A Level H2 Chemistry Practice Paper 5

TuitionGoWhere Practice Paper - Chemistry H2 A-Level

Answer Key and Marking Scheme

Subject: Chemistry H2 Level: A-Level Paper: Practice Paper 5 (Acids, Bases & Salts) Version: 5

Section A: Multiple Choice & Short Structured Questions

1. B – H₂PO₄⁻ can donate a proton to form HPO₄²⁻ (acting as an acid) and can accept a proton to form H₃PO₄ (acting as a base). [1 mark]

2.

• pH = 2.90, so [H⁺] = 10⁻²·⁹⁰ = 1.26 × 10⁻³ mol dm⁻³ [1 mark]
• Kₐ = [H⁺]² / [HA] = (1.26 × 10⁻³)² / 0.10 = 1.59 × 10⁻⁵ mol dm⁻³ [1 mark]
• Accept 1.6 × 10⁻⁵ mol dm⁻³ (2 s.f.)

3.

• NH₄Cl is a salt of a weak base (NH₃) and a strong acid (HCl). [1 mark]
• NH₄⁺ undergoes hydrolysis: NH₄⁺(aq) + H₂O(l) ⇌ NH₃(aq) + H₃O⁺(aq), producing H₃O⁺ ions, making the solution acidic. [1 mark]
• NaCl is a salt of a strong acid (HCl) and a strong base (NaOH). Neither Na⁺ nor Cl⁻ undergoes hydrolysis, so [H⁺] = [OH⁻] and the solution is neutral. [1 mark]

4. (a) CH₃COOH(aq) + NaOH(aq) → CH₃COONa(aq) + H₂O(l) [1 mark]

(b)

• Moles of CH₃COOH initially = 0.20 × 0.0500 = 0.0100 mol [1 mark]
• Moles of NaOH added = 0.10 × 0.0250 = 0.00250 mol
• After reaction: moles of CH₃COOH remaining = 0.0100 − 0.00250 = 0.00750 mol
• Moles of CH₃COO⁻ formed = 0.00250 mol
• Total volume = 75.0 cm³ = 0.0750 dm³ [1 mark]
• [CH₃COOH] = 0.00750 / 0.0750 = 0.100 mol dm⁻³
• [CH₃COO⁻] = 0.00250 / 0.0750 = 0.0333 mol dm⁻³
• [H⁺] = Kₐ × [CH₃COOH] / [CH₃COO⁻] = (1.8 × 10⁻⁵) × 0.100 / 0.0333 = 5.41 × 10⁻⁵ mol dm⁻³
• pH = −log(5.41 × 10⁻⁵) = 4.27 [1 mark]

5. (a) HCl is a monoprotic strong acid, fully dissociating to give one H⁺ per molecule. H₂SO₄ is a diprotic strong acid, fully dissociating in the first step and partially in the second, giving more H⁺ ions per molecule. Hence, at the same concentration, H₂SO₄ has a lower pH. [2 marks]

(b) CH₃COOH is a weak acid and only partially dissociates in water, producing a lower concentration of H⁺ ions compared to the strong acid HCl at the same concentration. [1 mark]

Section B: Structured Questions

6. (a) Titrations 2 and 3 are concordant as their titres (24.30 cm³) differ by 0.00 cm³. Titration 1 (24.50 cm³) differs by 0.20 cm³, which is greater than the acceptable 0.10 cm³ difference. [2 marks]

(b) Mean titre = (24.30 + 24.30) / 2 = 24.30 cm³ [1 mark]

(c)

• n(HCl) = 0.100 × 0.02430 = 2.43 × 10⁻³ mol [1 mark]
• NaOH + HCl → NaCl + H₂O, so n(NaOH) = n(HCl) = 2.43 × 10⁻³ mol
• [NaOH] = 2.43 × 10⁻³ / 0.0250 = 0.0972 mol dm⁻³ [1 mark]

(d)

• Phenolphthalein changes from pink (in alkali) to colourless (in acid) at the end-point. [1 mark]
• The titration is between a strong acid and a strong base; the pH at the equivalence point is approximately 7, which falls within the pH range of phenolphthalein (8.3–10.0). The indicator changes colour sharply at the end-point. [1 mark]

7. (a) n(NaOH) = 0.0500 × 0.02240 = 1.12 × 10⁻³ mol [1 mark]

(b) From the equation, 1 mol H₂X reacts with 2 mol NaOH. n(H₂X) in 25.0 cm³ = 1.12 × 10⁻³ / 2 = 5.60 × 10⁻⁴ mol [1 mark]

(c)

• n(H₂X) in 250.0 cm³ = 5.60 × 10⁻⁴ × 10 = 5.60 × 10⁻³ mol [1 mark]
• Mass of pure H₂X = n × M. (Molar mass of H₂X is not given; assume a value or leave expression. For marking, accept: mass = 5.60 × 10⁻³ × M(H₂X) g) [1 mark]
• Note: If a specific molar mass is assumed in the question, substitute accordingly. In this generic answer, the method is key.

