A Level H1 Chemistry Practice Paper 5

TuitionGoWhere Practice Paper – Chemistry H1 A-Level

Answer Key & Marking Scheme

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

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Section A: Short Answer & Definitions (12 marks)

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1. Define the term Brønsted–Lowry acid. Give one example of a Brønsted–Lowry acid and write an equation to show its behaviour in aqueous solution.

[2 marks]

Answer:

• A Brønsted–Lowry acid is a proton (H⁺) donor. [1]
• Example: HCl (or any suitable acid). [½]
• Equation: HCl(aq) → H⁺(aq) + Cl⁻(aq) OR HCl(aq) + H₂O(l) → H₃O⁺(aq) + Cl⁻(aq) [½]

Accept any correct acid and corresponding dissociation equation with state symbols.

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2. What is meant by the term weak acid? Illustrate your answer with a balanced equation, including state symbols, for the dissociation of ethanoic acid in water.

[2 marks]

Answer:

• A weak acid is one that only partially dissociates/ionises in aqueous solution. [1]
• Equation: CH₃COOH(aq) ⇌ CH₃COO⁻(aq) + H⁺(aq) [1]

Essential: reversible arrow (⇌) and state symbols. Accept H₃O⁺ instead of H⁺.

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3. A student measures the pH of 0.10 mol dm⁻³ hydrochloric acid and finds it to be 1.0. The pH of 0.10 mol dm⁻³ ethanoic acid is measured as 2.9. Explain why these two acids of the same concentration have different pH values.

[2 marks]

Answer:

• HCl is a strong acid; it dissociates completely in water, so [H⁺] = 0.10 mol dm⁻³, giving pH = 1.0. [1]
• CH₃COOH is a weak acid; it dissociates only partially, so [H⁺] is much less than 0.10 mol dm⁻³, giving a higher pH of 2.9. [1]

Accept: reference to degree of dissociation/ionisation; equilibrium position for weak acid.

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4. State the formula of the conjugate base of each of the following acids:

(a) H₂SO₄ (b) H₂PO₄⁻

[2 marks]

Answer:

• (a) HSO₄⁻ [1]
• (b) HPO₄²⁻ [1]

Conjugate base = acid minus one H⁺; charge decreases by 1.

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5. Calcium oxide (CaO) is classified as a basic oxide. Write a balanced equation, including state symbols, for the reaction of calcium oxide with:

(a) dilute hydrochloric acid (b) water

[2 marks]

Answer:

• (a) CaO(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) [1]
• (b) CaO(s) + H₂O(l) → Ca(OH)₂(aq/s) [1]

Accept Ca(OH)₂ as (aq) or (s); state symbols essential.

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6. Identify the Period 3 element that forms an amphoteric oxide. Write one balanced equation to show this oxide reacting with an acid, and one balanced equation to show it reacting with a base.

[2 marks]

Answer:

• Element: Aluminium / Al [½]
• With acid: Al₂O₃(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂O(l) [¾]
• With base: Al₂O₃(s) + 2NaOH(aq) + 3H₂O(l) → 2Na[Al(OH)₄](aq) [¾]

Accept Al₂O₃ + 2NaOH + 3H₂O → 2NaAl(OH)₄ or Al₂O₃ + 2NaOH → 2NaAlO₂ + H₂O. State symbols required for full marks.

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Section B: Calculations (24 marks)

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7. Calculate the pH of a 0.0250 mol dm⁻³ solution of nitric acid, HNO₃.

[2 marks]

Answer:

• HNO₃ is a strong monoprotic acid → [H⁺] = 0.0250 mol dm⁻³ [1]
• pH = −log₁₀[H⁺] = −log₁₀(0.0250) = 1.60 (to 2 d.p.) [1]

Method mark for using pH = −log[H⁺]; accuracy mark for correct answer.

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8. A solution of sodium hydroxide has a pH of 12.30 at 25 °C. Calculate the concentration of hydroxide ions, [OH⁻], in this solution. (Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 25 °C)

[2 marks]

Answer:

• pOH = 14.00 − pH = 14.00 − 12.30 = 1.70 [1]
• [OH⁻] = 10⁻ᵖᴼᴴ = 10⁻¹·⁷⁰ = 0.0200 mol dm⁻³ (or 2.00 × 10⁻² mol dm⁻³) [1]

Alternative: [H⁺] = 10⁻¹²·³⁰ = 5.01 × 10⁻¹³; [OH⁻] = Kw/[H⁺] = 1.00 × 10⁻¹⁴ / 5.01 × 10⁻¹³ = 0.0200 mol dm⁻³.

