A Level H2 Chemistry Practice Paper 4

TuitionGoWhere Practice Paper - Chemistry H2 A-Level

Answer Key and Marking Scheme

Version 4 of 5

Section A: Multiple Choice and Short Structured Questions

1. C. It is a proton donor. [1]

Explanation: A Brønsted–Lowry acid is defined as a proton (H⁺) donor. Option A describes a Brønsted–Lowry base. Options B and D describe Lewis acids and bases respectively.

2. B. 2.00 [1]

Explanation: For a strong monoprotic acid, [H⁺] = concentration of acid = 0.010 mol dm⁻³. pH = −log₁₀(0.010) = 2.00.

3. C. H₂PO₄⁻ and HPO₄²⁻ [1]

Explanation: A conjugate acid–base pair differs by one proton (H⁺). H₂PO₄⁻ (acid) donates a proton to form HPO₄²⁻ (conjugate base). In A, H₃O⁺ and OH⁻ differ by 2H⁺ and 1O. In B, HCl and NaOH are not a conjugate pair. In D, H₂SO₄ and SO₄²⁻ differ by 2H⁺.

4. A. The ethanoate ions react with H⁺ to form ethanoic acid. [1]

Explanation: In the ethanoic acid/ethanoate buffer, the ethanoate ions (CH₃COO⁻) are the basic component that neutralises added H⁺: CH₃COO⁻ + H⁺ → CH₃COOH. This removes the added H⁺ and minimises pH change.

5. B. 7.00 [1]

Explanation: In pure water, [H⁺] = [OH⁻]. Kw = [H⁺][OH⁻] = [H⁺]² = 1.0 × 10⁻¹⁴. Therefore [H⁺] = 1.0 × 10⁻⁷ mol dm⁻³, and pH = 7.00.

6.

(a) Titrations 1, 2, and 3 are concordant. [1] The volumes are 24.10, 23.90, and 24.00 cm³. The range is 24.10 − 23.90 = 0.20 cm³, which is within the acceptable concordancy range of ≤0.20 cm³. The rough titration (24.50 cm³) is excluded as it is an approximation. [1]

(b) Mean titre = (24.10 + 23.90 + 24.00) ÷ 3 = 24.00 cm³. [1]

(c) n(HCl) = c × V = 0.100 × (24.00/1000) = 0.00240 mol. [1] NaOH + HCl → NaCl + H₂O; mole ratio 1:1. n(NaOH) = 0.00240 mol in 25.0 cm³. [NaOH] = 0.00240 ÷ (25.0/1000) = 0.0960 mol dm⁻³. [1]

7.

(a) Kb = [NH₄⁺][OH⁻] / [NH₃] [1]

(b) NH₃ + H₂O ⇌ NH₄⁺ + OH⁻ Let [OH⁻] = x. Kb = x² / (0.200 − x) ≈ x² / 0.200 = 1.8 × 10⁻⁵. [1] x = √(1.8 × 10⁻⁵ × 0.200) = √(3.6 × 10⁻⁶) = 1.90 × 10⁻³ mol dm⁻³. [1] pOH = −log₁₀(1.90 × 10⁻³) = 2.72. pH = 14.00 − 2.72 = 11.28 (or 11.3). [1]

Accept pH 11.3. Marks for: correct Kb expression substitution, correct [OH⁻] calculation, correct pH.

8.

(a) n(sodium ethanoate) = mass / Mr = 4.10 / 82.0 = 0.0500 mol. [1] Concentration = n / V = 0.0500 / (250/1000) = 0.200 mol dm⁻³. [1]

(b) pH = pKa + log₁₀([CH₃COO⁻] / [CH₃COOH]) pKa = −log₁₀(1.8 × 10⁻⁵) = 4.74. [1] pH = 4.74 + log₁₀(0.200 / 0.200) = 4.74 + log₁₀(1) = 4.74. [1]

Accept 4.74 or 4.7.

Section B: Data Interpretation and Structured Questions

9.

(a) HCl is a strong acid and dissociates completely in water: HCl → H⁺ + Cl⁻. [H⁺] = 0.100 mol dm⁻³, so pH = 1.0. [1] CH₃COOH is a weak acid and dissociates partially: CH₃COOH ⇌ H⁺ + CH₃COO⁻. [H⁺] < 0.100 mol dm⁻³, so pH > 1.0 (specifically 2.9). [1]

(b) [H⁺] = 10⁻²·⁹ = 1.26 × 10⁻³ mol dm⁻³. [1]

Accept 1.3 × 10⁻³ mol dm⁻³.

