A Level Chemistry H3 Practice Paper 1

A-Level Chemistry H3 Exam - Acids Bases Salts - Answer Key

Total Marks: 50

Section A: Multiple-Choice Questions (10 marks)

1. B. PO₄³⁻

   • Explanation: The conjugate base is formed by removing a proton (H⁺) from the acid. HPO₄²⁻ - H⁺ = PO₄³⁻.
   • Marking: 1 mark for correct answer.

2. B. 2.0 × 10⁻⁵

   • Explanation:
     1. [H⁺] = 10⁻²·⁸⁵ = 1.41 × 10⁻³ mol dm⁻³
     2. For a weak acid, [H⁺] = √(Ka × [HA]), so Ka = [H⁺]² / [HA]
     3. Ka = (1.41 × 10⁻³)² / 0.10 = 1.99 × 10⁻⁶ / 0.10 = 1.99 × 10⁻⁵ ≈ 2.0 × 10⁻⁵ mol dm⁻³
   • Marking: 1 mark for correct answer.

3. C. NH₄Cl

   • Explanation: NH₄Cl is the salt of a weak base (NH₃) and a strong acid (HCl). The ammonium ion (NH₄⁺) hydrolyses in water to produce H⁺ ions, making the solution acidic.
   • Marking: 1 mark for correct answer.

4. A. 0.050 mol dm⁻³

   • Explanation: For a strong base, [OH⁻] = concentration of base = 0.050 mol dm⁻³. The pH of 12.70 confirms this, as pOH = 14.00 - 12.70 = 1.30, and [OH⁻] = 10⁻¹·³⁰ = 0.050 mol dm⁻³.
   • Marking: 1 mark for correct answer.

5. C. The pH of a buffer solution is independent of the ratio of the concentrations of the weak acid and its conjugate base.

   • Explanation: The pH of a buffer does depend on the ratio of [conjugate base] to [weak acid], as shown by the Henderson-Hasselbalch equation: pH = pKa + log([A⁻]/[HA]).
   • Marking: 1 mark for correct answer.

6. B. 7.0

   • Explanation: Equal volumes and concentrations of a strong acid (HCl) and a strong base (NaOH) will neutralise each other completely, forming a neutral salt (NaCl) and water. The pH of a neutral solution at 25°C is 7.0.
   • Marking: 1 mark for correct answer.

7. A. 2.87

   • Explanation:
     1. [H⁺] = √(Ka × [HA]) = √(1.8 × 10⁻⁵ × 0.10) = √(1.8 × 10⁻⁶) = 1.34 × 10⁻³ mol dm⁻³
     2. pH = -log(1.34 × 10⁻³) = 2.87
   • Marking: 1 mark for correct answer.

8. A. Kw = [H⁺][OH⁻]

   • Explanation: The ionic product of water is the equilibrium constant for the self-ionisation of water: 2H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq). The expression is Kw = [H⁺][OH⁻].
   • Marking: 1 mark for correct answer.

9. B. 2.0 × 10⁻⁵

   • Explanation:
     1. pOH = 14.00 - 11.30 = 2.70
     2. [OH⁻] = 10⁻²·⁷⁰ = 2.00 × 10⁻³ mol dm⁻³
     3. For a weak base, [OH⁻] = √(Kb × [B]), so Kb = [OH⁻]² / [B]
     4. Kb = (2.00 × 10⁻³)² / 0.20 = 4.0 × 10⁻⁶ / 0.20 = 2.0 × 10⁻⁵ mol dm⁻³
   • Marking: 1 mark for correct answer.

10. C. KCl

    • Explanation: KCl is the salt of a strong acid (HCl) and a strong base (KOH). Neither the K⁺ nor the Cl⁻ ion hydrolyses appreciably in water, so the solution remains neutral.
    • Marking: 1 mark for correct answer.

