slowly. If it is added quickly, a brown suspension of MnO2 is formed. 7. No external indicator is required for KMnO4 titration because KMnO4 can act as it own indicator. The end point is reached when one drop of KMnO4 solution produces a light pink colour in the reaction mixture which should last for more than 35 seconds. 6
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Example 4 To find the mass of FeSO4.7H2O in 250 cm3 solution A sample of iron(II) sulphate crystals, FeSO4.7H2O, was dissolved in dilute sulphuric acid and made up to 250 cm3 with distilled water in a standard flask. 25.0 cm3 of this solution needed 25.80 cm3 of KMnO4 for complete reaction. (a) Write the ionic equation for the reaction between iron(II) sulphate and KMnO4. (b) If the concentration of KMnO4 solution is 3.16 g dm–3, calculate the concentration (in g dm–3) of (i) FeSO4, (ii) FeSO4.7H2O. (c) What was the mass of FeSO4.7H2O dissolved in 250 cm3 solution? Solution (a) 5Fe2+ + MnO4– + 8H+ → 5Fe3+ + Mn2+ + 4H2O (b) (i) Step 1 Calculate the concentration of KMnO4 in mol dm–3 Relative molecular mass of KMnO4 = 158 3.16 Concentration of KMnO4 = ——— = 0.020 mol dm–3 158 Calculate the concentration of Fe2+
Step 2
(M1V1)Fe2+ 5 — — — — — — — — — — = — (M2V2)KMnO 1 4
0.020 25.8 — — — — — — — — — — — — — — — — — — — — — — — — — = 0.103 mol dm–3 Concentration of Fe2+ = 5 — 25 Relative molecular mass of FeSO4 = 152 Concentration of FeSO4 = 0.103 152 = 15.7 g dm–3 (ii) Relative molecular mass of FeSO4.7H2O = 278 Concentration of FeSO4.7H2O = 0.103 278 = 28.6 g dm–3 (c) Mass of FeSO4.7H2O dissolved in 250 cm3 solution 28.6 = — — — — — = 7.15 g 4 Example 5 To find the concentration of Na2C2O4 F1 is a solution containing oxalic acid, H2C2O4 and sodium oxalate, Na2C2O4. F2 is 0.10 mol dm–3 sodium hydroxide solution. F3 is 0.024 mol dm–3 KMnO4 solution. 25.0 cm3 of F1 required 18.50 cm3 of F2 for complete neutralisation. 25.0 cm3 of F1 required 33.60 cm3 of F3 for complete redox reaction. (a) Calculate the concentration (in g dm–3) of H2C2O4 in F1 solution (b) Calculate the concentration (in mol dm–3) of C2O42– ions in F1 solution. (c) Hence, calculate the concentration (in g dm–3) of Na2C2O4 in F1 solution. Solution (a) H2C2O4 + 2NaOH → Na2C2O4 + 2H2O (M1V1)H C O 1 2 2 4 — — — — — — — — — — — — = — 2 (M2V2)NaOH
1 0.1 18.5 — — — — — — — — — = 0.037 mol dm–3 Concentration of H2C2O4 = — — 2 25
Relative molecular mass of H2C2O4 = 90 Concentration of H2C2O4 = 0.037 90 = 3.33 g dm–3 (b) 5C2O42– + 2MnO4– + 16H+ → 2Mn2+ + 10CO2 + 8H2O (M1V1)C O 2– 5 2 4 — — — — — — — — — — — —= — 2 (M2V2)KMnO 4