(c) NH₃ + H₂O ⇌ NH₄⁺ + OH⁻ [1] Ammonia is a weak base; it only partially dissociates in water, producing a lower concentration of OH⁻ ions than a strong base of the same concentration. Therefore, [OH⁻] < 0.100 mol dm⁻³, pOH > 1.0, and pH < 13.0. [1]

(d) The resulting solution will have pH = 7.0 (neutral). [1] HCl and NaOH react in a 1:1 mole ratio: HCl + NaOH → NaCl + H₂O. Since equal volumes of equal concentrations are mixed, the amounts are exactly stoichiometric. The products are NaCl (a neutral salt) and water, so the solution is neutral. [1]

10.

(a) n(HA) = 1.20 / 60.0 = 0.0200 mol. [1] [HA] = 0.0200 / (250/1000) = 0.0800 mol dm⁻³. [1]

(b) [H⁺] = 10⁻²·⁸⁵ = 1.41 × 10⁻³ mol dm⁻³. [1]

Accept 1.4 × 10⁻³ mol dm⁻³.

(c) HA ⇌ H⁺ + A⁻. Ka = [H⁺][A⁻] / [HA]. [H⁺] = [A⁻] = 1.41 × 10⁻³ mol dm⁻³. [HA] at equilibrium ≈ 0.0800 − 1.41 × 10⁻³ ≈ 0.0786 mol dm⁻³. [1] Ka = (1.41 × 10⁻³)² / 0.0786 = 2.53 × 10⁻⁵ mol dm⁻³. [1]

Accept 2.5 × 10⁻⁵ mol dm⁻³. Award marks for correct substitution and calculation.

(d) The pH would be less than 3.85. [1] For a weak acid, dilution by a factor of 10 does not increase pH by exactly 1 unit because the degree of dissociation increases upon dilution (Le Chatelier's principle). More of the acid dissociates, so [H⁺] is greater than one-tenth of the original, resulting in a pH increase of less than 1 unit. [1]

11.

(a) Before the equivalence point, the solution contains a mixture of unreacted weak base (NH₃) and its conjugate acid (NH₄⁺) formed from the reaction with added HCl. This mixture acts as a buffer solution. [1] The pH decreases gradually because the buffer resists changes in pH as acid is added. The pH is governed by the Henderson–Hasselbalch equation for a basic buffer. [1]

(b) A suitable indicator is methyl red (pH range 4.2–6.3) or bromocresol green (pH range 3.8–5.4). [1] The equivalence point of a weak base–strong acid titration occurs at pH < 7 (approximately pH 5 for this titration). The indicator must change colour within the steep portion of the curve, which includes the equivalence point. Methyl red or bromocresol green have pH ranges that fall within this steep region. Phenolphthalein would be unsuitable as it changes colour at pH 8.2–10.0, which is before the equivalence point. [1]

(c) At the half-equivalence point, [NH₄⁺] = [NH₃], so pH = pKa of NH₄⁺. [1] pKa = 9.3, so Ka(NH₄⁺) = 10⁻⁹·³ = 5.01 × 10⁻¹⁰ mol dm⁻³. Kb(NH₃) = Kw / Ka(NH₄⁺) = 1.0 × 10⁻¹⁴ / 5.01 × 10⁻¹⁰ = 2.00 × 10⁻⁵ mol dm⁻³. [1]

Accept 2.0 × 10⁻⁵ mol dm⁻³.

Section C: Extended Structured Questions

12.

(a)(i) Ksp = [Ag⁺(aq)][Cl⁻(aq)] [1]

(a)(ii) Let solubility = s mol dm⁻³. AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq). [Ag⁺] = s, [Cl⁻] = s. Ksp = s² = 1.8 × 10⁻¹⁰. [1] s = √(1.8 × 10⁻¹⁰) = 1.34 × 10⁻⁵ mol dm⁻³. [1]

Accept 1.3 × 10⁻⁵ mol dm⁻³.

(a)(iii) In 500 cm³ (0.500 dm³): n(AgCl) = 1.34 × 10⁻⁵ × 0.500 = 6.70 × 10⁻⁶ mol. [1] Mass = n × Mr = 6.70 × 10⁻⁶ × 143.5 = 9.61 × 10⁻⁴ g (or 0.961 mg). [1]

Accept 9.6 × 10⁻⁴ g.