Section B: Short-Answer Questions (20 marks)

11. (a) A Bronsted-Lowry acid is a proton (H⁺) donor. [1 mark]

(b) The conjugate acid-base pairs are: [2 marks] - NH₃ (base) and NH₄⁺ (conjugate acid) - H₂O (acid) and OH⁻ (conjugate base)

Marking: 1 mark for each correctly identified pair.

12. [3 marks] 1. [H⁺] = 10⁻²·⁹⁵ = 1.12 × 10⁻³ mol dm⁻³ 2. For a weak acid, HA ⇌ H⁺ + A⁻, so [H⁺] = [A⁻] = 1.12 × 10⁻³ mol dm⁻³ 3. Ka = [H⁺][A⁻] / [HA] = (1.12 × 10⁻³)² / (0.15 - 1.12 × 10⁻³) 4. Assuming [HA] ≈ 0.15 (since dissociation is small): Ka ≈ (1.25 × 10⁻⁶) / 0.15 = 8.3 × 10⁻⁶ mol dm⁻³

Marking: 1 mark for calculating [H⁺], 1 mark for correct Ka expression, 1 mark for final answer.

13. (a) pH = pKa + log([A⁻]/[HA]) [1 mark]

(b) [4 marks] 1. Moles of ethanoic acid = (50.0/1000) × 0.20 = 0.010 mol 2. Moles of sodium ethanoate = (30.0/1000) × 0.10 = 0.0030 mol 3. Total volume = 50.0 + 30.0 = 80.0 cm³ = 0.080 dm³ 4. [HA] = 0.010 / 0.080 = 0.125 mol dm⁻³ 5. [A⁻] = 0.0030 / 0.080 = 0.0375 mol dm⁻³ 6. pKa = -log(1.8 × 10⁻⁵) = 4.74 7. pH = 4.74 + log(0.0375 / 0.125) = 4.74 + log(0.30) = 4.74 + (-0.52) = 4.22

Marking: 1 mark for calculating moles, 1 mark for calculating concentrations, 1 mark for pKa, 1 mark for final pH.

14. [3 marks] - Ammonium chloride (NH₄Cl) is a salt formed from a weak base (NH₃) and a strong acid (HCl). - In water, the ammonium ion (NH₄⁺) acts as a weak acid and donates a proton to water: NH₄⁺(aq) + H₂O(l) ⇌ NH₃(aq) + H₃O⁺(aq). - The chloride ion (Cl⁻) is the conjugate base of a strong acid and does not hydrolyse. - The production of H₃O⁺ ions results in an acidic solution (pH < 7).

Marking: 1 mark for identifying the salt type, 1 mark for the hydrolysis equation, 1 mark for explaining the production of H⁺.

15. [2 marks] 1. [OH⁻] = 0.010 mol / 1.0 dm³ = 0.010 mol dm⁻³ 2. pOH = -log(0.010) = 2.00 3. pH = 14.00 - 2.00 = 12.00

Marking: 1 mark for calculating [OH⁻], 1 mark for final pH.

16. (a) The equivalence point should be labelled at the centre of the steep vertical section of the curve, at approximately pH 8.5. [1 mark]

(b) [2 marks] - The salt formed at the equivalence point is the conjugate base of the weak acid (e.g., CH₃COONa). - This salt hydrolyses in water to produce OH⁻ ions: CH₃COO⁻(aq) + H₂O(l) ⇌ CH₃COOH(aq) + OH⁻(aq). - The presence of OH⁻ ions makes the solution basic (pH > 7).

Marking: 1 mark for identifying the salt hydrolysis, 1 mark for explaining the production of OH⁻.

(c) At the half-equivalence point, [HA] = [A⁻], so pH = pKa. Therefore, the Ka of the weak acid can be determined by reading the pH at the half-equivalence point. [1 mark]

Section C: Structured Questions (20 marks)

17. (a) [3 marks] - Measure the pH of each solution using a pH meter. - The solution with pH 1.0 is HCl (strong acid). - The solution with pH around 2.9 is CH₃COOH (weak acid). - The solution with pH 13.0 is NaOH (strong base).