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5 0.024 33.6 — — — — — — — — — — — — = 0.0806 mol dm–3 Concentration of C2O42– = — — 2 25 (c) Total concentration of C2O42– = 0.0806 mol dm–3 Concentration of H2C2O4 = 0.037 mol dm–3 Concentration of Na2C2O4 = 0.0806 – 0.037 = 0.0436 mol dm–3 Relative molecular mass of Na2C2O4 = 134 Concentration of Na2C2O4 = 0.0436 134 = 5.84 g dm–3 Example 6 Acidified hydroxyammonium ion, NH3OH+, reduces iron(III) ion, Fe3+, to iron(II) ion, Fe2+. In the following experiment, you are asked to determine the chemical equation for the reaction between hydroxyammonium ion and iron(III) ion. F4 is a solution prepared by boiling 1.56 g of hydroxyammonium sulphate, NH3OH+HSO4– ,with excess iron(III) ammonium sulphate and dilute sulphuric acid. The reaction mixture is then made up to 250 cm3 with distilled water. F5 is potassium manganate(VII) solution containing 1.58 g of KMnO4 per 500 cm3. In a titration experiment, 25.0 cm3 of F4 required 24.4 cm3 of F5 for complete reaction. (a) Calculate the concentration of (i) NH3OH+HSO4–, (ii) KMnO4 in mol dm–3. (b) Calculate the concentration (in mol dm-3) of Fe2+ ions produced in F4. (c) Hence, calculate the number of moles of Fe3+ that react with 1 mol of NH3OH+. (d) The half-equation for the oxidation of hydroxylamine to nitrogen is: 2NH2OH → N2O + H2O + 4H+ + 4e– Hence, write a balanced redox equation between NH3OH+ ions and Fe3+ ions. Solution (a) (i) Relative molecular mass of NH3OH+HSO4– = 131 1.56 4 Concentration = — — — — — — — — = 0.0476 mol dm–3 131 (ii) Relative molecular mass of KMnO4 = 158 1.58 2 Concentration = — — — — — — — — = 0.020 mol dm–3 158 (b) 5Fe2+ + MnO4– + 8H+ → 5Fe3+ + Mn2+ + 4H2O (M1V1)Fe2+ 5 — — — — — — — — — — — —= — 1 (M2V2)MnO – 4
5 0.02 24.4 — — — — — — — — — — — — — — — = 0.0976 mol dm–3 Concentration of Fe2+ = — 25 (c) Fe3+ + e– → Fe2+ 1 mol of Fe2+ is produced from 1 mol of Fe3+. Number of moles of Fe3+ that have reacted with NH3OH+ = 0.0976 NH3OH+ : Fe3+ = 0.0476 : 0.0976 = 1 : 2 (d) 2NH2OH → N2O + H2O + 4H+ + 4e– Fe3+ + e– → Fe2+ 2NH2OH + 4Fe3+ → 4Fe2+ + N2O + H2O + 4H+
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1.4 Iodine-sodium thiosulphate (Na2S2O3) titration 1. The half-equations for the redox reaction between iodine and thiosulphate are 2S2O32– → S4O62– + 2e– … (1) I2 + 2e– → 2I– … (2) 2. The ionic equation for the reaction between sodium thiosulphate and iodine is 2S2O32– + I2 → S4O62– + 2I– … (3) This reaction is used to determine the concentration of iodine by titration against sodium thiosulphate. In some cases, the reaction is used to determine the concentration of an oxidising agent. For example, the oxidising agent is added to potassium iodide solution. The iodine liberated is titrated against sodium thiosulphate. From this, the concentration of the oxidising agent can be deduced indirectly. 3. Iodine has a low solubility in water but dissolves readily in potassium iodide solution. I3–(aq) + K+(aq)
I2(s) + KI(aq)
In most direct titrations with iodine, a solution of iodine in potassium iodide is used. For simplicity, equation (3) is commonly used for calculation.
