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A Level H1 Chemistry Practice Paper 5
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TuitionGoWhere Practice Paper - Chemistry H1 A-Level (Answers)
TuitionGoWhere Practice Paper (AI) – Version 5 of 5
Section A: Multiple Choice (20 marks)
| Question | Answer | Explanation |
|---|---|---|
| 1 | B | A Brønsted–Lowry base is defined as a proton (H⁺) acceptor. Option A describes a Brønsted–Lowry acid. Option C describes an Arrhenius base. Option D describes a Lewis base. |
| 2 | B | For a strong monoprotic acid, [H⁺] = concentration of acid. If the acid were strong, pH = –log(0.010) = 2.0. Since the measured pH is 3.0, [H⁺] = 1.0 × 10⁻³ mol dm⁻³, which is less than the acid concentration, indicating partial dissociation – characteristic of a weak acid. |
| 3 | B | A buffer requires a weak acid and its conjugate base in significant amounts. Option B: excess CH₃COOH with NaOH produces a mixture of CH₃COOH (weak acid) and CH₃COONa (conjugate base). The pH of an acidic buffer is less than 7. Option D produces a basic buffer (pH > 7). |
| 4 | B | pKa = –log₁₀(Ka) = –log₁₀(1.8 × 10⁻⁵) = 4.74 |
| 5 | B | NH₄Cl dissolves to give NH₄⁺ and Cl⁻. NH₄⁺ is the conjugate acid of the weak base NH₃ and undergoes hydrolysis: NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺, producing an acidic solution. The other salts are from strong acid-strong base (neutral) or strong base-weak acid (basic). |
| 6 | B | n(HCl) = 0.100 × 20.0/1000 = 0.00200 mol. NaOH + HCl → NaCl + H₂O (1:1 ratio). n(NaOH) = 0.00200 mol. c(NaOH) = 0.00200 / (25.0/1000) = 0.0800 mol dm⁻³. |
| 7 | C | Weak acid-strong base titrations have an equivalence point pH > 7 (typically 8–10). Phenolphthalein changes colour in the pH range 8.3–10.0, which brackets the equivalence point. Methyl orange changes at pH 3.1–4.4, which is too acidic. |
| 8 | B | The conjugate base of an acid is the species formed when the acid donates one proton. H₂PO₄⁻ → HPO₄²⁻ + H⁺. Therefore, HPO₄²⁻ is the conjugate base. |
| 9 | C | The buffer contains CH₃COOH and CH₃COO⁻. Added H⁺ reacts with CH₃COO⁻: CH₃COO⁻ + H⁺ → CH₃COOH. The [H⁺] remains almost unchanged, so pH remains almost constant. |
| 10 | D | A strong acid fully dissociates, producing a high [H⁺], which means a low pH, not a high pH. All other options are correct properties of strong acids. |
| 11 | C | NaOH is a strong base: [OH⁻] = 0.0010 mol dm⁻³. pOH = –log(0.0010) = 3.0. pH = 14 – pOH = 14 – 3.0 = 11.0. |
| 12 | B | Al₂O₃ is amphoteric. With NaOH, it forms sodium aluminate: Al₂O₃ + 2NaOH + 3H₂O → 2Na[Al(OH)₄]. This is the correct balanced equation for the reaction in aqueous solution. |
| 13 | C | n(NaOH) = 0.50 / 40.0 = 0.0125 mol. Volume = 250 cm³ = 0.250 dm³. c = 0.0125 / 0.250 = 0.0500 mol dm⁻³. |
| 14 | B | For a weak acid: [H⁺] = √(Ka × c) = √(1.0 × 10⁻⁵ × 0.10) = √(1.0 × 10⁻⁶) = 1.0 × 10⁻³ mol dm⁻³. pH = –log(1.0 × 10⁻³) = 3.0. |