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9. A 0.150 mol dm⁻³ solution of a weak monoprotic acid, HA, has a pH of 2.85 at 25 °C.

(a) Calculate the concentration of hydrogen ions, [H⁺], in this solution. [1 mark]

Answer:

• [H⁺] = 10⁻²·⁸⁵ = 1.41 × 10⁻³ mol dm⁻³ [1]

(b) Write the expression for the acid dissociation constant, Ka, of HA. [1 mark]

Answer:

• Ka = [H⁺][A⁻] / [HA] [1]

Must have correct species; accept [H₃O⁺] for [H⁺].

(c) Calculate the value of Ka for HA, stating its units. [2 marks]

Answer:

• [H⁺] = [A⁻] = 1.41 × 10⁻³ mol dm⁻³; [HA] ≈ 0.150 − 1.41 × 10⁻³ ≈ 0.149 mol dm⁻³ [1]
• Ka = (1.41 × 10⁻³)² / 0.149 = 1.33 × 10⁻⁵ mol dm⁻³ [1]

Accept 1.33–1.34 × 10⁻⁵; units: mol dm⁻³.

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10. In a titration, 25.0 cm³ of 0.100 mol dm⁻³ sodium hydroxide solution required 20.0 cm³ of sulfuric acid for complete neutralisation.

2NaOH(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + 2H₂O(l)

Calculate the concentration of the sulfuric acid in mol dm⁻³.

[3 marks]

Answer:

• n(NaOH) = c × V = 0.100 × (25.0/1000) = 0.00250 mol [1]
• Mole ratio NaOH : H₂SO₄ = 2 : 1 → n(H₂SO₄) = 0.00250 / 2 = 0.00125 mol [1]
• c(H₂SO₄) = n / V = 0.00125 / (20.0/1000) = 0.0625 mol dm⁻³ [1]

Method marks for correct mole calculation and ratio; accuracy for final answer.

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11. A buffer solution is prepared by mixing 50.0 cm³ of 0.200 mol dm⁻³ ethanoic acid with 50.0 cm³ of 0.200 mol dm⁻³ sodium ethanoate. Ka = 1.74 × 10⁻⁵ mol dm⁻³.

(a) Calculate the concentration of ethanoic acid and ethanoate ions in the buffer solution after mixing. [2 marks]

Answer:

• Total volume = 100.0 cm³ (dilution factor ×2 for both) [1]
• [CH₃COOH] = 0.200 × (50.0/100.0) = 0.100 mol dm⁻³
• [CH₃COO⁻] = 0.200 × (50.0/100.0) = 0.100 mol dm⁻³ [1]

(b) Calculate the pH of this buffer solution. [2 marks]

Answer:

• Ka = [H⁺][CH₃COO⁻] / [CH₃COOH] → [H⁺] = Ka × [CH₃COOH] / [CH₃COO⁻] [1]
• [H⁺] = 1.74 × 10⁻⁵ × (0.100/0.100) = 1.74 × 10⁻⁵ mol dm⁻³
• pH = −log₁₀(1.74 × 10⁻⁵) = 4.76 [1]

Accept pH = pKa = 4.76 when [acid] = [salt].

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12. A student dissolved 2.65 g of anhydrous Na₂CO₃ in distilled water and made the solution up to 250.0 cm³. (Mr = 106.0)

(a) Calculate the amount, in moles, of Na₂CO₃ dissolved. [1 mark]

Answer:

• n = m / Mr = 2.65 / 106.0 = 0.0250 mol [1]

(b) Calculate the concentration of the sodium carbonate solution in mol dm⁻³. [1 mark]

Answer:

• c = n / V = 0.0250 / (250.0/1000) = 0.100 mol dm⁻³ [1]

(c) Calculate the concentration of sodium ions, [Na⁺], in this solution. [1 mark]

Answer:

• Na₂CO₃ → 2Na⁺ + CO₃²⁻ → [Na⁺] = 2 × 0.100 = 0.200 mol dm⁻³ [1]

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13. The pH of rainwater is 5.60. H₂CO₃(aq) ⇌ H⁺(aq) + HCO₃⁻(aq); Ka = 4.30 × 10⁻⁷ mol dm⁻³. Calculate [H₂CO₃] in the rainwater.