Important equations for iodine-sodium thiosulphate titration Note: a = number of moles of the first reactant; b = number of moles for thiosulphate Ionic equation
a Ratio of — b
2S2O32– + I2 → S4O62– + 2I–
1 — 2 In the following reactions, iodine produced by the reaction is titrated against sodium thiosulphate solution. H2O2 + 2H+ + 2I– → 2H2O + I2 (H2O2 ≡ I2 ≡ 2S2O32–)
1 — 2
IO3– + 6H+ + 5I– → 3H2O + 3I2 (IO3– ≡ 3I2 ≡ 6S2O32–)
1 — 6
2Cu2+ + 4I– → Cu2I2 + I2 (2Cu2+ ≡ I2 ≡ 2S2O32–)
1 — 1
Precautions in iodine-sodium thiosulphate titrations 1. Iodine is a volatile substance. Hence the iodine solution prepared should be titrated immediately and quickly to avoid loss of iodine due to evaporation. 2. In I2 – S2O32– titration, starch is used as an indicator. The starch solution should not be added at the beginning of the titration where there is a high concentration of iodine. This is because iodine is adsorbed onto the starch molecule and may remain adsorbed even at the end point. 3. The starch solution should be added towards the end of the titration when the reaction mixture turns pale yellow. The reaction mixture in the conical flask should be shaken vigorously but carefully as the titration proceeds. 4. The starch solution will produce a dark blue colour with iodine and when the end point is reached, the solution turns colourless abruptly. 5. When the solution in the conical flask is left aside after titration, the solution becomes blue again. Ignore this, because iodine is formed due to the atmospheric oxidation of excess potassium iodide in the reaction mixture. 6. Starch solution is unstable and should be prepared fresh for each titration.
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Example 1 To determine the concentration of iodine solution F1 contains 20.0 g of Na2S2O3.5H2O per dm3. F2 is an iodine solution. 25.0 cm3 of F2 required 22.5 cm3 of F1 for complete reaction. Calculate (a) the concentration (in mol dm–3) of iodine in solution F2, (b) the mass of iodine required to prepare 200 cm3 of F2. Solution (a) Relative molecular mass of Na2S2O3.5H2O = 248 20.0 Concentration of Na2S2O3.5H2O = — — — — = 0.0806 mol dm–3 248 2S2O32– + I2 → S4O62– + 2I– (M1V1)I 1 2 — — — — — — — — — — — —= — 2 (M2V2)S O 2– 2 3
1 0.0806 22.5 Concentration of I2 = — ————————— = 0.0363 mol dm–3 2 25.0 (b) Relative molecular mass of I2 = 254 200 Mass of iodine = (0.0363 254) — — — — = 1.84 g 1000 Example 2 To determine the percentage purity of hydrated sodium sulphite, Na2SO3.7H2O F3 is a solution containing 11.1 g of hydrated sodium sulphite, Na2SO3.7H2O per dm3. F4 is 0.050 mol dm–3 iodine. F5 is 0.025 mol dm–3 sodium thiosulphate solution. 25.0 cm3 of F4 is mixed with 25.0 cm3 of F3. The resulting solution required 18.90 cm3 of F5 for complete reaction. Sulphite ion reacts with iodine as represented by the equation SO32– + I2 + H2O
Solution Step 1 Calculate excess iodine
SO42– + 2HI
0.025 18.90 Number of moles of S2O32– that reacted with excess I2 = — — — — — — — — — — — — — — = 4.725 10–4 1000 2S2O32– + I2 → S4O62– + 2I– 1 Number of moles of excess I2 = — 4.725 10–4 = 2.36 10–4 2
Step 2 Calculate amount of iodine reacted with SO32– 0.050 25 — — — — — — — — — — = 1.25 10–3 Number of moles of I2 added = — 1000 Number of moles of I2 reacted = 1.25 10–3 – 2.36 10–4 = 1.014 10–3 Step 3 Calculate mass of pure Na2SO3.7H2O SO32– + I2 + H2O SO42– + 2HI Number of moles of pure Na2SO3.7H2O = 1.014 10–3 Relative molecular mass of Na2SO3.7H2O = 252 Mass of Na2SO3.7H2O in 25.0 cm3 solution = (1.014 10–3) 252 = 0.256 g 1000 Mass of Na2SO3.7H2O in 1000 cm3 solution = 0.256 — — — — — = 10.24 g 25 10
