OxfordAQA A-Level Chemistry: Acids and bases (9620) – 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 3.1.10, Acids and bases (3.1.10.1 to 3.1.10.6), of the OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards, including required practical 7. All questions are on International A2 material (International A-level only), assessed in Unit 3, with Unit 5 able to draw on any topic. Use Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 25 °C unless told otherwise, and give pH to 2 decimal places.

If a method is unfamiliar, go back to the study guide or the revision notes. The course hub, specification checklist and our diagnostics cover other topics.

Questions

1. This question is about Brønsted–Lowry theory.

(a) State what is meant by a Brønsted–Lowry acid. [1] (b) CH₃NH₃⁺ + CN⁻ ⇌ CH₃NH₂ + HCN. Name the Brønsted–Lowry acid on each side of this equation. [2]

2. Work out:

(a) the pH when [H⁺] = 2.6 × 10⁻⁵ mol dm⁻³ [1] (b) the hydrogen ion concentration at pH 12.35 [1] (c) the pH of 0.185 mol dm⁻³ hydrochloric acid [1] (d) the pH of a solution made by dissolving 0.81 g of hydrogen bromide (Mr = 80.9) in water and making it up to 250 cm³. [2]

3. Water ionises slightly: H₂O ⇌ H⁺ + OH⁻.

(a) Explain how the expression Kw = [H⁺][OH⁻] is obtained from the equilibrium constant for the dissociation of water. [2] (b) Barium hydroxide, Ba(OH)₂, is a strong base. Calculate the pH of 0.0225 mol dm⁻³ Ba(OH)₂ at 25 °C. [2] (c) At 75 °C, Kw = 2.0 × 10⁻¹³ mol² dm⁻⁶. What is the pH of pure water at 75 °C? Say whether it is acidic, neutral or alkaline. [2]

4. HY is a weak monoprotic acid. Ka = 3.40 × 10⁻⁵ mol dm⁻³.

(a) Write an expression for Ka for HY. [1] (b) Calculate the pH of a 0.0850 mol dm⁻³ solution of HY. [3] (c) State the two approximations you used in (b). [2]

5. A 0.0640 mol dm⁻³ solution of a weak monoprotic acid has pH 2.62. Calculate Ka, with its units, and pKa. [4]

6. Three weak acids have these values: acid R, pKa = 2.86; acid S, Ka = 6.6 × 10⁻⁴ mol dm⁻³; acid T, pKa = 3.40. Calculate Ka for R and pKa for S, then list the three acids from strongest to weakest. [3]

7. A student dissolves 1.22 g of a solid monoprotic acid and makes 250 cm³ of solution in a volumetric flask. 25.0 cm³ portions are titrated with 0.0480 mol dm⁻³ NaOH. The titres are 21.05 cm³ (rough), 20.70 cm³, 20.80 cm³ and 20.75 cm³. Calculate the relative molecular mass of the acid. [5]

8. 35.0 cm³ of 0.120 mol dm⁻³ nitric acid is mixed with 20.0 cm³ of 0.150 mol dm⁻³ NaOH at 25 °C. Calculate the pH of the mixture. [4]

9. Indicator ranges: bromocresol green 3.8–5.4; phenol red 6.4–8.0; thymolphthalein 9.3–10.5.

(a) Methanoic acid is titrated with KOH. Explain why the pH at the equivalence point is above 7. [2] (b) Choose an indicator from the list for the methanoic acid–KOH titration and explain your choice. [2] (c) Choose an indicator from the list for titrating hydrochloric acid with aqueous ammonia. [1] (d) Explain why none of the indicators suits a titration of methanoic acid with aqueous ammonia. [1]

10. In required practical 7, a student adds 0.100 mol dm⁻³ NaOH from a burette to 20.0 cm³ of a weak monoprotic acid HW and records the pH. The equivalence point is at 18.60 cm³. At 9.30 cm³ of NaOH, the pH is 4.98.