| 15 | C | The equivalence point is when the amounts of acid and base are stoichiometrically equivalent (moles of H⁺ = moles of OH⁻). The pH at equivalence depends on the strengths of the acid and base. The indicator end point should coincide with the equivalence point, but they are not the same concept. |
| 16 | A | [H⁺] = 10⁻ᵖᴴ = 10⁻⁴·⁵ = 3.16 × 10⁻⁵ ≈ 3.2 × 10⁻⁵ mol dm⁻³. |
| 17 | C | Al₂O₃ is amphoteric – it reacts with both acids and bases. Na₂O and MgO are basic oxides. SO₂ is an acidic oxide. |
| 18 | A | From Ka = [H⁺][A⁻]/[HA], rearranging: [H⁺] = Ka × [HA]/[A⁻]. This is the Henderson-Hasselbalch relationship. |
| 19 | A | CH₃COOH + KOH → CH₃COOK + H₂O. The salt is potassium ethanoate. |
| 20 | B | Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K. The units are mol² dm⁻⁶ because it is the product of two concentrations. |
Section B: Structured Questions (40 marks)
Question 21
(a) A weak acid is an acid that partially dissociates in water, existing in equilibrium with its ions. [1]
(b) H₂CO₃(aq) ⇌ HCO₃⁻(aq) + H⁺(aq) [1] [Award 1 mark for correct equation with equilibrium arrow and state symbols. Accept H₂CO₃(aq) + H₂O(l) ⇌ HCO₃⁻(aq) + H₃O⁺(aq)]
(c) Ka₁ = [HCO₃⁻][H⁺] / [H₂CO₃] [1] [Award 1 mark for correct expression. Do not penalise if [H₃O⁺] is used instead of [H⁺]. Do not accept inclusion of [H₂O].]
(d) H₂CO₃ ⇌ H⁺ + HCO₃⁻ Let [H⁺] = x mol dm⁻³ [1] Ka₁ = x² / (0.050 – x) ≈ x² / 0.050 (since Ka is very small, x << 0.050) [1] x² = 4.3 × 10⁻⁷ × 0.050 = 2.15 × 10⁻⁸ x = √(2.15 × 10⁻⁸) = 1.47 × 10⁻⁴ mol dm⁻³ [1] pH = –log₁₀(1.47 × 10⁻⁴) = 3.83 [1]
[Award marks as follows: 1 mark for setting up Ka expression; 1 mark for stating/using the assumption; 1 mark for correct [H⁺]; 1 mark for correct pH. Accept 3.83 or 3.8.]
(e) The first dissociation involves removing H⁺ from a neutral H₂CO₃ molecule. [1] The second dissociation involves removing H⁺ from the negatively charged HCO₃⁻ ion. It is more difficult to remove a positively charged proton from a negatively charged ion due to stronger electrostatic attraction. Therefore, Ka₂ is much smaller than Ka₁. [1]
[Total: 8 marks]
Question 22
(a) MgO(s) + 2HCl(aq) → MgCl₂(aq) + H₂O(l) [2] [1 mark for correct formulae; 1 mark for correct state symbols and balancing]
(b)(i) n(MgO) = 0.40 / 40.3 = 0.00993 mol ≈ 0.0099 mol [1]
(b)(ii) n(HCl) = 0.50 × 50.0/1000 = 0.0250 mol [1]
(b)(iii) From equation: 1 mol MgO reacts with 2 mol HCl. 0.00993 mol MgO requires 0.00993 × 2 = 0.01986 mol HCl. [1] Available HCl = 0.0250 mol, which is greater than 0.01986 mol. Therefore, HCl is in excess. [1]
(b)(iv) MgO is the limiting reagent. n(MgCl₂) = n(MgO) = 0.00993 mol [1] Mass of MgCl₂ = 0.00993 × 95.3 = 0.946 g ≈ 0.95 g [1]
(c) HCl is a strong acid that fully dissociates, giving a high concentration of H⁺ ions. [1] Ethanoic acid is a weak acid that partially dissociates, giving a lower concentration of H⁺ ions. [1] The lower [H⁺] in ethanoic acid means fewer effective collisions per unit time (slower rate) and less heat released as the acid dissociates during reaction. [1]
[Total: 10 marks]
Question 23
