A Level H2 Chemistry Practice Paper 2

TuitionGoWhere Practice Paper – Chemistry H2 A-Level – ANSWER KEY

TuitionGoWhere Exam Practice (AI)

Paper: PRACTICE – Topic: Acids, Bases & Salts | Version: 2 of 5 | Total Marks: 60

Section A: Short Answer & Qualitative Analysis (20 marks)

1. (a) Ammonia, NH₃ [1]

(b) NH₄⁺(aq) + OH⁻(aq) → NH₃(g) + H₂O(l) [2]

• Award 1 mark for correct reactants and products with state symbols.
• Award 1 mark for correct balancing.
• Accept NH₄Cl(s) + NaOH(aq) → NaCl(aq) + NH₃(g) + H₂O(l) for 1 mark only (molecular equation, not ionic).

2. Table completion [6 marks – 1 mark per correct cell]

Marking notes:

• Al³⁺ with NaOH: Must state "white ppt." AND "soluble in excess" for 1 mark.
• Al³⁺ with NH₃: Must state "white ppt." AND "insoluble in excess" for 1 mark.
• Cu²⁺ with NaOH: Must state "blue ppt." AND "insoluble in excess" for 1 mark.
• Cu²⁺ with NH₃: Must state "blue ppt." AND "soluble in excess" AND "deep blue solution" for 1 mark.
• Zn²⁺ with NaOH: Must state "white ppt." AND "soluble in excess" for 1 mark.
• Zn²⁺ with NH₃: Must state "white ppt." AND "soluble in excess" for 1 mark.
• Accept "gelatinous" as additional descriptor for Al³⁺ and Zn²⁺ precipitates.
• Do not penalise omission of "colourless solution" for Al³⁺ and Zn²⁺ if "soluble in excess" is stated.

3. (a) A white precipitate is formed / limewater turns milky/cloudy. [1]

(b) The white precipitate (calcium carbonate) reacts with excess carbon dioxide and water to form soluble calcium hydrogencarbonate. [1] Equation: CaCO₃(s) + CO₂(g) + H₂O(l) → Ca(HCO₃)₂(aq) [1]

• Accept CaCO₃(s) + CO₂(g) + H₂O(l) → Ca²⁺(aq) + 2HCO₃⁻(aq) for ionic equation.
• State symbols required for the equation mark.

4. (a) Damp litmus paper is bleached / turns white (or red litmus turns white). [1]

• Do not accept "turns colourless" alone without reference to bleaching.
• Accept "red litmus is first turned blue then bleached" if the sequence is described.

(b) Chlorine reacts with water to form hypochlorous acid (HOCl/HClO), which is a strong oxidising agent that bleaches the dye in litmus by oxidation. [1] Hydrogen chloride gas dissolves in water to form hydrochloric acid (HCl), which is not an oxidising agent and does not bleach; it only turns blue litmus red due to its acidic nature. [1]

• Award 1 mark for identifying the bleaching species (HOCl or Cl₂ in water).
• Award 1 mark for contrasting with HCl (no bleaching, only acidic behaviour).
• Relevant equations: Cl₂(g) + H₂O(l) ⇌ HCl(aq) + HOCl(aq); HCl(g) → H⁺(aq) + Cl⁻(aq).

5. Sequence of tests [5 marks]:

Test 1 – Carbonate: Add dilute nitric acid (or dilute HCl) to a portion of the solid. [1]

• Observation: Effervescence / bubbles of gas evolved. [0.5]
• Confirmatory test: Pass gas through limewater → white precipitate forms. [0.5]
• If no effervescence, carbonate is absent.

Test 2 – Sulfate: To a fresh portion of the solution (dissolve solid in dilute HNO₃ if not already in solution), add aqueous barium nitrate (or barium chloride) solution. [1]

• Observation: White precipitate of BaSO₄ forms if sulfate is present. [0.5]
• The precipitate is insoluble in dilute acids.

Test 3 – Chloride: To a fresh portion of the solution, add aqueous silver nitrate solution, followed by dilute nitric acid. [1]

• Observation: White precipitate of AgCl forms if chloride is present. [0.5]
• The precipitate is insoluble in dilute HNO₃ but soluble in dilute/aqueous ammonia.