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Step 4 Calculate percentage purity of Na2SO3.7H2O 10.24 % purity = — — — — — — 100 = 92.3 11.1 Alternative method Step 1 Calculate the volume of iodine that did not react with the sulphite ion 2S2O32– + I2 → S4O62– + 2I– (M1V1)I 1 2 — — — — — — — — — — — — — — — — —= — 2 (M2V2)S O 2– 2 3
1 0.025 18.9 — — — — — — — — — — — — Volume of I2 = — — 2 0.05 = 4.73 cm3 Step 2 Calculate the volume of iodine that reacted with the sulphite ion 25.0 – 4.73 = 20.3 cm3 Step 3 Calculate the concentration of Na2SO3.7H2O SO32– + I2 + H2O
SO42– + 2HI
(M1V1)SO 2– 1 3 — — — — — — — — — — — — — — — —= — (M2V2)I 1 2
0.050 20.3 — — — — — — — — — — — — = 0.0406 Concentration of SO32– = — 25 Relative molecular mass of Na2SO3.7H2O = 252 Concentration of Na2SO3.7H2O = 0.0406 252 = 10.23 g dm–3 Step 4 Calculate the percentage purity of hydrated sodium sulphite 10.23 % purity = — — — — — — — 100 = 92.2 % 11.1
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SECTION 2 Physical chemistry Thermochemistry Important formula ∆H = mc∆T where m = mass of solution, c = specific heat capacity (4.2 J g–1 K–1) ∆T = maximum rise or fall in temperature Example 1 To determine the heat of reaction between a metal hydrogen carbonate and dilute hydrochloric acid. F1 is 1.0 mol dm–3 hydrochloric acid. F2 is a metal hydrogen carbonate (MHCO3). 50.0 cm3 of F1 is placed in a polystyrene cup. 4.7 g of F2 is added to F1. The mixture is constantly stirred and the temperature recorded. Temperature of F1 before mixing
30.0 °C
Lowest temperature reached after mixing
25.0 °C
Temperature change
5.0 °C
Take Note 1. The density of solution is usually assumed to be 1.0 g cm–3. Hence, the mass of solution = volume of solution used. 2. The mass of solution does not include the mass of solid added to water. For example, when 1.0 g of NaOH is added to 100 cm3 of water, the mass of solution is 100 g and not 101 g.
(a) (b) (c) (d)
Calculate the amount of heat absorbed in the experiment. Calculate the number of moles of HCl used. Calculate the number of moles of MHCO3 used. Calculate the number of moles of MHCO3 reacted. (The relative molecular mass of the metal hydrogen carbonate = 100) (e) Calculate the enthalpy change for the reaction. Solution (a) ∆H = mc∆T = 50.0 4.2 5.0 = 1050 J = 1.05 kJ 1.0 50.0 (b) Number of moles of HCl used = — — — — — — — — — — 1000 = 0.05 4.7 (c) Number of moles of MHCO3 used = — — — 100 = 0.047 (d) MHCO3(aq) + HCl(aq) → MCl(aq) + H2O(l) + CO2(g); ∆H (+)ve HCl is in excess. Number of moles of MHCO3 reacted = 0.047 1.05 (e) ∆H = — — — — — = 22.3 kJ mol–1 0.047 Example 2 To determine the partition of an organic acid, HOOC(CH2)nCOOH between water and ether. F3 is a solution of an organic acid, HOOC(CH2)nCOOH. F4 is 0.1 mol dm–3 NaOH solution. F5 is 0.01 mol dm–3 NaOH solution. 40.0 cm3 of F3 is added to a bottle, followed by 60.0 cm3 of ether. The bottle is tightly closed and shaken vigorously. After 30 minutes, 10.0 cm3 of the ether layer is pipetted out and titrated with F5. Then, 10.0 cm3 of aqueous layer is pipetted out and titrated with F4. 12
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The experiment is repeated using (a) 50.0 cm3 of ether and 50.0 cm3 of F3. (b) 40.0 cm3 of ether and 60.0 cm3 of F3. Results Titration results Experiment
Volume of solution used
Volume of F4 (V1)
Volume of F5 (V2)
1
40 cm3 F3 + 60 cm3 ether
12.5 cm3
18.25 cm3
2
50 cm3 F3 + 50 cm3 ether
15.5 cm3
23.4 cm3
3
60 cm3 F3 + 40 cm3 ether
18.8 cm3
28.1 cm3
(a) Plot a graph of V1 (on the y-axis) against V2 (on the x-axis). (b) Based on the graph, determine the partition coefficient for the organic acid between water and ether. Solution Solution
