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A Level H2 Chemistry Practice Paper 2
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TuitionGoWhere Practice Paper - Chemistry H2 A-Level
Answer Key and Marking Scheme (Version 2)
Section A: Structured Questions
Question 1: Acid-Base Definitions and pH
(a) Define a Brønsted-Lowry acid and a Brønsted-Lowry base. [2 marks]
Answer:
- Brønsted-Lowry acid: a proton (H⁺) donor. [1 mark]
- Brønsted-Lowry base: a proton (H⁺) acceptor. [1 mark]
(b) Calculate the pH of pure water at 298 K. [2 marks]
Answer: K_w = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ In pure water, [H⁺] = [OH⁻] = √(1.0 × 10⁻¹⁴) = 1.0 × 10⁻⁷ mol dm⁻³ [1 mark] pH = −log₁₀(1.0 × 10⁻⁷) = 7.00 [1 mark]
(c) Explain why pure water remains neutral at this temperature despite the pH being less than 7. [2 marks]
Answer: At 313 K, K_w is larger than at 298 K because the autoionisation of water is endothermic. [1 mark] Neutrality is defined by [H⁺] = [OH⁻], not by pH = 7. Since [H⁺] = [OH⁻] in pure water at any temperature, the water remains neutral. [1 mark]
Question 2: Buffer Solutions
(a) Write an expression for the K_a of ethanoic acid. [1 mark]
Answer: K_a = [H⁺][CH₃COO⁻] / [CH₃COOH] [1 mark]
(b) Calculate the pH of this buffer solution. [3 marks]
Answer: [CH₃COOH] = 0.0500 mol dm⁻³; [CH₃COO⁻] = 0.0500 mol dm⁻³ K_a = [H⁺][CH₃COO⁻] / [CH₃COOH] → [H⁺] = K_a × [CH₃COOH] / [CH₃COO⁻] [1 mark] [H⁺] = 1.8 × 10⁻⁵ × (0.0500 / 0.0500) = 1.8 × 10⁻⁵ mol dm⁻³ [1 mark] pH = −log₁₀(1.8 × 10⁻⁵) = 4.74 [1 mark]
(c) Explain, with the aid of an equation, how the buffer solution resists a change in pH. [2 marks]
Answer: Added H⁺ reacts with the conjugate base: CH₃COO⁻ + H⁺ → CH₃COOH [1 mark] This removes added H⁺ from solution, so [H⁺] and pH remain approximately constant. [1 mark]
(d) Calculate the new pH after adding 0.0050 mol of HCl to 1.00 dm³ of buffer. [2 marks]
Answer: Added H⁺ reacts: CH₃COO⁻ + H⁺ → CH₃COOH New [CH₃COO⁻] = 0.0500 − 0.0050 = 0.0450 mol dm⁻³ New [CH₃COOH] = 0.0500 + 0.0050 = 0.0550 mol dm⁻³ [1 mark] [H⁺] = 1.8 × 10⁻⁵ × (0.0550 / 0.0450) = 2.20 × 10⁻⁵ mol dm⁻³ pH = −log₁₀(2.20 × 10⁻⁵) = 4.66 [1 mark] Assumption: volume change is negligible; all added H⁺ reacts with CH₃COO⁻.