Marking notes:

• Award marks for correct sequence and reagents.
• Must include confirmatory observations for each test.
• Accept alternative valid tests (e.g., lead(II) nitrate for chloride).
• Deduct 0.5 marks if acidification steps are omitted where necessary (e.g., for sulfate and chloride tests to eliminate carbonate interference).
• Full marks require all three tests described with expected positive and negative outcomes.

Section B: Calculations & Quantitative Analysis (25 marks)

6. (a) Mᵣ of NaOH = 23.0 + 16.0 + 1.0 = 40.0 g mol⁻¹ [0.5] Moles of NaOH = mass / Mᵣ = 4.00 / 40.0 = 0.100 mol [0.5] Concentration = moles / volume (dm³) = 0.100 / (250/1000) = 0.400 mol dm⁻³ [1] Answer: 0.400 mol dm⁻³ [2 marks total]

(b) Moles of NaOH used = c × V = 0.400 × (20.00/1000) = 0.00800 mol [1] From equation: 2 mol NaOH ≡ 1 mol H₂SO₄ Moles of H₂SO₄ = 0.00800 / 2 = 0.00400 mol [1] Concentration of H₂SO₄ = moles / volume (dm³) = 0.00400 / (25.0/1000) = 0.160 mol dm⁻³ [1] Answer: 0.160 mol dm⁻³ [3 marks total]

7. (a) Concordant titrations are those within 0.10 cm³ of each other. Titrations 1, 2, and 3 all show 23.30 cm³. [1] The rough titration (24.50 cm³) is excluded. Mean titre = (23.30 + 23.30 + 23.30) / 3 = 23.30 cm³ [1] Answer: 23.30 cm³ [3 marks total – 1 for identifying concordant results, 1 for excluding rough, 1 for correct mean]

(b) Moles of NaOH used = 0.100 × (23.30/1000) = 0.00233 mol [1] CH₃COOH + NaOH → CH₃COONa + H₂O (1:1 ratio) Moles of CH₃COOH in 25.0 cm³ of diluted vinegar = 0.00233 mol [0.5] Concentration of diluted vinegar = 0.00233 / (25.0/1000) = 0.0932 mol dm⁻³ [0.5] Dilution factor = 10, so original concentration = 0.0932 × 10 = 0.932 mol dm⁻³ [1] Answer: 0.932 mol dm⁻³ [3 marks total]

(c) Concentration from label: 5.0 g per 100 cm³ = 50 g dm⁻³ [0.5] Concentration in mol dm⁻³ = 50 / 60.0 = 0.833 mol dm⁻³ [0.5] Student's result: 0.932 mol dm⁻³ Percentage difference = [(0.932 – 0.833) / 0.833] × 100 = 11.9% [0.5] The student's result is higher than the label claim by approximately 12%, which is not consistent (or: the result is reasonably close but shows some experimental error). [0.5] Answer: Not fully consistent; student's value is ~12% higher. [2 marks total]

• Accept any reasoned comparison with appropriate justification.

8. (a) Moles of NaOH = c × V = 0.100 × (18.50/1000) = 0.00185 mol [1] Answer: 0.00185 mol [1 mark]

(b) From equation: 2 mol NaOH ≡ 1 mol H₂A Moles of H₂A = 0.00185 / 2 = 0.000925 mol [1] Answer: 9.25 × 10⁻⁴ mol [1 mark]

(c) Moles of H₂A in 100 cm³ = 0.000925 × (100/25.0) = 0.00370 mol [1] Mass of pure H₂A = moles × Mᵣ = 0.00370 × 90.0 = 0.333 g [1] Percentage purity = (mass of pure / mass of sample) × 100 = (0.333 / 0.500) × 100 = 66.6% [1] Answer: 66.6% [3 marks total]

9. (a) Kₐ = [H⁺][CH₃COO⁻] / [CH₃COOH] [1]

• Accept Kₐ = [H₃O⁺][CH₃COO⁻] / [CH₃COOH].