(a) A buffer solution is a solution that resists changes in pH when small amounts of acid or base are added. [1] It consists of a weak acid and its conjugate base (or a weak base and its conjugate acid) in significant concentrations. [1]
(b)(i) Total volume = 30.0 + 20.0 = 50.0 cm³ [CH₃COOH] = (0.20 × 30.0/1000) / (50.0/1000) = 0.12 mol dm⁻³ [1]
(b)(ii) [CH₃COO⁻] = (0.15 × 20.0/1000) / (50.0/1000) = 0.060 mol dm⁻³ [1]
(b)(iii) [H⁺] = Ka × [CH₃COOH] / [CH₃COO⁻] [1] [H⁺] = (1.8 × 10⁻⁵) × (0.12 / 0.060) = 3.6 × 10⁻⁵ mol dm⁻³ [1] pH = –log₁₀(3.6 × 10⁻⁵) = 4.44 [1]
(c) The buffer contains CH₃COOH (weak acid) and CH₃COO⁻ (conjugate base). [1] When OH⁻ is added: OH⁻ + CH₃COOH → CH₃COO⁻ + H₂O. The added OH⁻ is neutralised by the weak acid. [1] The ratio [CH₃COOH]/[CH₃COO⁻] changes only slightly, so [H⁺] and pH remain almost constant. [1]
[Total: 10 marks]
Question 24
(a) HA is a weak acid. [1] Evidence: The initial pH is approximately 3.0, which is higher than expected for a strong acid of similar concentration (if HA were 0.10 mol dm⁻³ strong acid, pH would be 1.0). Also, the equivalence point pH is approximately 8.5 (>7), indicating the salt undergoes hydrolysis, characteristic of a weak acid-strong base titration. [1]
(b) At equivalence point: n(NaOH) = 0.100 × 25.0/1000 = 0.00250 mol [1] Since HA is monoprotic, n(HA) = 0.00250 mol. c(HA) = 0.00250 / (25.0/1000) = 0.100 mol dm⁻³ [1]
(c) At the equivalence point, all the weak acid HA has been converted to its conjugate base A⁻. [1] A⁻ undergoes hydrolysis: A⁻ + H₂O ⇌ HA + OH⁻, producing OH⁻ ions and making the solution alkaline (pH > 7). [1]
(d) Phenolphthalein (pH range 8.3–10.0). [1] The equivalence point pH is approximately 8.5, which falls within the colour change range of phenolphthalein, ensuring a sharp colour change at the end point. [1]
(e) The curve should show:
- Initial pH ≈ 1.0 (for 0.100 mol dm⁻³ strong acid) [1]
- Gradual rise, then very sharp vertical rise from pH ~3 to ~11 around 25.0 cm³ [1]
- Equivalence point at 25.0 cm³, pH = 7.0 (clearly labelled) [1]
- Final pH approaching ~12.5
[Total: 12 marks]
Section C: Free Response Questions (20 marks)
Question 25
(a)(i) H₂SO₃(aq) ⇌ HSO₃⁻(aq) + H⁺(aq) [1] [Accept H₂SO₃(aq) + H₂O(l) ⇌ HSO₃⁻(aq) + H₃O⁺(aq)]
(a)(ii) H₂SO₃ ⇌ H⁺ + HSO₃⁻ Let [H⁺] = x mol dm⁻³ [1] Ka₁ = [H⁺][HSO₃⁻] / [H₂SO₃] = x² / (0.010 – x) [1] Since Ka₁ is relatively large (1.5 × 10⁻²), the assumption x << 0.010 may not be valid. However, using the quadratic formula or approximation: If we assume x << 0.010: x² = 1.5 × 10⁻² × 0.010 = 1.5 × 10⁻⁴; x = 0.0122 mol dm⁻³ (which is >0.010, so assumption invalid). [1] Solving quadratic: x² + 1.5 × 10⁻²x – 1.5 × 10⁻⁴ = 0 x = [–0.015 + √(0.015² + 4 × 1.5 × 10⁻⁴)] / 2 = [–0.015 + √(2.25 × 10⁻⁴ + 6.0 × 10⁻⁴)] / 2 x = [–0.015 + √(8.25 × 10⁻⁴)] / 2 = [–0.015 + 0.0287] / 2 = 0.00685 mol dm⁻³ [1] pH = –log₁₀(0.00685) = 2.16 [1]
[Award marks: 1 for Ka expression; 1 for recognising assumption may be invalid; 1 for correct method (quadratic or successive approximation); 1 for correct pH. Accept 2.16 or 2.2.]