20
10
V1 (cm3)
10
0
15
10
20
30
V2 (cm3)
Concentration of acid in aqueous layer (b) K = — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — Concentration of acid in ether layer Concentration of acid is proportional to the number of moles of alkali used. V1 (20 – 10) — —= — — — — — — — = 0.67 (30 – 15) V2 V1 0.1 ∴K=— — — — — — — — V2 0.01 0.67 0.1 =— — — — — — — — 0.01 = 6.67
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SECTION 3 Qualitative analysis Table 1 Reactions with dilute hydrochloric acid (I) Gas released Observation Gas
Colour
Odour
Action on blue litmus paper
Specific test
Inference
CO2
Colourless Odourless
Little effect
Limewater turns milky
CO32– or HCO3– present
SO2
Colourless Pungent
Blue → red
• Acidified K2Cr2O7 Orange → Green • Acidified KMnO4 Decolorisation
SO32– or S2O32– present
NO2
Brown
Pungent
Blue → red
–
NO2– present
CH3COOH Colourless Vinegar
Blue → red
–
CH3COO– present
(II) Precipitate formed Observation
Inference
Explanation
2+
White precipitate, soluble in concentrated HCl
Pb present
PbCl2 formed. Solubility in concentrated HCl due to formation of H2PbCl4
White precipitate, turns grey on exposure to sunlight
Ag+ present
AgCl formed. On exposure to sunlight, Ag deposited
Yellow precipitate
S2O32– present
Sulphur formed
White precipitate, dissolves when solution is heated. On cooling, yellow crystals formed
Pb2+ present
PbCl2 formed. PbCl2 is insoluble in water but dissolves readily in hot water
Table 2 Reactions with aqueous sodium hydroxide (I) Gas liberated Inference
Observation Red litmus → blue
Explanation
NH4+ present
NH3 liberated NH3 + HCl → NH4Cl
White fumes with concentrated HCl vapour
(II) Precipitate formed Observation
Inference 2+
2+
Explanation 2+
White precipitate insoluble in excess Mg , Ca or Ba NaOH present 2+ White precipitate soluble in excess NaOH Pb , Zn2+ or Al3+ present White precipitate turns rapidly to brown; insoluble in excess NaOH Mn2+ present Brown precipitate, insoluble in excess NaOH Fe3+ present Dirty green precipitate, insoluble in Fe2+ present excess NaOH Green precipitate, insoluble in excess Ni2+ present NaOH Blue precipitate, insoluble in excess NaOH Cu2+ present Bluish-green precipitate, soluble in excess NaOH to form a green solution Cr3+ present 14
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The basic hydroxides, M(OH)2 are formed The amphoteric hydroxides are formed Mn(OH)2 is formed; oxidised by air to Mn(OH)3 The basic hydroxide, Fe(OH)3 is formed The basic hydroxide, Fe(OH)2 is formed, Some Fe(OH)2 is oxidised to brown Fe(OH)3 The basic hydroxide, Ni(OH)2 is formed The basic hydroxide, Cu(OH)2 is formed The amphoteric hydroxide, Cr(OH)3 is formed
Table 3 Reactions with aqueous ammonia Note: The asterisk (*) shows the cation that will form a precipitate with NH3(aq) but the precipitate will dissolve when NH4Cl is added. Observation
Take Note
Explanation
Inference
White precipitate, insoluble in excess NH3(aq)
Pb2+, *Mg2+, Al3+ present
The metal hydroxides are formed
White precipitate, soluble in excess NH3(aq)
*Zn2+ present
Zn(OH)2 is soluble in NH3 due to the formation of [Zn(NH3)4]2+
White precipitate, turns rapidly to brown colour; insoluble in excess NH3(aq)
*Mn2+ present
Mn(OH)2 is oxidised easily to Mn(OH)3
Dirty green precipitate, insoluble in excess NH3(aq)
*Fe2+ present
Fe(OH)2 formed and oxidised by air to brown Fe(OH)3
Brown precipitate, insoluble in excess NH3(aq)
Fe3+ present
Fe(OH)3 formed
Bluish-green precipitate, insoluble in excess NH3(aq)
Cr3+ present
Cr(OH)3 formed