Question 3: Titration Curves and Indicators
(a) Write an equation for the reaction. [1 mark]
Answer: NH₃(aq) + HCl(aq) → NH₄Cl(aq) [or NH₃ + H⁺ → NH₄⁺] [1 mark]
(b) Calculate the pH at the equivalence point. Explain why the pH is not 7. [4 marks]
Answer: At equivalence, all NH₃ has been converted to NH₄⁺. Volume at equivalence = 25.0 cm³ (equal concentrations, equal volumes). Total volume = 50.0 cm³. [NH₄⁺] = (0.100 × 0.0250) / 0.0500 = 0.0500 mol dm⁻³ [1 mark] NH₄⁺ is a weak acid: K_a(NH₄⁺) = K_w / K_b(NH₃) = 1.0 × 10⁻¹⁴ / 1.8 × 10⁻⁵ = 5.56 × 10⁻¹⁰ mol dm⁻³ [1 mark] [H⁺] = √(K_a × c) = √(5.56 × 10⁻¹⁰ × 0.0500) = 5.27 × 10⁻⁶ mol dm⁻³ pH = 5.28 [1 mark] The pH is less than 7 because NH₄⁺ undergoes hydrolysis, producing H⁺: NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺. [1 mark]
(c) Select the most suitable indicator. Explain. [2 marks]
Answer: Methyl orange (pH range 3.1–4.4). [1 mark] The equivalence point pH (5.28) falls within the steep portion of the titration curve, and methyl orange changes colour within the pH range of the near-vertical rise. [1 mark]
(d) State the colour change at the end point. [1 mark]
Answer: Yellow to red (or orange-red). [1 mark]
Question 4: Solubility Product
(a) Write the K_sp expression for AgCl, stating its units. [2 marks]
Answer: AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq) K_sp = [Ag⁺][Cl⁻] [1 mark] Units: mol² dm⁻⁶ [1 mark]
(b) Calculate the solubility of AgCl in water. [3 marks]
Answer: Let solubility = s mol dm⁻³. Then [Ag⁺] = s, [Cl⁻] = s. K_sp = s² = 1.8 × 10⁻¹⁰ [1 mark] s = √(1.8 × 10⁻¹⁰) = 1.34 × 10⁻⁵ mol dm⁻³ [1 mark] M_r(AgCl) = 143.4; solubility in g dm⁻³ = 1.34 × 10⁻⁵ × 143.4 = 1.92 × 10⁻³ g dm⁻³ [1 mark]
(c) Calculate the minimum [Ag⁺] to initiate precipitation. [2 marks]
Answer: Precipitation occurs when [Ag⁺][Cl⁻] ≥ K_sp. [Ag⁺] = K_sp / [Cl⁻] = 1.8 × 10⁻¹⁰ / 0.010 = 1.8 × 10⁻⁸ mol dm⁻³ [2 marks]
(d) State and explain whether AgCl is more or less soluble in 0.10 mol dm⁻³ NaCl(aq). [1 mark]
Answer: Less soluble. The common ion effect: the presence of Cl⁻ from NaCl shifts the equilibrium AgCl(s) ⇌ Ag⁺ + Cl⁻ to the left, reducing solubility. [1 mark]
Section B: Data-Based and Applied Questions
Question 5: Acid Rain and Environmental Chemistry
(a) Write equations for the two dissociations of H₂SO₃. Identify conjugate pairs. [3 marks]
Answer: First dissociation: H₂SO₃ + H₂O ⇌ H₃O⁺ + HSO₃⁻ Conjugate acid-base pairs: H₂SO₃/HSO₃⁻ and H₃O⁺/H₂O [1 mark for equation, 1 mark for pairs] Second dissociation: HSO₃⁻ + H₂O ⇌ H₃O⁺ + SO₃²⁻ [1 mark]
(b) Calculate [H⁺] in acid rain of pH 4.20. [1 mark]
Answer: [H⁺] = 10⁻⁴·²⁰ = 6.31 × 10⁻⁵ mol dm⁻³ [1 mark]
(c) Calculate total acid concentration from titration data. [3 marks]
