A Level H2 Chemistry Practice Paper 1

TuitionGoWhere Practice Paper - Chemistry H2 A-Level

Answer Key and Marking Scheme

Paper: Practice Paper 1 (Acids, Bases & Salts) Version: 1 of 5 Total Marks: 50

Section A: Structured Questions (20 marks)

Question 1

(a) Mᵣ(Na₂CO₃) = (2 × 23.0) + 12.0 + (3 × 16.0) = 106.0 [1] n(Na₂CO₃) = 2.650 / 106.0 = 0.0250 mol Concentration = 0.0250 / 0.2500 = 0.100 mol dm⁻³ [1]

(b) n(Na₂CO₃) = 0.100 × (22.40/1000) = 0.00224 mol [1] From equation: 1 mol Na₂CO₃ reacts with 2 mol HCl n(HCl) = 2 × 0.00224 = 0.00448 mol [HCl] = 0.00448 / 0.0250 = 0.179 mol dm⁻³ [1]

(c) Titrations 3 and 4 should be used. [1] Titration 1 is a rough titration and should be excluded. Titration 2 (22.40 cm³) differs from titrations 3 and 4 (both 22.30 cm³) by 0.10 cm³, which is at the limit of concordancy. Titrations 3 and 4 are concordant (within 0.10 cm³ of each other). [1]

Question 2

(a) Sulfate ions, SO₄²⁻, and ammonium ions, NH₄⁺. [1 for both]

• Test (i): White precipitate with Ba(NO₃)₂ in acidic medium confirms SO₄²⁻ (BaSO₄ is insoluble in acid). [1]
• Test (ii): Gas evolved turns damp red litmus blue, confirming NH₃ gas, hence NH₄⁺ ions present. [1]
• Test (iii): No precipitate with AgNO₃/HNO₃ confirms absence of Cl⁻, Br⁻, I⁻.

(b) Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s) [1]

Question 3

(a) Kₐ = [H⁺][A⁻] / [HA] [1]

(b) [H⁺] = 10⁻²·⁸⁸ = 1.32 × 10⁻³ mol dm⁻³ [1] Since HA is monoprotic, [A⁻] = [H⁺] = 1.32 × 10⁻³ mol dm⁻³ [HA] at equilibrium ≈ 0.100 − 1.32 × 10⁻³ = 0.0987 mol dm⁻³ [1] Kₐ = (1.32 × 10⁻³)² / 0.0987 = 1.76 × 10⁻⁵ mol dm⁻³ [1]

(c) n(HA) initially = 0.100 × 0.0500 = 0.00500 mol n(NaOH) added = 0.100 × 0.0250 = 0.00250 mol [1] NaOH reacts with HA: HA + OH⁻ → A⁻ + H₂O n(HA) remaining = 0.00500 − 0.00250 = 0.00250 mol n(A⁻) formed = 0.00250 mol [1] Total volume = 75.0 cm³ [HA] = [A⁻] in the mixture, so pH = pKₐ = −log(1.76 × 10⁻⁵) = 4.75 [1]

Question 4

(a) Yield of CO₂ increases. [1] The forward reaction is endothermic (ΔH positive). Increasing temperature shifts the equilibrium position in the endothermic direction to absorb the added heat, favouring the forward reaction and producing more CO₂. [1]

(b) Equilibrium shifts to the left (yield of CO₂ decreases). [1] Increasing pressure favours the side with fewer gaseous moles. The reactant side has 0 gaseous moles, the product side has 1 gaseous mole (CO₂).

Section B: Data Interpretation and Application (18 marks)

Question 5

(a) 25.0 cm³ [1]

(b) At the equivalence point, all the ethanoic acid has been neutralised. The solution contains sodium ethanoate, CH₃COONa. [1] The ethanoate ion, CH₃COO⁻, is the conjugate base of a weak acid and undergoes hydrolysis: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻, producing OH⁻ ions and making the solution alkaline (pH > 7). [1]

(c) At half-neutralisation, [CH₃COOH] = [CH₃COO⁻]. [1] From Kₐ = [H⁺][CH₃COO⁻] / [CH₃COOH], when [CH₃COOH] = [CH₃COO⁻], Kₐ = [H⁺]. Therefore, pKₐ = pH = 4.76, so Kₐ = 10⁻⁴·⁷⁶ = 1.74 × 10⁻⁵ mol dm⁻³. [1]

(d) Phenolphthalein. [1] The pH at the equivalence point is approximately 8.7, which falls within the pH range of phenolphthalein (8.2–10.0). Methyl orange and bromothymol blue change colour at pH values below the equivalence point.

