A Level H2 Chemistry Acids Bases Salts Quiz

A-Level Chemistry H2 Quiz - Acids Bases Salts: ANSWER KEY

Total Marks: 50

Section A: Short Answer and Structured Questions (20 marks)

1. [2 marks]

• A Brønsted–Lowry acid is a proton (H⁺) donor. [1 mark]
• A Brønsted–Lowry base is a proton (H⁺) acceptor. [1 mark]
• Accept any correct example for each (e.g., HCl as acid, NH₃ as base).

2. [2 marks]

• Equation: 2H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq) OR H₂O(l) ⇌ H⁺(aq) + OH⁻(aq) [1 mark]
• K_w = 1.00 × 10⁻¹⁴ [½ mark]
• Units: mol² dm⁻⁶ [½ mark]

3. [3 marks]

• [H⁺] = 10⁻²·⁸⁷ = 1.35 × 10⁻³ mol dm⁻³ [1 mark]
• K_a = [H⁺]² / [HA] = (1.35 × 10⁻³)² / 0.100 = 1.82 × 10⁻⁵ mol dm⁻³ [1 mark]
• Assumptions: [H⁺] ≈ [A⁻]; dissociation is small so [HA]ₑq ≈ [HA]ᵢₙᵢₜᵢₐₗ; [H⁺] from water autoionisation is negligible. [1 mark for at least two valid assumptions]

4. [3 marks]

• NH₄Cl dissociates completely: NH₄Cl(s) → NH₄⁺(aq) + Cl⁻(aq) [1 mark]
• NH₄⁺ is the conjugate acid of the weak base NH₃ and undergoes hydrolysis: NH₄⁺(aq) + H₂O(l) ⇌ NH₃(aq) + H₃O⁺(aq) [1 mark]
• The production of H₃O⁺ ions makes the solution acidic (pH < 7). [1 mark]

5. [5 marks total]

(a) [1 mark]

• Moles of CH₃COOH = 0.200 × 0.0500 = 0.0100 mol [½ mark]
• Moles of NaOH = 0.200 × 0.0250 = 0.00500 mol [½ mark]

(b) [2 marks]

• Equation: CH₃COOH(aq) + NaOH(aq) → CH₃COONa(aq) + H₂O(l) [1 mark]
• Moles of CH₃COOH remaining = 0.0100 − 0.00500 = 0.00500 mol [½ mark]
• Moles of CH₃COO⁻ formed = 0.00500 mol [½ mark]

(c) [2 marks]

• Total volume = 75.0 cm³ = 0.0750 dm³
• [CH₃COOH] = 0.00500 / 0.0750 = 0.0667 mol dm⁻³
• [CH₃COO⁻] = 0.00500 / 0.0750 = 0.0667 mol dm⁻³ [1 mark for correct concentrations]
• pH = pK_a + log([CH₃COO⁻]/[CH₃COOH]) = −log(1.74 × 10⁻⁵) + log(0.0667/0.0667) = 4.76 + 0 = 4.76 [1 mark]

Section B: Data Interpretation and Calculations (15 marks)

6. [2 marks] (a) Red/pink [1 mark] (b) Yellow [1 mark]

7. [3 marks]

• CH₃COONa dissociates completely: CH₃COONa(s) → CH₃COO⁻(aq) + Na⁺(aq) [1 mark]
• CH₃COO⁻ is the conjugate base of the weak acid CH₃COOH and undergoes hydrolysis: CH₃COO⁻(aq) + H₂O(l) ⇌ CH₃COOH(aq) + OH⁻(aq) [1 mark]
• The production of OH⁻ ions makes the solution alkaline (pH > 7). [1 mark]

8. [6 marks total]

(a) [1 mark]

(b) [2 marks]

• Concordant titrations: 2 and 3 (differ by ≤0.10 cm³) [1 mark]
• Mean volume = (23.70 + 23.70) / 2 = 23.70 cm³ [1 mark]

(c) [1 mark]

• HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l) [1 mark]

(d) [2 marks]

• n(HCl) = 0.100 × 0.02370 = 2.37 × 10⁻³ mol [1 mark]
• n(NaOH) = n(HCl) = 2.37 × 10⁻³ mol
• [NaOH] = 2.37 × 10⁻³ / 0.0250 = 0.0948 mol dm⁻³ [1 mark]

9. [9 marks total]

(a) [2 marks]

• pH at equivalence point ≈ 5.0–5.5 (accept 5.0–5.5) [1 mark]
• Explanation: At equivalence point, the solution contains NH₄⁺ ions (from NH₃ + HCl → NH₄Cl). NH₄⁺ is a weak acid and undergoes hydrolysis producing H₃O⁺, making the solution acidic. [1 mark]

