AQA A-Level Chemistry: Acids and bases (7405) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 3.1.12 Acids and bases of the AQA AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for exams from June 2016 onwards: 3.1.12.1 to 3.1.12.6 and Required practical 9. All of it is A-level only and is placed on A-level Paper 1; Paper 3 can test any content.

Learn the ideas first in the acids and bases study guide and the acids and bases revision notes. The course hub is AQA A-Level Chemistry, and the printable checklist lists every specification point. Use Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K unless a question says otherwise. Give pH values to 2 decimal places.

Questions

1. Ammonia reacts with water: NH₃ + H₂O ⇌ NH₄⁺ + OH⁻. Identify the species acting as a Brønsted–Lowry acid in the forward reaction and in the reverse reaction. [2]

2. Calculate:

(a) the pH of a solution with [H⁺] = 4.5 × 10⁻⁴ mol dm⁻³ [1] (b) [H⁺] in a solution of pH 9.72 [1] (c) the pH of 0.0063 mol dm⁻³ hydrochloric acid. [1]

3. This question is about the ionic product of water.

(a) Write an expression for Kw. [1] (b) Calculate the pH of 0.0350 mol dm⁻³ KOH at 298 K. [2] (c) At a temperature above 298 K, Kw = 3.80 × 10⁻¹⁴ mol² dm⁻⁶. Calculate the pH of pure water at this temperature and state whether the water is acidic, neutral or alkaline. [2] (d) Use the data in (c) to deduce whether the dissociation of water is exothermic or endothermic. [1]

4. HA is a weak monoprotic acid with Ka = 6.30 × 10⁻⁵ mol dm⁻³.

(a) Write an expression for Ka. [1] (b) Calculate the pH of a 0.250 mol dm⁻³ solution of HA. [2] (c) State one assumption you made in (b). [1]

5. A 0.0450 mol dm⁻³ solution of a weak acid HB has pH 3.62. Calculate Ka, with units, and pKa for HB. [4]

6. Acid X has pKa = 3.86. Acid Y has Ka = 2.4 × 10⁻⁵ mol dm⁻³. Calculate Ka for X and pKa for Y, and state which is the stronger acid. [3]

7. A student titrates 25.0 cm³ of a weak monoprotic acid with 0.0850 mol dm⁻³ NaOH. The titres are 22.60 cm³ (rough), 22.15 cm³, 22.10 cm³ and 22.20 cm³.

(a) Calculate the mean titre. [1] (b) Calculate the concentration of the acid. [3]

8. Three indicators change colour over these ranges: methyl orange pH 3.1–4.4; bromothymol blue pH 6.0–7.6; phenolphthalein pH 8.3–10.0.

(a) Choose an indicator for titrating a weak acid with a strong base, and explain your choice. [2] (b) Choose an indicator for titrating a strong acid with a weak base, and explain your choice. [2] (c) Explain why none is suitable for a weak acid–weak base titration. [2]

9. In Required practical 9, a student adds 0.100 mol dm⁻³ NaOH to 25.0 cm³ of a weak acid HX and plots a pH curve. The equivalence point is at 24.0 cm³. The pH at 12.0 cm³ is 4.20.

(a) Describe how the student should collect the data for the pH curve. [3] (b) Calculate Ka for HX. [2] (c) Explain why the pH at 12.0 cm³ is equal to pKa. [2]

10. 30.0 cm³ of 0.200 mol dm⁻³ NaOH is mixed with 25.0 cm³ of 0.180 mol dm⁻³ HCl at 298 K. Calculate the pH of the mixture. [5]

11. This question is about buffers.

(a) A buffer contains ethanoic acid and sodium ethanoate. Explain how it resists a fall in pH when a small amount of acid is added. [2] (b) Explain how the same buffer resists a rise in pH when a small amount of alkali is added. [2] (c) A basic buffer contains ammonia and ammonium chloride. Write an equation for the reaction that removes added H⁺ and one for the reaction that removes added OH⁻. [2]

12. A weak acid HA has Ka = 1.50 × 10⁻⁵ mol dm⁻³. 15.0 cm³ of 0.200 mol dm⁻³ NaOH is added to 40.0 cm³ of 0.250 mol dm⁻³ HA.