(b)(i) Acid rain: [H⁺] = 10⁻⁴·⁰ = 1.0 × 10⁻⁴ mol dm⁻³ [1] Unpolluted rain: [H⁺] = 10⁻⁵·⁶ = 2.51 × 10⁻⁶ mol dm⁻³ Ratio = (1.0 × 10⁻⁴) / (2.51 × 10⁻⁶) = 39.8 ≈ 40 [1] [The acid rain has approximately 40 times the hydrogen ion concentration of unpolluted rainwater.]
(b)(ii) CaCO₃(s) + H₂SO₄(aq) → CaSO₄(aq) + H₂O(l) + CO₂(g) [2] [1 mark for correct formulae; 1 mark for correct state symbols and balancing]
(c) Environmental impacts:
- Lowered pH of lakes and rivers kills fish and other aquatic organisms [1]
- Mobilises toxic metal ions (e.g., Al³⁺) from soils, which damage fish gills [1]
- Disrupts aquatic food chains and reduces biodiversity
Method to reduce acid rain formation:
- Install scrubbers in factory chimneys to remove SO₂ from emissions before release (e.g., using limestone slurry or alkaline solutions) [1] [Accept other valid methods: catalytic converters to reduce NOₓ emissions, use of low-sulfur fuels, alternative energy sources]
[Total: 20 marks]
Question 26
(a)(i) CH₃COOH(aq) ⇌ CH₃COO⁻(aq) + H⁺(aq) [1] [Accept CH₃COOH(aq) + H₂O(l) ⇌ CH₃COO⁻(aq) + H₃O⁺(aq)]
(a)(ii) CH₃COOH ⇌ CH₃COO⁻ + H⁺ Let [H⁺] = x mol dm⁻³ [1] Ka = [CH₃COO⁻][H⁺] / [CH₃COOH] = x² / (0.10 – x) [1] Assumption: x << 0.10 since Ka is very small (1.8 × 10⁻⁵). [1] x² ≈ 1.8 × 10⁻⁵ × 0.10 = 1.8 × 10⁻⁶ x = √(1.8 × 10⁻⁶) = 1.34 × 10⁻³ mol dm⁻³ pH = –log₁₀(1.34 × 10⁻³) = 2.87 [1]
[Award marks: 1 for Ka expression; 1 for stating assumption; 1 for correct [H⁺]; 1 for correct pH. Accept 2.87 or 2.9.]
(b)(i) CH₃COOH(aq) + NaOH(aq) → CH₃COONa(aq) + H₂O(l) [1]
(b)(ii) n(CH₃COOH) = 0.10 × 25.0/1000 = 0.00250 mol [1] From equation, n(NaOH) = 0.00250 mol. Volume of NaOH = 0.00250 / 0.20 = 0.0125 dm³ = 12.5 cm³ [1]
(c)(i) The mixture contains a weak acid (CH₃COOH) and its conjugate base (CH₃COO⁻ from CH₃COONa). [1] Added H⁺ reacts with CH₃COO⁻: CH₃COO⁻ + H⁺ → CH₃COOH. Added OH⁻ reacts with CH₃COOH: CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O. In both cases, the pH remains almost unchanged. [1]
(c)(ii) n(CH₃COONa) = 0.82 / 82.0 = 0.0100 mol [1] Volume of solution = 100 cm³ = 0.100 dm³ [CH₃COO⁻] = 0.0100 / 0.100 = 0.100 mol dm⁻³ [1] [CH₃COOH] = 0.10 mol dm⁻³ [H⁺] = Ka × [CH₃COOH] / [CH₃COO⁻] = (1.8 × 10⁻⁵) × (0.10 / 0.100) = 1.8 × 10⁻⁵ mol dm⁻³ [1] pH = –log₁₀(1.8 × 10⁻⁵) = 4.74 [1]
(d) Comparison of 0.10 mol dm⁻³ CH₃COOH and 0.10 mol dm⁻³ HCl:
| Property | CH₃COOH | HCl | Explanation |
|---|---|---|---|
| pH | ~2.9 | ~1.0 | HCl fully dissociates; CH₃COOH partially dissociates, giving lower [H⁺] |
| Electrical conductivity | Lower | Higher | Fewer ions in CH₃COOH solution due to partial dissociation |
| Rate of reaction with Mg | Slower | Faster | Lower [H⁺] in CH₃COOH means fewer effective collisions per unit time |
[3 marks – 1 for each correct comparison with explanation]
[Total: 20 marks]
END OF ANSWER KEY