(b)(i) Total volume after mixing = 100.0 cm³. [Ag⁺] = (0.0100 × 50.0/1000) / (100.0/1000) = 5.00 × 10⁻³ mol dm⁻³. [1] [Cl⁻] = (0.0100 × 50.0/1000) / (100.0/1000) = 5.00 × 10⁻³ mol dm⁻³. [1]

(b)(ii) Ionic product = [Ag⁺][Cl⁻] = (5.00 × 10⁻³) × (5.00 × 10⁻³) = 2.50 × 10⁻⁵. [1] Since ionic product (2.50 × 10⁻⁵) > Ksp (1.8 × 10⁻¹⁰), a precipitate of AgCl will form. [1]

13.

(a)(i) Ka = [H⁺][HCO₃⁻] / [H₂CO₃] [1]

(a)(ii) pH = pKa + log₁₀([HCO₃⁻] / [H₂CO₃]) 7.4 = 6.1 + log₁₀([HCO₃⁻] / [H₂CO₃]) [1] log₁₀([HCO₃⁻] / [H₂CO₃]) = 1.3 [HCO₃⁻] / [H₂CO₃] = 10¹·³ = 20 (or 19.95). [1]

Accept 20:1 or 20.

(b)(i) pKa = −log₁₀(1.8 × 10⁻⁵) = 4.74. [1]

Accept 4.74 or 4.7.

(b)(ii) pH = pKa + log₁₀([CH₃COO⁻] / [CH₃COOH]) 4.50 = 4.74 + log₁₀([CH₃COO⁻] / [CH₃COOH]) [1] log₁₀([CH₃COO⁻] / [CH₃COOH]) = −0.24 [CH₃COO⁻] / [CH₃COOH] = 10⁻⁰·²⁴ = 0.575 (or 0.58). [1]

Accept 0.58:1 or 0.58.

(b)(iii) Using the buffer solution: [CH₃COOH] = 0.500 mol dm⁻³ (given). [CH₃COO⁻] required = 0.575 × 0.500 = 0.288 mol dm⁻³. [1] For 250 cm³: n(CH₃COO⁻) = 0.288 × 0.250 = 0.0720 mol. Mass of sodium ethanoate = 0.0720 × 82.0 = 5.90 g. [1] Method: Dissolve 5.90 g of solid sodium ethanoate in approximately 200 cm³ of 0.500 mol dm⁻³ ethanoic acid in a 250 cm³ volumetric flask. Stir to dissolve completely, then make up to the mark with more 0.500 mol dm⁻³ ethanoic acid. Mix thoroughly. [1]

Accept alternative valid preparation methods. Marks for: correct [CH₃COO⁻] calculation, correct mass, valid preparation description.

14.

(a)(i) Mg(OH)₂ + 2HCl → MgCl₂ + 2H₂O [1] HCl + NaOH → NaCl + H₂O [1]

(a)(ii) n(HCl) originally = c × V = 0.500 × (50.0/1000) = 0.0250 mol. [1]

(a)(iii) n(NaOH) used in titration = 0.100 × (18.50/1000) = 0.00185 mol. This neutralised the HCl in 25.0 cm³ of the 250 cm³ solution. [1] n(HCl) remaining in 250 cm³ = 0.00185 × (250/25.0) = 0.0185 mol. [1]

(a)(iv) n(HCl) reacted with Mg(OH)₂ = 0.0250 − 0.0185 = 0.00650 mol. [1] From equation, n(Mg(OH)₂) = ½ × n(HCl) reacted = 0.00650 / 2 = 0.00325 mol. [1] Mass of pure Mg(OH)₂ = 0.00325 × 58.3 = 0.1895 g. Percentage purity = (0.1895 / 0.500) × 100 = 37.9%. [1]

Accept 37.9% or 38%. Marks for: correct HCl reacted, correct Mg(OH)₂ moles, correct percentage.

(b) Magnesium hydroxide is insoluble in water, so it cannot be titrated directly with a standard acid using an indicator. [1] In a back titration, the sample is reacted with a known excess of acid, which dissolves the Mg(OH)₂. The unreacted acid is then titrated with standard alkali. This allows the amount of Mg(OH)₂ to be determined indirectly. [1]

END OF ANSWER KEY