(b) [2 marks]

• Suitable indicator: Methyl orange [1 mark]
• Justification: The pH range of methyl orange (3.1–4.4) falls within the steep vertical portion of the pH curve near the equivalence point (pH ~5). The indicator changes colour completely within the rapid pH change region. [1 mark]

(c) [3 marks]

• For weak base: [OH⁻] = √(K_b × c) = √(1.8 × 10⁻⁵ × 0.100) = √(1.8 × 10⁻⁶) = 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 ≈ 11.1 [1 mark]

(d) [2 marks]

• At half-equivalence point: [NH₃] = [NH₄⁺] [1 mark]
• pOH = pK_b = −log(1.8 × 10⁻⁵) = 4.74
• pH = 14 − 4.74 = 9.26 [1 mark]

10. [2 marks]

• A strong acid dissociates completely in water, e.g., HCl → H⁺ + Cl⁻. [1 mark]
• A weak acid dissociates partially in water, establishing an equilibrium, e.g., CH₃COOH ⇌ CH₃COO⁻ + H⁺. [1 mark]

Section C: Qualitative Analysis and Applications (15 marks)

11. [8 marks total]

(i) [1 mark]

• Carbon dioxide, CO₂ [1 mark]

(ii) [2 marks]

• Cation: Al³⁺ [1 mark]
• Reasoning: White precipitate with NaOH(aq) that dissolves in excess indicates an amphoteric hydroxide (Al(OH)₃ or Zn(OH)₂ or Pb(OH)₂). White precipitate with NH₃(aq) that is insoluble in excess NH₃ eliminates Zn²⁺ (which forms soluble complex) and Pb²⁺ (also forms soluble complex with excess NaOH but not NH₃). Al³⁺ gives white ppt. with NH₃, insoluble in excess. [1 mark]

(iii) [2 marks]

• Anion: SO₄²⁻ [1 mark]
• Reasoning: Effervescence with dilute HCl producing CO₂ (turns limewater milky) indicates carbonate, CO₃²⁻. However, white precipitate with BaCl₂ that is insoluble in dilute HCl indicates sulfate, SO₄²⁻. The effervescence with HCl could be from a carbonate impurity, but the definitive test is the BaCl₂/HCl test. [Note: Accept CO₃²⁻ if student argues the BaCl₂ test shows sulfate but the HCl test shows carbonate — in practice, the solid may be a mixture. Primary evidence from BaCl₂/HCl points to SO₄²⁻.] [1 mark]

(iv) [1 mark]

• Formula: Al₂(SO₄)₃ [1 mark]

(v) [2 marks]

• Al³⁺(aq) + 3OH⁻(aq) → Al(OH)₃(s) [1 mark]
• Al(OH)₃(s) + OH⁻(aq) → [Al(OH)₄]⁻(aq) [1 mark]
• Accept combined equation: Al³⁺(aq) + 4OH⁻(aq) → [Al(OH)₄]⁻(aq)

12. [7 marks total]

(a) [1 mark]

• CuO(s) + H₂SO₄(aq) → CuSO₄(aq) + H₂O(l) [1 mark]

(b) [1 mark]

• To ensure all the sulfuric acid is completely neutralised/reacted, maximising the yield of copper(II) sulfate. [1 mark]

(c) [3 marks]

• Filter the mixture to remove unreacted/excess CuO(s). [1 mark]
• Heat the filtrate to evaporate some water until the solution is saturated / until crystallisation point is reached. [1 mark]
• Allow the solution to cool slowly; filter the crystals formed; wash with a little cold distilled water; dry between filter papers or in a desiccator. [1 mark for cooling, filtering, washing, drying — at least 3 steps]

(d) [2 marks]

• n(H₂SO₄) = 1.00 × 0.0500 = 0.0500 mol [½ mark]
• Theoretical moles of CuSO₄·5H₂O = 0.0500 mol (1:1 ratio)
• Theoretical mass = 0.0500 × 249.7 = 12.485 g [½ mark]
• Percentage yield = (4.98 / 12.485) × 100 = 39.9% [1 mark]

13. [2 marks]

• The solubility product, K_sp, is the equilibrium constant for the dissolution of a sparingly soluble salt, representing the product of the concentrations of the constituent ions, each raised to the power of its stoichiometric coefficient in the equilibrium equation. [1 mark]
• For AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq), K_sp = [Ag⁺][Cl⁻] [1 mark]

14. [2 marks]

• The solubility of Ca(OH)₂ decreases due to the common ion effect. [1 mark]
• Ca(OH)₂(s) ⇌ Ca²⁺(aq) + 2OH⁻(aq). The addition of NaOH increases [OH⁻], shifting the equilibrium to the left, thus decreasing solubility. [1 mark]