Marking: 1 mark for measuring pH, 1 mark for identifying the strong acid, 1 mark for identifying the weak acid and strong base.

(b) [3 marks] - HCl is a strong acid, fully dissociated, so [H⁺] = 0.10 mol dm⁻³, pH = 1.0. - CH₃COOH is a weak acid, partially dissociated, so [H⁺] < 0.10 mol dm⁻³, pH ≈ 2.9. - NaOH is a strong base, fully dissociated, so [OH⁻] = 0.10 mol dm⁻³, pOH = 1.0, pH = 13.0.

Marking: 1 mark for each correct explanation.

18. (a) [3 marks] 1. [HA] = 0.25 mol / 1.0 dm³ = 0.25 mol dm⁻³ 2. [A⁻] = 0.15 mol / 1.0 dm³ = 0.15 mol dm⁻³ 3. pKa = -log(1.3 × 10⁻⁵) = 4.89 4. pH = 4.89 + log(0.15 / 0.25) = 4.89 + log(0.60) = 4.89 + (-0.22) = 4.67

Marking: 1 mark for calculating concentrations, 1 mark for pKa, 1 mark for final pH.

(b) [4 marks] 1. Initial moles of HA = 0.25 mol, initial moles of A⁻ = 0.15 mol. 2. Added HCl (0.020 mol) reacts with A⁻: A⁻ + H⁺ → HA. 3. Moles of HA after addition = 0.25 + 0.020 = 0.27 mol. 4. Moles of A⁻ after addition = 0.15 - 0.020 = 0.13 mol. 5. Volume remains 1.0 dm³, so [HA] = 0.27 mol dm⁻³, [A⁻] = 0.13 mol dm⁻³. 6. pH = 4.89 + log(0.13 / 0.27) = 4.89 + log(0.481) = 4.89 + (-0.32) = 4.57

Marking: 1 mark for calculating new moles, 1 mark for calculating new concentrations, 1 mark for correct pH calculation, 1 mark for final answer.

19. (a) [3 marks] 1. [H⁺] = √(Ka × [HA]) = √(4.5 × 10⁻⁶ × 0.20) = √(9.0 × 10⁻⁷) = 9.49 × 10⁻⁴ mol dm⁻³ 2. pH = -log(9.49 × 10⁻⁴) = 3.02

Marking: 1 mark for correct formula, 1 mark for calculating [H⁺], 1 mark for final pH.

(b) [4 marks] 1. [H⁺] = 10⁻³·⁵⁰ = 3.16 × 10⁻⁴ mol dm⁻³ 2. For a weak acid, [H⁺] = √(Ka × [HA]), so [HA] = [H⁺]² / Ka 3. [HA] = (3.16 × 10⁻⁴)² / (4.5 × 10⁻⁶) = (1.0 × 10⁻⁷) / (4.5 × 10⁻⁶) = 0.022 mol dm⁻³

Marking: 1 mark for calculating [H⁺], 1 mark for correct formula, 1 mark for calculating [HA], 1 mark for final answer.

20. (a) Kw = [H⁺][OH⁻] [1 mark]

(b) [2 marks] - The self-ionisation of water is an endothermic process: 2H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq) ΔH > 0. - According to Le Chatelier's principle, increasing the temperature favours the endothermic reaction, shifting the equilibrium to the right. - This increases the concentrations of H⁺ and OH⁻ ions, thus increasing the value of Kw.

Marking: 1 mark for identifying the reaction as endothermic, 1 mark for explaining the shift in equilibrium.

(c) [3 marks] 1. In pure water, [H⁺] = [OH⁻]. 2. Kw = [H⁺]², so [H⁺] = √Kw = √(5.5 × 10⁻¹⁴) = 2.35 × 10⁻⁷ mol dm⁻³ 3. pH = -log(2.35 × 10⁻⁷) = 6.63

Marking: 1 mark for recognising [H⁺] = [OH⁻], 1 mark for calculating [H⁺], 1 mark for final pH.

END OF ANSWER KEY