(d) Percentage purity = (mass of pure H₂X / 0.500) × 100% [1 mark]

8. (a) Kₛₚ = [Ca²⁺][OH⁻]². Units: mol³ dm⁻⁹. [2 marks]

(b)

• Molar mass of Ca(OH)₂ = 40.1 + 2(16.0 + 1.0) = 74.1 g mol⁻¹ [1 mark]
• Solubility in mol dm⁻³ = (0.113 g / 74.1 g mol⁻¹) / 0.100 dm³ = 0.01525 mol dm⁻³ [1 mark]
• [Ca²⁺] = 0.01525 mol dm⁻³; [OH⁻] = 2 × 0.01525 = 0.0305 mol dm⁻³
• Kₛₚ = 0.01525 × (0.0305)² = 1.42 × 10⁻⁵ mol³ dm⁻⁹ [1 mark]

(c) A white precipitate of Ca(OH)₂ would form. [1 mark] Adding CaCl₂ increases [Ca²⁺]; by Le Chatelier's principle, the equilibrium Ca(OH)₂(s) ⇌ Ca²⁺(aq) + 2OH⁻(aq) shifts left, precipitating Ca(OH)₂. [1 mark]

9. (a) Ethanol < Phenol < Ethanoic acid (increasing acid strength) [1 mark]

(b)

• Ethanoic acid has the carboxyl group (−COOH); the conjugate base (CH₃COO⁻) is stabilised by resonance delocalisation of the negative charge over two oxygen atoms. [1 mark]
• Ethanol has the hydroxyl group (−OH); the conjugate base (C₂H₅O⁻) has the negative charge localised on one oxygen atom with no resonance stabilisation. [1 mark]
• The greater stability of the ethanoate ion makes ethanoic acid more willing to donate a proton, hence a stronger acid. [1 mark]

(c)

• The phenoxide ion (C₆H₅O⁻) is stabilised by delocalisation of the negative charge into the benzene ring (resonance). [1 mark]
• This stabilisation is less effective than in the ethanoate ion (where charge is delocalised over two electronegative oxygen atoms) but greater than in the ethoxide ion (no delocalisation). Hence, phenol is intermediate in acid strength. [1 mark]

Section C: Data-Based & Extended Questions

10. (a) At X (half-neutralisation point), the solution contains significant amounts of NH₃ and NH₄⁺ (a buffer mixture). [2 marks]

(b) At the equivalence point, the solution contains NH₄Cl. NH₄⁺ is a weak acid and undergoes hydrolysis: NH₄⁺(aq) + H₂O(l) ⇌ NH₃(aq) + H₃O⁺(aq), producing H₃O⁺ ions, so pH < 7. [2 marks]

(c) Methyl orange (pH range 3.1–4.4) is suitable. [1 mark] The equivalence point pH is around 5–6 (acidic), which falls within or near the colour change range of methyl orange. Phenolphthalein would change colour too early (pH 8.3–10.0). [1 mark]

(d)

• K_b for NH₃ = 1.8 × 10⁻⁵ mol dm⁻³
• [OH⁻] = √(K_b × c) = √(1.8 × 10⁻⁵ × 0.10) = 1.34 × 10⁻³ mol dm⁻³ [1 mark]
• pOH = −log(1.34 × 10⁻³) = 2.87 [1 mark]
• pH = 14 − 2.87 = 11.13 (or 11.1) [1 mark]

11. (a) pH = pKₐ + log([CH₃COO⁻]/[CH₃COOH]) pKₐ = −log(1.8 × 10⁻⁵) = 4.74 [1 mark] 4.50 = 4.74 + log([CH₃COO⁻]/[CH₃COOH]) log([CH₃COO⁻]/[CH₃COOH]) = −0.24 [CH₃COO⁻]/[CH₃COOH] = 10⁻⁰·²⁴ = 0.575 (or 0.58) [1 mark]

(b)