(a) Calculate the pH of the buffer formed. [4] (b) 1.00 cm³ of 0.500 mol dm⁻³ HCl is added to the buffer. Calculate the new pH. [3] (c) Calculate the pH if the same HCl were added to 55.0 cm³ of water instead, and comment on the difference. [2]

Answers

1. Forward: H₂O donates a proton to NH₃ [1]. Reverse: NH₄⁺ donates a proton to OH⁻ [1]. [2] Examiner insight: Each direction is a separate mark, so naming only water earns half the credit.

2. (a) pH = −log₁₀(4.5 × 10⁻⁴) = 3.35 [1] (b) [H⁺] = 10^(−9.72) = 1.91 × 10⁻¹⁰ mol dm⁻³ [1] (c) HCl is fully dissociated, so [H⁺] = 0.0063; pH = 2.20 [1] Examiner insight: A pH given as 2.2 rather than 2.20 can lose the mark when 2 decimal places are asked for.

3. (a) Kw = [H⁺][OH⁻] [1] (b) [H⁺] = 1.00 × 10⁻¹⁴ / 0.0350 = 2.86 × 10⁻¹³ mol dm⁻³ [1]; pH = 12.54 [1] (c) [H⁺] = √(3.80 × 10⁻¹⁴) = 1.95 × 10⁻⁷; pH = 6.71 [1]. Neutral, because [H⁺] = [OH⁻] [1] (d) Kw increases with temperature, so the equilibrium shifts right on heating: endothermic [1] Examiner insight: “Neutral” without the reason [H⁺] = [OH⁻] is unlikely to score the second mark in (c).

4. (a) Ka = [H⁺][A⁻] / [HA] [1] (b) [H⁺] = √(6.30 × 10⁻⁵ × 0.250) = 3.97 × 10⁻³ mol dm⁻³ [1]; pH = 2.40 [1] (c) Any one: [H⁺] = [A⁻]; or [HA] at equilibrium = 0.250 mol dm⁻³ because dissociation is slight [1] Examiner insight: An expression with [H₂O] on the bottom, or with square brackets missing, loses (a).

5. [H⁺] = 10^(−3.62) = 2.40 × 10⁻⁴ mol dm⁻³ [1] Ka = [H⁺]² / [HB] = (2.40 × 10⁻⁴)² / 0.0450 [1] Ka = 1.28 × 10⁻⁶ mol dm⁻³ [1] pKa = −log₁₀(1.28 × 10⁻⁶) = 5.89 [1] Examiner insight: Write [H⁺] down before squaring it; if it is wrong, the Ka and pKa steps here can still earn method credit.

6. Ka(X) = 10^(−3.86) = 1.38 × 10⁻⁴ mol dm⁻³ [1] pKa(Y) = −log₁₀(2.4 × 10⁻⁵) = 4.62 [1] X has the larger Ka (smaller pKa), so X is stronger [1] Examiner insight: The comparison mark needs a reason tied to Ka or pKa, not just the name.

7. (a) Mean of concordant titres = (22.15 + 22.10 + 22.20) / 3 = 22.15 cm³ [1] (b) n(NaOH) = 0.0850 × 22.15/1000 = 1.88 × 10⁻³ mol [1] n(acid) = 1.88 × 10⁻³ mol (1 : 1) [1] concentration = 1.88 × 10⁻³ / 0.0250 = 0.0753 mol dm⁻³ [1] Examiner insight: Including the rough titre in the mean costs the mark in (a), though later marks can follow through.

8. (a) Phenolphthalein [1]; its range lies within the steep section, which is in the alkaline region (equivalence above pH 7) [1] (b) Methyl orange [1]; its range lies within the steep section, which is in the acidic region (equivalence below pH 7) [1] (c) The curve has no steep (vertical) section [1], so any indicator changes colour gradually over a large volume and no sharp end point is seen [1] Examiner insight: “Changes colour at the right pH” is too vague; link the range to the steep section.