Green precipitate, soluble *Ni2+ present in excess NH3(aq) to form light blue solution
Ni(OH)2 formed which dissolves in NH3(aq) to form the complex ion [Ni(NH3)6]2+
Blue precipitate, soluble *Cu2+ present in excess NH3(aq) to form dark blue solution
Cu(OH)2 formed which dissolves in NH3(aq) to form the complex ion, [Cu(NH3)4]2+
1. Ammonia is a weak base. In aqueous solution, it undergoes partial dissociation to form NH4+ and OH– ions. NH3(aq) + H2O(l) NH4+(aq) + OH–(aq) 2. Ca2+(aq) forms a precipitate with NaOH(aq) but it does not form a precipitate with NH3(aq). This is because the concentration of OH– ions in aqueous ammonia is very low. The ionic product of Ca(OH)2 for ammonia is lower than the solubility product (Ksp) of Ca(OH)2. Hence, precipitation of Ca(OH)2 cannot occur. 3. Some metal hydroxides, such as Mg(OH)2, dissolve in NH4Cl, because of the common ion effect. In the presence of NH4Cl, the concentration of OH– is decreased due to the common ion, NH4+. The metal hydroxide dissolves because the ionic product is less than the solubility product.
Table 4 Reactions with aqueous iron(III) chloride, FeCl3(aq) (I) Formation of precipitate Observation
Inference
Explanation
–
Buff-coloured (yellowish-brown) precipitate
C6H5COO (benzoate) present
Formation of iron(III) benzoate, (C6H5COO)3Fe
Brown precipitate; CO2 liberated
CO32– present
Formation of iron(III) carbonate; CO2 liberated because FeCl3(aq) undergoes hydrolysis to produce H+ ions
(II) Colour change without precipitation Observation Solution turns red; Brown precipitate formed on heating
Inference –
Explanation –
CH3COO or HCOO present
(CH3COO)3Fe or (HCOO)3Fe formed. On heating, (CH3COO)3Fe or (HCOO)3Fe undergoes hydrolysis to form the basic salt, e.g. (CH3COO)Fe(OH)2
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Observation
Explanation
Inference –
Solution turns brownish-red; black deposit may also formed
I present
FeCl3 reduces I– to I2
Solution turns deep purple/ violet
Phenol, C6H5OH, or derivatives of phenol, e.g. HOC6H4COOH present
Formation of complex ion
Table 5 Reactions with silver nitrate solution Observation
Inference –
Explanation
White precipitate, insoluble in HNO3 but soluble in dilute ammonia solution
Cl present
AgCl formed. Dissolves in NH3(aq) because of complex ion formation, [Ag(NH3)2]+(aq)
Cream-coloured precipitate; insoluble in HNO3; insoluble in dilute ammonia solution
Br – present
AgBr formed
Yellow precipitate; insoluble in HNO3; insoluble in dilute ammonia solution
I– present
AgI formed
White precipitate, soluble in HNO3
SO32–, NO2– , C2O42–, CH3COO– or HCOO– present
Ag+ ions react with these anions to form insoluble salts. Insoluble salts of weak acids are soluble in HNO3
White precipitate, soluble in NH3 and hot water
C6H5COO– present
C6H5COOAg is insoluble in cold water but soluble in hot water
Table 6 Reactions with potassium chromate(VI) solution Explanation
Inference
Observation 2+
2+
Yellow precipitate, soluble in HNO3
Ba or Pb present
BaCrO4 or PbCrO4 precipitated
Yellow precipitate, soluble in HCl
Ba2+ present
BaCrO4 precipitated
Solution turns green
Reducing agent, such as SO32– or NO2– present
Reducing agent reduces CrO42– to Cr3+ (green)
Table 7 Reactions with potassium iodide solution Observation
Inference
Explanation
Pale yellow precipitate in brown solution
Cu2+ present
2Cu2+ + 4KI → 2CuI(s) + I2 + 4K+
Yellow precipitate, soluble in hot water. Yellow crystals formed on cooling
Pb2+ present
PbI2 formed. PbI2 is soluble in hot water but insoluble in cold water
Solution turns brown; black deposit formed
Oxidising agent present
Oxidising agents such as Fe3+, CrO42–, Cr2O72– or NO2– oxidises KI to I2.