(b) pH = 4.50, so [H⁺] = 10⁻⁴·⁵⁰ = 3.16 × 10⁻⁵ mol dm⁻³ [1] Kₐ = [H⁺][CH₃COO⁻] / [CH₃COOH] 1.74 × 10⁻⁵ = (3.16 × 10⁻⁵) × [CH₃COO⁻] / [CH₃COOH] [1] [CH₃COO⁻] / [CH₃COOH] = (1.74 × 10⁻⁵) / (3.16 × 10⁻⁵) = 0.551 [1] Answer: 0.551 (or 0.55) [3 marks total]

(c) Let volume of CH₃COONa required = V cm³. Moles of CH₃COOH = 0.200 × (50.0/1000) = 0.0100 mol Moles of CH₃COONa = 0.200 × (V/1000) = 0.000200V mol Since both are in the same total volume, ratio of moles = ratio of concentrations. [CH₃COO⁻] / [CH₃COOH] = (0.000200V) / 0.0100 = 0.551 [1] 0.000200V = 0.0100 × 0.551 = 0.00551 V = 0.00551 / 0.000200 = 27.55 cm³ [1] Answer: 27.6 cm³ (or 27.5 cm³) [2 marks total]

Section C: Structured & Data Interpretation (15 marks)

10. (a) From the graph, the equivalence point is at 25.0 cm³ of NaOH added, where the pH is approximately 8.5 (accept 8.0–9.0). [1] A suitable indicator is phenolphthalein (pH range 8.2–10.0, colourless to pink). [1] Explanation: The pH at the equivalence point (~8.5) falls within the pH range over which phenolphthalein changes colour, so the endpoint will coincide closely with the equivalence point. Methyl orange (pH range 3.1–4.4) would change colour before the equivalence point and is therefore unsuitable. [1] Answer: pH ~8.5; phenolphthalein [3 marks total]

(b) At half-neutralisation, [HA] = [A⁻], so pH = pKₐ. [1] pKₐ = 4.75 Kₐ = 10⁻⁴·⁷⁵ = 1.78 × 10⁻⁵ mol dm⁻³ [1] Answer: 1.78 × 10⁻⁵ mol dm⁻³ [2 marks total]

(c) At the equivalence point, all the weak acid HA has been neutralised to form its conjugate base A⁻. [1] The anion A⁻ undergoes hydrolysis with water: A⁻(aq) + H₂O(l) ⇌ HA(aq) + OH⁻(aq) [0.5] This produces hydroxide ions, making the solution alkaline (pH > 7). [0.5] Answer: Hydrolysis of the conjugate base produces OH⁻ ions. [2 marks total]

11. (a) Al₂O₃(s) + 2OH⁻(aq) + 3H₂O(l) → 2[Al(OH)₄]⁻(aq) [2]

• Award 1 mark for correct reactants and products.
• Award 1 mark for correct balancing and state symbols.
• Accept Al₂O₃(s) + 2OH⁻(aq) + 3H₂O(l) → 2Al(OH)₄⁻(aq).
• Do not accept Al₂O₃(s) + 2OH⁻(aq) → 2AlO₂⁻(aq) + H₂O(l) (incorrect aluminate formula for A-Level).

(b) Al₂O₃(s) + 6H⁺(aq) → 2Al³⁺(aq) + 3H₂O(l) [2]

• Award 1 mark for correct reactants and products.
• Award 1 mark for correct balancing and state symbols.

(c) Aluminium oxide (Al₂O₃) has a giant ionic lattice structure with strong electrostatic forces of attraction between Al³⁺ and O²⁻ ions throughout the lattice. [1] A large amount of energy is required to overcome these strong ionic bonds, resulting in a very high melting point (over 2000 °C). [1] Aluminium chloride (AlCl₃) exists as discrete covalent molecules (Al₂Cl₆ dimers or AlCl₃ monomers) with weak intermolecular forces (van der Waals' forces) between the molecules. [1] Only a small amount of energy is required to overcome these weak intermolecular forces, so AlCl₃ sublimes at a relatively low temperature (approximately 180 °C). [1] Answer: Al₂O₃ – giant ionic, strong ionic bonds; AlCl₃ – simple molecular, weak intermolecular forces. [4 marks total]

• Award marks for correct identification of structure type and bonding in each compound, and for linking structure to melting point.
• Accept reference to the polarising power of Al³⁺ causing covalent character in AlCl₃.

END OF ANSWER KEY

Total marks: 60