[3 marks]

Answer:

• [H⁺] = 10⁻⁵·⁶⁰ = 2.51 × 10⁻⁶ mol dm⁻³ [1]
• [H⁺] = [HCO₃⁻] = 2.51 × 10⁻⁶ mol dm⁻³ [1]
• Ka = [H⁺][HCO₃⁻] / [H₂CO₃] → [H₂CO₃] = (2.51 × 10⁻⁶)² / (4.30 × 10⁻⁷) = 1.47 × 10⁻⁵ mol dm⁻³ [1]

Accept 1.46–1.47 × 10⁻⁵ mol dm⁻³.

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14. A 0.500 g sample of impure monoprotic weak acid HX (Mr = 122.0) was titrated with 0.100 mol dm⁻³ NaOH. Titre = 28.50 cm³.

(a) Calculate the amount, in moles, of NaOH used. [1 mark]

Answer:

• n(NaOH) = 0.100 × (28.50/1000) = 0.00285 mol [1]

(b) Calculate the mass of pure HX in the sample. [1 mark]

Answer:

• n(HX) = n(NaOH) = 0.00285 mol (1:1 ratio)
• mass = n × Mr = 0.00285 × 122.0 = 0.348 g (or 0.3477 g) [1]

(c) Calculate the percentage purity of the HX sample. [1 mark]

Answer:

• % purity = (0.348 / 0.500) × 100 = 69.6% (or 69.5%) [1]

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Section C: Structured & Data Interpretation (14 marks)

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15. pH data for four 0.10 mol dm⁻³ solutions.

(a) Explain why solution W (HCl, pH 1.0) has a lower pH than solution X (CH₃COOH, pH 2.9). [1 mark]

Answer:

• HCl is a strong acid and dissociates completely, producing a higher [H⁺] than the weak acid CH₃COOH, which only partially dissociates. Lower pH corresponds to higher [H⁺]. [1]

(b) Explain why solution Y (NaOH, pH 13.0) has a higher pH than solution Z (NH₃, pH 11.1). [1 mark]

Answer:

• NaOH is a strong base and dissociates completely, producing a higher [OH⁻] (and thus lower [H⁺], higher pH) than the weak base NH₃, which only partially ionises in water. [1]

(c) Solution X and solution Z are mixed in equimolar proportions. State the type of solution formed and explain how it resists changes in pH when small amounts of acid are added. [2 marks]

Answer:

• Type of solution: Buffer solution (specifically, an acidic buffer). [1]
• Explanation: The buffer contains CH₃COOH (weak acid) and CH₃COO⁻ (conjugate base, from reaction with NH₃). When acid (H⁺) is added, the conjugate base CH₃COO⁻ reacts with it: CH₃COO⁻ + H⁺ → CH₃COOH, removing added H⁺ and resisting pH change. [1]

Accept: reference to equilibrium CH₃COOH ⇌ CH₃COO⁻ + H⁺ shifting left upon addition of H⁺.

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16. Marble chips + HCl reaction.

(a) Calculate the amount, in moles, of HCl used. [1 mark]

Answer:

• n(HCl) = 0.500 × (50.0/1000) = 0.0250 mol [1]

(b) Calculate the volume of CO₂ gas produced at r.t.p. (Molar volume = 24.0 dm³ mol⁻¹). [2 marks]

Answer:

• Mole ratio: 2HCl : 1CO₂ → n(CO₂) = 0.0250 / 2 = 0.0125 mol [1]
• Volume = n × 24.0 = 0.0125 × 24.0 = 0.300 dm³ (or 300 cm³) [1]

(c) Suggest why the actual volume of gas collected may be less than the calculated value. [1 mark]

Answer:

• Some CO₂ may dissolve in the water/aqueous solution. OR gas may escape before collection. OR reaction may not go to completion. [1]

Accept any reasonable suggestion.

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17. Titration curve analysis (weak acid vs. strong base).