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Take Note NO2– (nitrite) can act as an oxidising agent or a reducing agent depending on the reagent added to it.
Table 8 Reactions of sodium ethanoate, CH3COONa Observation
Explanation
Inference
Solution turns red. On heating, a brown precipitate formed
Fe present
3CH3COO (aq) + Fe (aq) → (CH3COO)3Fe red solution On heating, (CH3COO)3Fe + 2H2O → (CH3COO)Fe(OH)2 + 2CH3COOH brown precipitate
No visible change. On heating, white precipitate formed
Al3+ present
3CH3COO–(aq) + Al3+(aq) → (CH3COO)3Al colourless
3+
–
3+
(CH3COO)3Al + 2H2O → (CH3COO)Al(OH)2 + 2CH3COOH white precipitate
Table 9 Reactions of sodium carbonate solution Observation
Inference
Explanation 2– 3
+
→ H2O + CO2
Effervescence. Gas turns limewater milky
Acid, such as C6H5COOH or H2C2O4 or acid salt, such as HSO4– present
2H + CO
Pungent gas liberated. Gas turns red litmus blue
NH3 given off. NH4+ present
Na2CO3 is a base. A base will react with NH4+ to give NH3(g)
White precipitate formed
Ba2+, Ca2+, Mg2+, Pb2+ or Zn2+ present
These cations react with CO32–(aq) to form insoluble metal carbonates
White precipitate, turns brown Mn2+ present
MnCO3 precipitate is oxidised to Mn(III) salt
3+
Brown precipitate, gas liberated turns limewater milky
Fe present
Brown precipitate is Fe(OH)3. Fe2(CO3)3 is unstable and undergoes rapid hydrolysis to form Fe(OH)3. Fe3+(aq) is acidic and reacts with CO32– to give CO2
Green precipitate
Fe2+, Cr3+ or Ni2+ present
These cations react with CO32–(aq) to form insoluble metal carbonates
Table 10 Identification of functional groups in organic compounds Reagent
Observation
Inference
Explanation
Cl2 water
White precipitate
C6H5OH or C6H5NH2 present
Precipitation of 2,4,6–trichlorophenol or 2,4,6–trichlorophenylamine
Sodium chlorate(I), NaClO
Solution turns purple
C6H5NH2 present
Formation of organic complex
Br2 water
(a) Decolourisation
Unsaturated organic compound present
C = C + Br2 → –C – C– Br Br colourless
(b) Decolourisation and white precipitate formed
C6H5OH or C6H5NH2 present
Precipitation of 2,4,6–trichlorophenol or 2,4,6–trichlorophenylamine
NaNO2 in HCl at 5 °C followed by phenol
Red dye
C6H5NH2 present
Diazotisation followed by coupling reaction to form coloured dye
Alcohol + a few drops of concentrated H2SO4, then heat
Sweet-smelling odour
RCOOH present
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