Question 6

(a) n(NaHCO₃) = 2.10 / 84.0 = 0.0250 mol [NaHCO₃] = 0.0250 / 0.250 = 0.100 mol dm⁻³ [1] n(Na₂CO₃) = 1.06 / 106.0 = 0.0100 mol [Na₂CO₃] = 0.0100 / 0.250 = 0.0400 mol dm⁻³ [1] [CO₃²⁻] = 0.0400 mol dm⁻³, [HCO₃⁻] = 0.100 mol dm⁻³ [1]

(b) CO₃²⁻(aq) + H⁺(aq) → HCO₃⁻(aq) [1]

(c) pKₐ = −log(4.7 × 10⁻¹¹) = 10.33 [1] pH = pKₐ + log([CO₃²⁻]/[HCO₃⁻]) = 10.33 + log(0.0400/0.100) = 10.33 + log(0.400) = 10.33 − 0.398 = 9.93 [1]

(d) Dilution decreases the concentrations of both HCO₃⁻ and CO₃²⁻ by the same factor. [1] The buffer capacity depends on the absolute amounts of the acid-base pair present. With lower concentrations, fewer moles of H⁺ or OH⁻ can be neutralised before the pH changes significantly. [1]

Question 7

(a) Kₛₚ = [Mg²⁺][OH⁻]² [1] If solubility = s = 1.2 × 10⁻⁴ mol dm⁻³, then [Mg²⁺] = s = 1.2 × 10⁻⁴ mol dm⁻³, [OH⁻] = 2s = 2.4 × 10⁻⁴ mol dm⁻³ [1] Kₛₚ = (1.2 × 10⁻⁴) × (2.4 × 10⁻⁴)² = 1.2 × 10⁻⁴ × 5.76 × 10⁻⁸ = 6.91 × 10⁻¹² mol³ dm⁻⁹ [1]

(b) After mixing, total volume = 100.0 cm³. [Mg²⁺] = (0.010 × 50.0/1000) / 0.100 = 0.0050 mol dm⁻³ [1] [OH⁻] = (0.010 × 50.0/1000) / 0.100 = 0.0050 mol dm⁻³ [1] Ionic product = [Mg²⁺][OH⁻]² = (0.0050) × (0.0050)² = 1.25 × 10⁻⁷ mol³ dm⁻⁹ Since 1.25 × 10⁻⁷ > 6.91 × 10⁻¹² (Kₛₚ), a precipitate of Mg(OH)₂ will form. [1]

Section C: Free-Response Questions (12 marks)

Question 8

Marking guidelines (6 marks): Award up to 3 marks for each context discussed. Marks are awarded for:

• Correct identification of acid-base chemistry (1 mark)
• Relevant chemical equation(s) (1 mark)
• Application of equilibrium principles (1 mark)

Context 1: Antacid tablets in the stomach

• Stomach acid is HCl(aq). Antacids contain bases such as Mg(OH)₂, Al(OH)₃, or CaCO₃.
• Equation: Mg(OH)₂(s) + 2HCl(aq) → MgCl₂(aq) + 2H₂O(l) [or equivalent]
• The neutralisation reaction removes excess H⁺, relieving acidity. Some antacids (e.g., CaCO₃) also produce CO₂ gas. The reaction is not an equilibrium but goes to completion; however, the buffer system of the stomach may be discussed.

Context 2: Maintaining blood pH

• Blood pH is maintained at approximately 7.4 by the H₂CO₃/HCO₃⁻ buffer system.
• Equation: H₂CO₃(aq) ⇌ H⁺(aq) + HCO₃⁻(aq)
• When excess acid enters the blood, HCO₃⁻ reacts with H⁺: HCO₃⁻ + H⁺ → H₂CO₃. When excess base enters, H₂CO₃ reacts with OH⁻: H₂CO₃ + OH⁻ → HCO₃⁻ + H₂O. The equilibrium shifts to minimise pH changes.

Context 3: Lime in agriculture

• Acidic soil contains excess H⁺ ions. Lime (CaO) is a basic oxide.
• Equation: CaO(s) + 2H⁺(aq) → Ca²⁺(aq) + H₂O(l) [or CaO + H₂O → Ca(OH)₂; Ca(OH)₂ + 2H⁺ → Ca²⁺ + 2H₂O]
• The base neutralises soil acidity, raising the pH. This is a stoichiometric neutralisation, not an equilibrium process, but the resulting pH depends on the acid-base balance in the soil.

Question 9

(a) n(H₂SO₄) = 0.100 × (21.60/1000) = 2.16 × 10⁻³ mol [1]

(b) From equation: 2 mol KOH react with 1 mol H₂SO₄ n(KOH) in 25.0 cm³ = 2 × 2.16 × 10⁻³ = 4.32 × 10⁻³ mol [1]

(c) n(KOH) in 250.0 cm³ = 4.32 × 10⁻³ × 10 = 0.0432 mol [1] Mᵣ(KOH) = 39.1 + 16.0 + 1.0 = 56.1 Mass of pure KOH = 0.0432 × 56.1 = 2.42 g [1]

(d) Percentage purity = (2.42 / 2.50) × 100 = 96.8% [1]

(e) Add dilute hydrochloric acid (or nitric acid) to the solid. [1] Effervescence (bubbling) indicates the presence of carbonate ions, as CO₂ gas is evolved. [or: Pass any gas evolved through limewater; a white precipitate confirms CO₂ and hence carbonate.]

END OF ANSWER KEY