(a) Give two steps the student should take to get reliable pH readings that define the curve well. [2] (b) Calculate the concentration of HW. [2] (c) Determine Ka for HW. [2] (d) Use your answers to (b) and (c) to calculate the pH of the HW solution before any NaOH was added. [2] (e) The student then titrates a weak base with a strong acid. State how the pH at equivalence would compare with 7. [1]

11. This question is about buffers.

(a) A basic buffer contains methylamine, CH₃NH₂, and methylammonium chloride. Explain, with equations, how it keeps its pH almost steady when a little acid or alkali is added. [3] (b) A buffer is made by dissolving 2.04 g of sodium methanoate (Mr = 68.0) in 200 cm³ of 0.250 mol dm⁻³ methanoic acid. Ka for methanoic acid is 1.80 × 10⁻⁴ mol dm⁻³. Calculate the pH of the buffer. [3] (c) 5.0 cm³ of 1.00 mol dm⁻³ NaOH is added to the buffer in (b). Calculate the new pH. [3] (d) Explain why adding water to the buffer in (b) does not change its pH. [1]

Answers

1. (a) A proton (H⁺) donor [1] (b) Forward: CH₃NH₃⁺ [1]; reverse: HCN [1] Examiner insight: Give the formulae exactly as written in the equation, not a vague “the acid on the right”.

2. (a) pH = −log₁₀(2.6 × 10⁻⁵) = 4.59 [1] (b) [H⁺] = 10^(−12.35) = 4.47 × 10⁻¹³ mol dm⁻³ [1] (c) [H⁺] = 0.185 mol dm⁻³, pH = 0.73 [1] (d) n(HBr) = 0.81 / 80.9 = 0.0100 mol, so [H⁺] = 0.0100 / 0.250 = 0.0400 mol dm⁻³ [1]; pH = 1.40 [1] Examiner insight: A pH below 1 is perfectly possible for a concentrated strong acid; do not “correct” it to a positive whole number.

3. (a) Kc = [H⁺][OH⁻] / [H₂O] [1]; so little water dissociates that [H₂O] is effectively constant, and Kw = Kc × [H₂O] = [H⁺][OH⁻] [1] (b) [OH⁻] = 2 × 0.0225 = 0.0450 mol dm⁻³ [1]; [H⁺] = 1.00 × 10⁻¹⁴ / 0.0450 = 2.22 × 10⁻¹³, pH = 12.65 [1] (c) [H⁺] = √(2.0 × 10⁻¹³) = 4.47 × 10⁻⁷, pH = 6.35 [1]; neutral, because [H⁺] = [OH⁻] [1] Examiner insight: In (b), forgetting the two OH⁻ per Ba(OH)₂ gives 12.35; in (c), give the reason for “neutral”, not just the word.

4. (a) Ka = [H⁺][Y⁻] / [HY] [1] (b) [H⁺]² = 3.40 × 10⁻⁵ × 0.0850 = 2.89 × 10⁻⁶ [1]; [H⁺] = 1.70 × 10⁻³ mol dm⁻³ [1]; pH = 2.77 [1] (c) [H⁺] = [Y⁻] (H⁺ from water ignored) [1]; [HY] at equilibrium equals 0.0850 mol dm⁻³, as dissociation is so slight [1] Examiner insight: Use square brackets for concentrations and include Y⁻ in the numerator.

5. [H⁺] = 10^(−2.62) = 2.40 × 10⁻³ mol dm⁻³ [1]; Ka = (2.40 × 10⁻³)² / 0.0640 [1]; Ka = 8.99 × 10⁻⁵ mol dm⁻³ [1]; pKa = 4.05 [1] Examiner insight: Carry the unrounded [H⁺] forward when you square it; rounding to 2.4 × 10⁻³ first shifts Ka in the third figure.

6. Ka(R) = 10^(−2.86) = 1.38 × 10⁻³ mol dm⁻³ [1]; pKa(S) = −log₁₀(6.6 × 10⁻⁴) = 3.18 [1]; strongest to weakest: R > S > T (lowest pKa is strongest) [1] Examiner insight: Convert all three to the same form before ranking, or the order often comes out reversed.

7. Mean of concordant titres = (20.70 + 20.80 + 20.75) / 3 = 20.75 cm³ [1]; n(NaOH) = 0.0480 × 20.75 / 1000 = 9.96 × 10⁻⁴ mol [1]; n(acid) in 25.0 cm³ = 9.96 × 10⁻⁴ mol (1 : 1) [1]; n(acid) in 250 cm³ = 9.96 × 10⁻³ mol [1]; Mr = 1.22 / 9.96 × 10⁻³ = 122 [1] Examiner insight: Including the rough titre, or forgetting to scale from 25.0 cm³ to 250 cm³, gives a wrong Mr.