• [CH₃COOH] = 0.50 mol dm⁻³ in 500 cm³, so n(CH₃COOH) = 0.50 × 0.500 = 0.25 mol [1 mark]
• [CH₃COO⁻] required = 0.575 × 0.50 = 0.2875 mol dm⁻³
• n(CH₃COONa) needed = 0.2875 × 0.500 = 0.14375 mol [1 mark]
• M(CH₃COONa) = 82.0 g mol⁻¹
• Mass = 0.14375 × 82.0 = 11.8 g (or 12 g to 2 s.f.) [1 mark]

(c)

• The buffer contains CH₃COOH and CH₃COO⁻. [1 mark]
• When H⁺ is added: CH₃COO⁻(aq) + H⁺(aq) → CH₃COOH(aq). The added H⁺ is removed by reaction with the ethanoate ions, so pH remains relatively constant. [1 mark]
• When OH⁻ is added: CH₃COOH(aq) + OH⁻(aq) → CH₃COO⁻(aq) + H₂O(l). The added OH⁻ is neutralised by ethanoic acid. [1 mark]

12. (a) pKₐ is the negative logarithm (base 10) of the acid dissociation constant, Kₐ: pKₐ = −log₁₀(Kₐ). It indicates the strength of an acid; the smaller the pKₐ, the stronger the acid. [1 mark]

(b) For an indicator to give a sharp colour change at the end-point, the pH at the equivalence point must lie within the range where the indicator changes colour (pKₐ ± 1). [1 mark] Within this range, the concentrations of HIn and In⁻ are comparable (ratio between 0.1 and 10), so the colour change is visible. Outside this range, the indicator is predominantly in one form and the colour appears constant. [1 mark]

(c) Bromothymol blue is suitable. [1 mark] The titration of a strong acid with a strong base has an equivalence point at pH 7, which falls within the pH range of bromothymol blue (6.0–7.6). Methyl orange changes colour below pH 4.4 (too early) and phenolphthalein above pH 8.3 (too late). [1 mark]

13. (a) With phenolphthalein, only the first stage of Na₂CO₃ neutralisation occurs (to NaHCO₃), giving a smaller titre. [1 mark] With methyl orange, both stages of Na₂CO₃ neutralisation and the neutralisation of any NaHCO₃ originally present occur, giving a larger titre. [1 mark]

(b) The titre with phenolphthalein corresponds to the conversion of Na₂CO₃ to NaHCO₃. n(HCl) with phenolphthalein = 0.100 × 0.01250 = 1.25 × 10⁻³ mol n(Na₂CO₃) in 25.0 cm³ = 1.25 × 10⁻³ mol [1 mark]

(c)

• Total n(HCl) with methyl orange = 0.100 × 0.03500 = 3.50 × 10⁻³ mol [1 mark]
• HCl used for Na₂CO₃ (both stages) = 2 × n(Na₂CO₃) = 2 × 1.25 × 10⁻³ = 2.50 × 10⁻³ mol
• HCl used for NaHCO₃ = 3.50 × 10⁻³ − 2.50 × 10⁻³ = 1.00 × 10⁻³ mol
• n(NaHCO₃) in 25.0 cm³ = 1.00 × 10⁻³ mol [1 mark]

(d)

• In 250.0 cm³: n(Na₂CO₃) = 1.25 × 10⁻³ × 10 = 0.0125 mol; n(NaHCO₃) = 1.00 × 10⁻³ × 10 = 0.0100 mol [1 mark]
• M(Na₂CO₃) = 106.0 g mol⁻¹; M(NaHCO₃) = 84.0 g mol⁻¹
• Mass of Na₂CO₃ = 0.0125 × 106.0 = 1.325 g
• Mass of NaHCO₃ = 0.0100 × 84.0 = 0.840 g [1 mark]
• % Na₂CO₃ = (1.325 / 1.00) × 100 = 132.5%? Wait, total mass is 1.00 g, but 1.325 + 0.840 = 2.165 g > 1.00 g. This indicates an error in the question design. In a real paper, values would be consistent. For marking, accept the method:
• % Na₂CO₃ = (mass of Na₂CO₃ / 1.00) × 100%; % NaHCO₃ = (mass of NaHCO₃ / 1.00) × 100% [1 mark for correct method]

14. (a) Thermal stability increases down Group 2. [1 mark] Down the group, the cation radius increases and charge density decreases. This reduces the polarising power of the cation, which distorts the carbonate ion less, making the C–O bond stronger and the carbonate more resistant to decomposition. [2 marks]

(b)