Answer: n(NaOH) = 0.0200 × 0.0185 = 3.70 × 10⁻⁴ mol [1 mark] Since both H₂SO₄ and HNO₃ are fully neutralised, n(H⁺) = n(OH⁻) = 3.70 × 10⁻⁴ mol [1 mark] [H⁺] = 3.70 × 10⁻⁴ / 0.0500 = 7.40 × 10⁻³ mol dm⁻³ [1 mark]
(d) Compare answers to (b) and (c). Suggest a reason for any difference. [2 marks]
Answer: The titration gives [H⁺] = 7.40 × 10⁻³ mol dm⁻³, while pH measurement gives 6.31 × 10⁻⁵ mol dm⁻³. The titration value is much larger. [1 mark] This suggests the acid rain contains weak acids (e.g., H₂SO₃, CO₂/H₂CO₃) that are not fully dissociated, so the total titratable acidity exceeds the free [H⁺] measured by pH. [1 mark]
(e) Write an ionic equation for CaCO₃ neutralising acid. Explain restoration of aquatic life. [3 marks]
Answer: CaCO₃(s) + 2H⁺(aq) → Ca²⁺(aq) + CO₂(g) + H₂O(l) [1 mark] This reaction removes H⁺ ions, raising the pH of the lake water. [1 mark] Aquatic organisms (fish, insects, plankton) are sensitive to low pH; raising the pH restores conditions suitable for their survival and reproduction. [1 mark]
Question 6: Polyprotic Acids and Fractional Composition
(a) Write equations for the three dissociations of H₃PO₄. [3 marks]
Answer: H₃PO₄ + H₂O ⇌ H₃O⁺ + H₂PO₄⁻ (K_a1) [1 mark] H₂PO₄⁻ + H₂O ⇌ H₃O⁺ + HPO₄²⁻ (K_a2) [1 mark] HPO₄²⁻ + H₂O ⇌ H₃O⁺ + PO₄³⁻ (K_a3) [1 mark]
(b) Explain why K_a1 > K_a2 > K_a3. [2 marks]
Answer: Each successive proton is removed from a more negatively charged species. [1 mark] The increasing negative charge makes it progressively more difficult to remove a positively charged proton due to stronger electrostatic attraction, so each successive K_a is smaller. [1 mark]
(c)(i) Show H₂PO₄⁻ as both acid and base. [2 marks]
Answer: As an acid: H₂PO₄⁻ + H₂O ⇌ HPO₄²⁻ + H₃O⁺ [1 mark] As a base: H₂PO₄⁻ + H₂O ⇌ H₃PO₄ + OH⁻ [1 mark]
(c)(ii) Explain why pH ≈ ½(pK_a1 + pK_a2). [3 marks]
Answer: H₂PO₄⁻ is an amphiprotic species. Its pH is given approximately by pH = ½(pK_a1 + pK_a2). [1 mark] pK_a1 = −log(7.1 × 10⁻³) = 2.15; pK_a2 = −log(6.3 × 10⁻⁸) = 7.20 [1 mark] pH = ½(2.15 + 7.20) = 4.68 ≈ 4.7, which matches the observed value. [1 mark]
(d) Calculate mass of Na₂HPO₄ needed. [3 marks]
Answer: Using Henderson-Hasselbalch: pH = pK_a2 + log([HPO₄²⁻]/[H₂PO₄⁻]) 7.40 = 7.20 + log([HPO₄²⁻]/[H₂PO₄⁻]) log([HPO₄²⁻]/[H₂PO₄⁻]) = 0.20 → [HPO₄²⁻]/[H₂PO₄⁻] = 10⁰·²⁰ = 1.58 [1 mark] [H₂PO₄⁻] = 0.100 mol dm⁻³, so [HPO₄²⁻] = 1.58 × 0.100 = 0.158 mol dm⁻³ [1 mark] n(HPO₄²⁻) in 500 cm³ = 0.158 × 0.500 = 0.0790 mol Mass = 0.0790 × 142.0 = 11.2 g [1 mark]
Section C: Free-Response Questions
Question 7: Acid-Base Theories and Comparative Analysis
(a) Compare and contrast Arrhenius, Brønsted-Lowry, and Lewis theories. [6 marks]
Answer: Arrhenius theory:
- Acid: produces H⁺ in water; Base: produces OH⁻ in water.
- Advantage: simple, explains neutralisation (H⁺ + OH⁻ → H₂O).
- Limitation: restricted to aqueous solutions; cannot explain basicity of NH₃ or acidity of CO₂.