8. n(H⁺) = 0.120 × 0.0350 = 4.20 × 10⁻³ mol; n(OH⁻) = 0.150 × 0.0200 = 3.00 × 10⁻³ mol [1]; excess H⁺ = 1.20 × 10⁻³ mol [1]; [H⁺] = 1.20 × 10⁻³ / 0.0550 = 0.0218 mol dm⁻³ [1]; pH = 1.66 [1] Examiner insight: Dividing by 35.0 cm³ instead of the total 55.0 cm³ gives pH 1.47; show the total volume clearly so the method is visible.

9. (a) At equivalence the solution contains methanoate ions, HCOO⁻ [1]; these are a base and accept protons from water, HCOO⁻ + H₂O ⇌ HCOOH + OH⁻, so [OH⁻] > [H⁺] [1] (b) Thymolphthalein [1]; its whole range (9.3–10.5) lies within the vertical section, which is in the alkaline region for a weak acid–strong base titration [1] (c) Bromocresol green [1] (d) A weak acid–weak base curve has no vertical section, so no indicator changes colour sharply at one drop [1] Examiner insight: In (b), say the whole range sits inside the steep section, not just “it changes in alkali”.

10. (a) Calibrate the pH meter with buffers of known pH [1]; add NaOH in small portions (0.10–0.20 cm³) near the end point, stirring before each reading [1] (b) n(NaOH) = 0.100 × 18.60 / 1000 = 1.86 × 10⁻³ mol = n(HW) [1]; [HW] = 1.86 × 10⁻³ / 0.0200 = 0.0930 mol dm⁻³ [1] (c) 9.30 cm³ is half the equivalence volume, so [HW] = [W⁻] and pH = pKa = 4.98 [1]; Ka = 10^(−4.98) = 1.05 × 10⁻⁵ mol dm⁻³ [1] (d) [H⁺] = √(1.05 × 10⁻⁵ × 0.0930) = 9.88 × 10⁻⁴ mol dm⁻³ [1]; pH = 3.01 [1] (e) Below 7 [1] Examiner insight: In (c), state why pH equals pKa at half-neutralisation; a bare Ka value without that link may lose the reasoning mark.

11. (a) Added acid: CH₃NH₂ + H⁺ → CH₃NH₃⁺ [1]; added alkali: CH₃NH₃⁺ + OH⁻ → CH₃NH₂ + H₂O [1]; both CH₃NH₂ and CH₃NH₃⁺ are present in large amounts, so their ratio and the pH change very little [1] (b) n(HCOO⁻) = 2.04 / 68.0 = 0.0300 mol [1]; n(HCOOH) = 0.250 × 0.200 = 0.0500 mol [1]; [H⁺] = 1.80 × 10⁻⁴ × 0.0500 / 0.0300 = 3.00 × 10⁻⁴, pH = 3.52 [1] (c) OH⁻ (5.0 × 10⁻³ mol) converts HCOOH to HCOO⁻: n(HCOOH) = 0.0450 mol, n(HCOO⁻) = 0.0350 mol [1]; [H⁺] = 1.80 × 10⁻⁴ × 0.0450 / 0.0350 = 2.31 × 10⁻⁴ [1]; pH = 3.64 [1] (d) Dilution lowers [HCOOH] and [HCOO⁻] by the same factor, so the ratio, and [H⁺], stay the same [1] Examiner insight: In (c), work in moles and change both amounts, not just one.

Where marks are usually lost

• Using a hydroxide’s formula concentration as [OH⁻] when each unit releases two OH⁻.
• Writing Kw with [H₂O] still in it, or with the wrong units.
• Using √(Ka × [HA]) for a buffer, where [H⁺] does not equal [A⁻].
• Finding the leftover concentration with one solution’s volume rather than the mixture’s.
• Choosing an indicator because one end of its range falls in the steep section.
• Leaving out the units of Ka (mol dm⁻³) when a question asks for them.

Next steps

• Revision notes for this section
• Study guide for this section
• Equilibria practice
• OxfordAQA A-Level Chemistry hub
• Printable OxfordAQA A-Level Chemistry checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards, section 3.1.10 Acids and bases (International A2), with required practical 7. Published by OxfordAQA.