• Mass of CO₂ lost = 2.50 − 1.62 = 0.88 g [1 mark]
• n(CO₂) = 0.88 / 44.0 = 0.0200 mol
• n(CaCO₃) = n(CO₂) = 0.0200 mol [1 mark]
• Mass of pure CaCO₃ = 0.0200 × 100.1 = 2.002 g
• Percentage purity = (2.002 / 2.50) × 100 = 80.1% (or 80%) [1 mark]

(c)

• The limewater turns milky (or a white precipitate forms). [1 mark]
• Ca(OH)₂(aq) + CO₂(g) → CaCO₃(s) + H₂O(l) [1 mark]

15. (a)

• Add aqueous NaOH dropwise, then in excess:
  • Al³⁺: White precipitate, soluble in excess NaOH → colourless solution. Al³⁺(aq) + 3OH⁻(aq) → Al(OH)₃(s); Al(OH)₃(s) + OH⁻(aq) → [Al(OH)₄]⁻(aq) [1 mark]
  • Zn²⁺: White precipitate, soluble in excess NaOH → colourless solution. Zn²⁺(aq) + 2OH⁻(aq) → Zn(OH)₂(s); Zn(OH)₂(s) + 2OH⁻(aq) → [Zn(OH)₄]²⁻(aq) [1 mark]
  • Cu²⁺: Blue precipitate, insoluble in excess NaOH. Cu²⁺(aq) + 2OH⁻(aq) → Cu(OH)₂(s) [1 mark]
• To distinguish Al³⁺ and Zn²⁺: Add aqueous NH₃ dropwise, then in excess:
  • Al³⁺: White precipitate, insoluble in excess NH₃. [1 mark]
  • Zn²⁺: White precipitate, soluble in excess NH₃ → colourless solution. Zn(OH)₂(s) + 4NH₃(aq) → [Zn(NH₃)₄]²⁺(aq) + 2OH⁻(aq) [1 mark]
• Cu²⁺ with NH₃: Blue precipitate, soluble in excess → deep blue solution. Cu(OH)₂(s) + 4NH₃(aq) → [Cu(NH₃)₄]²⁺(aq) + 2OH⁻(aq) [1 mark]

(b) Both Al³⁺ and Zn²⁺ form amphoteric hydroxides that dissolve in excess NaOH because their hydroxides can act as acids, reacting with OH⁻ to form complex ions. [1 mark] However, only Zn²⁺ forms a stable ammine complex with NH₃; Al(OH)₃ does not react with NH₃ because Al³⁺ has a higher charge density and prefers O-donor ligands over N-donor ligands. [1 mark]

16. (a) Pure water undergoes self-ionisation: 2H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq) (or H₂O(l) ⇌ H⁺(aq) + OH⁻(aq)). [1 mark] At 298 K, [H⁺] = [OH⁻] = 1.0 × 10⁻⁷ mol dm⁻³, so pH = −log(1.0 × 10⁻⁷) = 7.00. [1 mark]

(b) At 313 K, pH = 6.77, so [H⁺] = 10⁻⁶·⁷⁷ = 1.70 × 10⁻⁷ mol dm⁻³. [1 mark] In pure water, [H⁺] = [OH⁻], so K_w = [H⁺][OH⁻] = (1.70 × 10⁻⁷)² = 2.89 × 10⁻¹⁴ mol² dm⁻⁶. [1 mark]

(c) K_w increases with temperature (2.89 × 10⁻¹⁴ at 313 K vs. 1.0 × 10⁻¹⁴ at 298 K). [1 mark] By Le Chatelier's principle, increasing temperature favours the endothermic direction. Since K_w increases, the forward reaction (ionisation) is endothermic. [1 mark]

(d)

• [OH⁻] from NaOH = 0.010 mol dm⁻³ (fully dissociated)
• [H⁺] = K_w / [OH⁻] = 2.89 × 10⁻¹⁴ / 0.010 = 2.89 × 10⁻¹² mol dm⁻³ [1 mark]
• pH = −log(2.89 × 10⁻¹²) = 11.54 (or 11.5) [1 mark]

17. (a) M(Na₂CO₃) = 106.0 g mol⁻¹. n(Na₂CO₃) = 2.65 / 106.0 = 0.0250 mol. [1 mark] Concentration = 0.0250 / 0.250 = 0.100 mol dm⁻³. [1 mark]

(b)

• CO₃²⁻(aq) + H₂O(l) ⇌ HCO₃⁻(aq) + OH⁻(aq) [1 mark]
• HCO₃⁻(aq) + H₂O(l) ⇌ H₂CO₃(aq) + OH⁻(aq) [1 mark]

(c)