Brønsted-Lowry theory:
- Acid: proton donor; Base: proton acceptor.
- Advantage: broader scope; explains acid-base behaviour in non-aqueous solvents and gases; introduces conjugate pairs.
- Limitation: requires proton transfer; cannot explain Lewis acid-base reactions (e.g., BF₃ + NH₃).
Lewis theory:
- Acid: electron pair acceptor; Base: electron pair donor.
- Advantage: most general; explains reactions of metal ions, electron-deficient molecules (BF₃, AlCl₃).
- Limitation: so broad that many reactions are classified as acid-base; less useful for predicting strength trends.
Award marks for clear comparison addressing all three theories with at least one advantage and one limitation each. [up to 6 marks]
(b) Explain why AlCl₃ is a Lewis acid, with an equation. [2 marks]
Answer: AlCl₃ has an incomplete octet on Al (only 6 valence electrons), making it electron-deficient. [1 mark] It accepts an electron pair from a Lewis base, e.g.: AlCl₃ + Cl⁻ → AlCl₄⁻ [1 mark]
(c) Explain why BF₃ is a Lewis acid while NF₃ is a Lewis base. [2 marks]
Answer: BF₃: Boron has only 6 valence electrons (incomplete octet), so it can accept an electron pair → Lewis acid. [1 mark] NF₃: Nitrogen has a complete octet with a lone pair, which it can donate → Lewis base. [1 mark]
Question 8: Buffer Design and Evaluation
(a) Explain why NH₃/NH₄Cl is suitable for pH 9.25. [2 marks]
Answer: pK_a of NH₄⁺ = 14 − pK_b = 14 − 4.74 = 9.26. [1 mark] A buffer is most effective when pH ≈ pK_a, so the NH₃/NH₄⁺ system is ideal for pH 9.25. [1 mark]
(b) Calculate the ratio [NH₄⁺]/[NH₃] for pH 9.25. [3 marks]
Answer: pOH = 14 − 9.25 = 4.75 pOH = pK_b + log([NH₄⁺]/[NH₃]) 4.75 = 4.74 + log([NH₄⁺]/[NH₃]) [1 mark] log([NH₄⁺]/[NH₃]) = 0.01 [1 mark] [NH₄⁺]/[NH₃] = 10⁰·⁰¹ = 1.02 ≈ 1.0 [1 mark]
(c) Describe preparation of 250 cm³ of buffer. [5 marks]
Answer: Calculation: For pH 9.25, [NH₄⁺]/[NH₃] ≈ 1.0. Using 0.10 mol dm⁻³ NH₃: If [NH₃] = 0.10 mol dm⁻³, then [NH₄⁺] = 0.10 mol dm⁻³. n(NH₄Cl) needed in 250 cm³ = 0.10 × 0.250 = 0.0250 mol Mass of NH₄Cl = 0.0250 × 53.5 = 1.34 g [2 marks for correct calculations]
Procedure:
- Measure 125 cm³ of 0.10 mol dm⁻³ NH₃(aq) using a measuring cylinder or pipette and transfer to a 250 cm³ volumetric flask. [1 mark]
- Weigh 1.34 g of solid NH₄Cl accurately on a balance. Transfer quantitatively to the volumetric flask. [1 mark]
- Add distilled water, swirl to dissolve, and make up to the 250 cm³ mark with distilled water. Stopper and invert several times to mix thoroughly. [1 mark]
- Check pH with a calibrated pH meter and adjust if necessary by adding small amounts of dilute HCl or NaOH.
Alternative method using NH₃ and HCl:
- Measure 250 cm³ of 0.10 mol dm⁻³ NH₃. Add 125 cm³ of 0.10 mol dm⁻³ HCl to convert half the NH₃ to NH₄⁺, producing a 1:1 buffer. Dilute to appropriate volume if needed.
Award marks for a clear, practical procedure with correct calculations. [up to 5 marks]
END OF ANSWER KEY
Total: 75 marks