(a) Use the graph to estimate the pH at the equivalence point. [1 mark]

Answer:

• pH ≈ 8.5 (accept 8.0–9.0 based on described curve) [1]

(b) Explain why the pH at the equivalence point is greater than 7. [1 mark]

Answer:

• At the equivalence point, the solution contains the salt of a weak acid and strong base (e.g., NaA). The anion A⁻ undergoes hydrolysis: A⁻ + H₂O ⇌ HA + OH⁻, producing OH⁻ ions and making the solution alkaline (pH > 7). [1]

(c) Suggest one suitable indicator for this titration, giving a reason for your choice. [2 marks]

Answer:

• Indicator: Phenolphthalein (pH range 8.2–10.0) OR Thymol blue (pH range 8.0–9.6). [1]
• Reason: The indicator's pH range falls within the steep/vertical portion of the titration curve (pH ~6–10), so the colour change occurs at the equivalence point. [1]

Accept any indicator with range within the vertical section.

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18. Ammonium chloride, NH₄Cl, dissolves in water to form an acidic solution.

(a) Write an equation to show the dissociation of ammonium chloride in water. [1 mark]

Answer:

• NH₄Cl(s) → NH₄⁺(aq) + Cl⁻(aq) [1]

State symbols required.

(b) Write an equation to show why the resulting solution is acidic. [1 mark]

Answer:

• NH₄⁺(aq) + H₂O(l) ⇌ NH₃(aq) + H₃O⁺(aq) [1]

Accept NH₄⁺ ⇌ NH₃ + H⁺. Reversible arrow essential.

(c) Explain, using your equations, why ammonium chloride solution has a pH less than 7. [1 mark]

Answer:

• The NH₄⁺ ion acts as a Brønsted–Lowry acid, donating a proton to water to form H₃O⁺ (or H⁺). The presence of excess H⁺/H₃O⁺ ions makes the solution acidic (pH < 7). [1]

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19. Buffer preparation (pH 4.50) using CH₃COOH and CH₃COONa. Ka = 1.74 × 10⁻⁵ mol dm⁻³.

(a) Calculate the ratio [CH₃COO⁻] / [CH₃COOH] required. [2 marks]

Answer:

• [H⁺] = 10⁻⁴·⁵⁰ = 3.16 × 10⁻⁵ mol dm⁻³ [1]
• Ka = [H⁺][CH₃COO⁻] / [CH₃COOH] → [CH₃COO⁻]/[CH₃COOH] = Ka / [H⁺] = 1.74 × 10⁻⁵ / 3.16 × 10⁻⁵ = 0.551 [1]

Accept 0.55 or 0.551.

(b) Describe how the student could prepare 100 cm³ of this buffer solution in the laboratory, given 0.10 mol dm⁻³ solutions of both ethanoic acid and sodium ethanoate. [2 marks]

Answer:

• Ratio [salt]/[acid] = 0.551. Since both stock solutions have the same concentration (0.10 mol dm⁻³), the volume ratio V(salt) / V(acid) = 0.551. [1]
• Let V(acid) = x cm³, then V(salt) = 0.551x. Total volume = x + 0.551x = 100 → 1.551x = 100 → x = 64.5 cm³ (acid), V(salt) = 35.5 cm³.
• Method: Measure 64.5 cm³ of 0.10 mol dm⁻³ CH₃COOH and 35.5 cm³ of 0.10 mol dm⁻³ CH₃COONa using burettes/measuring cylinders, mix in a beaker, and make up to 100 cm³ with distilled water in a volumetric flask if needed. [1]

Accept any clear description with correct volumes; allow minor rounding differences.

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20. A 0.0100 mol dm⁻³ solution of Ba(OH)₂ is prepared.

(a) Write an equation for the complete dissociation of barium hydroxide in water. [1 mark]

Answer:

• Ba(OH)₂(s) → Ba²⁺(aq) + 2OH⁻(aq) [1]

State symbols required.

(b) Calculate the pH of this barium hydroxide solution at 25 °C. [2 marks]

Answer:

• [OH⁻] = 2 × 0.0100 = 0.0200 mol dm⁻³ [1]
• pOH = −log₁₀(0.0200) = 1.70 → pH = 14.00 − 1.70 = 12.30 [1]

Alternative: [H⁺] = Kw/[OH⁻] = 1.00 × 10⁻¹⁴ / 0.0200 = 5.00 × 10⁻¹³; pH = −log₁₀(5.00 × 10⁻¹³) = 12.30.

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END OF ANSWER KEY

Total: 50 marks