15. [3 marks]

• pH = 10.52, so pOH = 14 − 10.52 = 3.48 [1 mark]
• [OH⁻] = 10⁻³·⁴⁸ = 3.31 × 10⁻⁴ mol dm⁻³ [1 mark]
• Mg(OH)₂(s) ⇌ Mg²⁺(aq) + 2OH⁻(aq); [Mg²⁺] = ½[OH⁻] = 1.66 × 10⁻⁴ mol dm⁻³
• K_sp = [Mg²⁺][OH⁻]² = (1.66 × 10⁻⁴)(3.31 × 10⁻⁴)² = 1.82 × 10⁻¹¹ mol³ dm⁻⁹ [1 mark]

Section D: Advanced Concepts and Integrated Problems (15 marks)

16. [3 marks]

• The mixture contains a weak acid (CH₃COOH) and its conjugate base (CH₃COO⁻ from CH₃COONa). [1 mark]
• When a small amount of acid (H⁺) is added, it reacts with the conjugate base: CH₃COO⁻ + H⁺ → CH₃COOH. [1 mark]
• When a small amount of base (OH⁻) is added, it reacts with the weak acid: CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O. The pH remains relatively constant. [1 mark]

17. [3 marks]

• Initial pH of buffer: pH = pK_a + log([CH₃COO⁻]/[CH₃COOH]) = −log(1.74 × 10⁻⁵) + log(0.10/0.10) = 4.76 [1 mark]
• Moles of H⁺ added = 1.0 × 10⁻³ dm³ × 1.0 mol dm⁻³ = 1.0 × 10⁻³ mol
• Initial moles in 100 cm³ buffer: CH₃COOH = 0.010 mol, CH₃COO⁻ = 0.010 mol
• After addition: CH₃COOH = 0.011 mol, CH₃COO⁻ = 0.009 mol (volume change negligible) [1 mark]
• New pH = 4.76 + log(0.009/0.011) = 4.76 − 0.087 = 4.67
• pH change = 4.76 − 4.67 = 0.09 [1 mark]

18. [5 marks total]

(a) [1 mark]

• (COOH)₂(aq) + 2NaOH(aq) → (COONa)₂(aq) + 2H₂O(l) [1 mark]

(b) [2 marks]

• n(acid) = 0.100 × 0.0250 = 2.50 × 10⁻³ mol [1 mark]
• n(NaOH) required = 2 × 2.50 × 10⁻³ = 5.00 × 10⁻³ mol
• Volume NaOH = 5.00 × 10⁻³ / 0.100 = 0.0500 dm³ = 50.0 cm³ [1 mark]

(c) [2 marks]

• Sketch should show two equivalence points at 25.0 cm³ and 50.0 cm³. [1 mark]
• First equivalence point pH ~4–5 (suitable indicator: methyl orange); second equivalence point pH ~8–10 (suitable indicator: phenolphthalein). [1 mark]

19. [3 marks]

• Moles C₆H₅COOH = 0.200 × 0.0500 = 0.0100 mol
• Moles NaOH = 0.100 × 0.0500 = 0.00500 mol [1 mark]
• Reaction: C₆H₅COOH + NaOH → C₆H₅COONa + H₂O
• Moles C₆H₅COOH remaining = 0.00500 mol; moles C₆H₅COO⁻ formed = 0.00500 mol
• Total volume = 0.100 dm³; [C₆H₅COOH] = [C₆H₅COO⁻] = 0.0500 mol dm⁻³ [1 mark]
• pH = pK_a + log([C₆H₅COO⁻]/[C₆H₅COOH]) = −log(6.3 × 10⁻⁵) + log(1) = 4.20 [1 mark]

20. [4 marks]

• Acid rain is formed when sulfur dioxide (SO₂) and nitrogen oxides (NOₓ) from industrial emissions react with water and oxygen in the atmosphere to form sulfuric acid (H₂SO₄) and nitric acid (HNO₃). [1 mark]
• Equations: 2SO₂ + O₂ + 2H₂O → 2H₂SO₄; 4NO₂ + O₂ + 2H₂O → 4HNO₃. [1 mark]
• Effects on limestone buildings: CaCO₃(s) + H₂SO₄(aq) → CaSO₄(aq) + H₂O(l) + CO₂(g), causing erosion and structural damage. [1 mark]
• Effects on aquatic life: Acidification of lakes and rivers lowers pH, mobilising toxic metal ions (e.g., Al³⁺) and disrupting aquatic ecosystems, leading to reduced biodiversity and fish kills. [1 mark]

END OF ANSWER KEY