9. (a) Calibrate the pH meter with buffer solutions of known pH [1]. Add NaOH from a burette in small portions, swirling and recording the pH after each addition [1]. Use smaller additions near the end point, then plot pH against volume [1] (b) At 12.0 cm³ (half the equivalence volume), pH = pKa = 4.20 [1] Ka = 10^(−4.20) = 6.31 × 10⁻⁵ mol dm⁻³ [1] (c) Half of HX has been converted to X⁻, so [HX] = [X⁻] [1]. These cancel in Ka = [H⁺][X⁻]/[HX], so Ka = [H⁺] and pKa = pH [1] Examiner insight: Writing “pH = pKa” in (c) without saying [HX] = [X⁻] restates the answer and gains little.

10. n(OH⁻) = 0.200 × 0.0300 = 6.00 × 10⁻³ mol; n(H⁺) = 0.180 × 0.0250 = 4.50 × 10⁻³ mol [1] Excess OH⁻ = 1.50 × 10⁻³ mol [1] [OH⁻] = 1.50 × 10⁻³ / 0.0550 = 0.0273 mol dm⁻³ [1] [H⁺] = 1.00 × 10⁻¹⁴ / 0.0273 = 3.67 × 10⁻¹³ mol dm⁻³ [1] pH = 12.44 [1] Examiner insight: Dividing by 30.0 cm³ instead of the total 55.0 cm³ loses the third mark; the rest can follow through.

11. (a) The added H⁺ reacts with ethanoate ions: H⁺ + CH₃COO⁻ → CH₃COOH [1]. There is a large amount of CH₃COO⁻, so the ratio [CH₃COOH]/[CH₃COO⁻] changes little and pH stays nearly constant [1] (b) The added OH⁻ reacts with ethanoic acid: OH⁻ + CH₃COOH → CH₃COO⁻ + H₂O [1]. There is a large amount of CH₃COOH, so the ratio changes little [1] (c) NH₃ + H⁺ → NH₄⁺ [1]; NH₄⁺ + OH⁻ → NH₃ + H₂O [1] Examiner insight: Saying the buffer “neutralises” the acid without naming the species that reacts earns no credit.

12. (a) n(HA) start = 0.250 × 0.0400 = 0.0100 mol; n(OH⁻) = 0.200 × 0.0150 = 3.00 × 10⁻³ mol [1] HA left = 7.00 × 10⁻³ mol; A⁻ = 3.00 × 10⁻³ mol [1] [H⁺] = 1.50 × 10⁻⁵ × 7.00/3.00 = 3.50 × 10⁻⁵ mol dm⁻³ [1] pH = 4.46 [1] (b) n(H⁺) added = 0.500 × 0.00100 = 5.00 × 10⁻⁴ mol; HA = 7.50 × 10⁻³ mol, A⁻ = 2.50 × 10⁻³ mol [1] [H⁺] = 1.50 × 10⁻⁵ × 7.50/2.50 = 4.50 × 10⁻⁵ mol dm⁻³ [1] pH = 4.35 [1] (c) Total volume 56.0 cm³: [H⁺] = 5.00 × 10⁻⁴ / 0.0560 = 8.93 × 10⁻³; pH = 2.05 [1]. The buffer pH falls by only 0.11, but in water the pH falls to about 2 [1] Examiner insight: In (b), adding H⁺ to [H⁺] instead of changing the moles of HA and A⁻ scores only the first mark.

Where marks are usually lost

• Treating a weak acid as fully dissociated, so [H⁺] = [HA].
• Using √(Ka[HA]) for a buffer, where [H⁺] ≠ [A⁻].
• Dividing excess moles by one solution’s volume instead of the total.
• Forgetting Kw when OH⁻ is in excess.
• Omitting units for Ka, or giving Kw in mol dm⁻³.
• Including the rough titre in the mean.
• Choosing an indicator without linking its range to the steep section.
• Saying hot pure water is acidic.
• Writing buffer explanations without equations or without naming the reacting species.

Next steps

• Acids and bases revision notes
• Acids and bases study guide
• Mole practice questions
• AQA A-Level Chemistry hub
• Printable checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

AQA, AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams June 2016 onwards: section 3.1.12 Acids and bases (3.1.12.1 to 3.1.12.6, A-level only), with Required practical 9.