• For the first hydrolysis: K_b1 = K_w / Kₐ(HCO₃⁻) = 1.0 × 10⁻¹⁴ / 4.8 × 10⁻¹¹ = 2.08 × 10⁻⁴ mol dm⁻³ [1 mark]
• Since K_b1 >> K_b2, the second hydrolysis contributes negligibly to [OH⁻].
• [OH⁻] = √(K_b1 × c) = √(2.08 × 10⁻⁴ × 0.100) = 4.56 × 10⁻³ mol dm⁻³ [1 mark]
• pOH = −log(4.56 × 10⁻³) = 2.34 [1 mark]
• pH = 14 − 2.34 = 11.66 (or 11.7) [1 mark]

18. (a)

• Both curves start at the origin. [1 mark]
• The curve for 2.0 mol dm⁻³ HCl has a steeper initial gradient (faster rate). [1 mark]
• Both curves level off at the same final volume of CO₂. [1 mark] (Award marks for correctly labelled sketch.)

(b) With 2.0 mol dm⁻³ HCl, there are more H⁺ ions per unit volume. [1 mark] This increases the frequency of effective collisions between H⁺ ions and CaCO₃ particles per unit time, leading to a faster initial rate of reaction. [1 mark]

(c) The final volume of CO₂ would be the same. [1 mark] The amount of CaCO₃ (the limiting reagent) is the same in both experiments. The total amount of CO₂ produced depends on the amount of CaCO₃, not the concentration of HCl (provided HCl is in excess). [1 mark]

19. (a)

• Kₐ = [H⁺]² / [C₆H₅COOH]; assume [H⁺] << 0.050 [1 mark]
• [H⁺] = √(Kₐ × c) = √(6.3 × 10⁻⁵ × 0.050) = √(3.15 × 10⁻⁶) = 1.77 × 10⁻³ mol dm⁻³ [1 mark]
• pH = −log(1.77 × 10⁻³) = 2.75 (or 2.8) [1 mark]

(b) Sodium benzoate is the salt of a weak acid (benzoic acid) and a strong base. In acidic conditions (low pH), the benzoate ions react with H⁺ to form undissociated benzoic acid: C₆H₅COO⁻(aq) + H⁺(aq) → C₆H₅COOH(aq). [1 mark] The undissociated benzoic acid is the active preservative form, which can penetrate microbial cell membranes more effectively than the ionised form. [1 mark] In neutral or alkaline conditions, the equilibrium favours the ionised benzoate ion, which is less effective as a preservative. [1 mark]

(c) When equal volumes are mixed, [C₆H₅COOH] = [C₆H₅COO⁻] = 0.050 mol dm⁻³ (after dilution). [1 mark] pH = pKₐ + log([C₆H₅COO⁻]/[C₆H₅COOH]) = −log(6.3 × 10⁻⁵) + log(1) = 4.20 + 0 = 4.20. [1 mark]

20. (a)

• Test 1: Add dilute HCl (or water) to each solid.
  • NH₄Cl: Dissolves, no effervescence.
  • Na₂SO₄: Dissolves, no effervescence.
  • CaCO₃: Effervescence; gas produced turns limewater milky. CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g). [2 marks]
• Test 2: To the remaining two solutions, add aqueous NaOH and warm gently.
  • NH₄Cl: Pungent gas (NH₃) evolved, turns damp red litmus paper blue. NH₄⁺(aq) + OH⁻(aq) → NH₃(g) + H₂O(l). [1 mark]
  • Na₂SO₄: No reaction with NaOH. [1 mark]
• Test 3: To confirm Na₂SO₄, add aqueous BaCl₂ (or Ba(NO₃)₂) followed by dilute HCl.
  • White precipitate (BaSO₄) forms, insoluble in dilute HCl. Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s). [1 mark]
• (Alternative sequences are acceptable if logically sound.) [Total: 5 marks for tests + 1 mark for clarity = 6 marks]

(b)

• NH₄Cl with conc. H₂SO₄: White fumes of HCl gas evolved. NH₄Cl(s) + H₂SO₄(l) → NH₄HSO₄(s) + HCl(g). [1 mark]
• Na₂SO₄ with conc. H₂SO₄: No visible reaction at room temperature (or white solid remains). Na₂SO₄ is already the sulfate salt; no further reaction. [1 mark]
• CaCO₃ with conc. H₂SO₄: Effervescence initially, but reaction slows/stops as a layer of insoluble CaSO₄ forms, coating the CaCO₃. CaCO₃(s) + H₂SO₄(aq) → CaSO₄(s) + H₂O(l) + CO₂(g). [2 marks]

END OF ANSWER KEY