OCR A Level Chemistry A: Acids, bases and buffers (H432) – 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.

This set covers section 5.1.3, Acids, bases and buffers, from the H432 specification for OCR A Level Chemistry A (Version 3.1, May 2026; first assessment 2017). The questions reach every outcome from 5.1.3(a) to (o) in Module 5 of the full H432 A level. Questions 11 and 12 are extended problems.

If you get stuck, see the study guide or the revision notes.

Take Kw as 1.00 × 10⁻¹⁴ mol² dm⁻⁶. Every solution is at 298 K unless a question gives another temperature.

Questions

1. Methanoic acid reacts with hydrogenphosphate ions in this equilibrium: HCOOH + HPO₄²⁻ ⇌ HCOO⁻ + H₂PO₄⁻

(a) Define the term Brønsted–Lowry acid. [1] (b) Identify the two conjugate acid–base pairs, labelling the acid in each pair. [2]

2. State the basicity of each acid.

(a) H₃PO₄ [1] (b) ethanedioic acid, HOOC–COOH [1] (c) methanoic acid, HCOOH [1]

3. Dilute hydrochloric acid is added separately to four substances. For each one, give the ionic equation, including state symbols.

(a) iron filings, forming Fe²⁺ ions [1] (b) aqueous potassium carbonate [1] (c) solid magnesium oxide [1] (d) aqueous lithium hydroxide [1]

4. A technician pipettes 20.0 cm³ of 0.240 mol dm⁻³ nitric acid into a 500 cm³ volumetric flask and fills it to the mark using distilled water. Calculate the pH of the solution now in the flask. [3]

5. (a) Calculate the pH of 0.0315 mol dm⁻³ barium hydroxide, Ba(OH)₂. [3] (b) At 40 °C, Kw = 2.92 × 10⁻¹⁴ mol² dm⁻⁶. Find the pH of a sample of pure water held at 40 °C, and explain whether the sample is acidic, neutral or alkaline. [3]

6. Benzoic acid, C₆H₅COOH, is a weak monobasic acid with Ka = 6.46 × 10⁻⁵ mol dm⁻³.

(a) Calculate the pH of 0.0380 mol dm⁻³ benzoic acid. [3] (b) State the two approximations you used. [2]

7. A weak monobasic acid HR is made up at 0.160 mol dm⁻³, and the solution has pH 2.84. Calculate Ka and pKa for HR. [4]

8. A weak monobasic acid HW has Ka = 3.3 × 10⁻³ mol dm⁻³. A solution of HW has concentration 0.0200 mol dm⁻³.

(a) Use the usual weak acid approximations to calculate the pH. [2] (b) Calculate the percentage of HW that your answer to (a) implies has dissociated. Use it to explain why the approximation is unreliable for this acid. [3] (c) State whether the true pH is higher or lower than your answer to (a). [1]

9. A student adds 35.0 cm³ of 0.100 mol dm⁻³ sodium hydroxide to 60.0 cm³ of 0.150 mol dm⁻³ propanoic acid (Ka = 1.35 × 10⁻⁵ mol dm⁻³).

(a) Calculate the pH of the buffer solution formed. [4] (b) Explain how this buffer minimises the pH change when a small amount of dilute hydrochloric acid is added. [2]

10. A buffer of pH 5.15 is wanted. Sodium butanoate, C₃H₇COONa (M = 110.0 g mol⁻¹), is to be dissolved in 250 cm³ of 0.120 mol dm⁻³ butanoic acid (Ka = 1.48 × 10⁻⁵ mol dm⁻³). Assume the volume does not change. Calculate the mass of sodium butanoate needed. [4]

11. A pH probe follows the titration of 30.0 cm³ of 0.0960 mol dm⁻³ butanoic acid (Ka = 1.48 × 10⁻⁵ mol dm⁻³) with 0.160 mol dm⁻³ NaOH.

Indicator	pH range	Colour in acid	Colour in alkali
methyl orange	3.1–4.4	red	yellow
phenol red	6.8–8.4	yellow	red
phenolphthalein	8.3–10.0	colourless	pink

(a) Calculate the volume of NaOH needed to reach the equivalence point. [1] (b) Calculate the pH before any NaOH is added. [2] (c) Calculate the pH after 9.00 cm³ of NaOH has been added. [2] (d) Choose the most suitable indicator from the table and explain your choice. [2] (e) Treating the indicator as a weak acid, HIn, explain the colour change at the end point. [2] (f) The NaOH is replaced by aqueous ammonia of the same concentration. Describe how the curve changes after equivalence, and explain why no indicator in the table is then suitable. [2]

12. Blood plasma contains the carbonic acid–hydrogencarbonate buffer. For H₂CO₃, take Ka = 7.9 × 10⁻⁷ mol dm⁻³.

(a) Calculate [H⁺] in blood plasma at pH 7.40. [1] (b) Calculate the ratio [HCO₃⁻] / [H₂CO₃] at pH 7.40. [2] (c) Explain how this buffer responds if a little base enters the plasma. [2] (d) A student measures the pH of a plasma sample with a pH meter. Describe how she should calibrate the meter and take a reliable reading. [3]

Answers

1. (a) A proton (H⁺) donor [1] (b) HCOOH (acid) and HCOO⁻ (base) [1]; H₂PO₄⁻ (acid) and HPO₄²⁻ (base) [1] Examiner insight: each pair must differ by exactly one H⁺; pairing the two reactants loses both marks.

2. (a) tribasic [1] (b) dibasic [1] (c) monobasic [1] Examiner insight: basicity counts donatable protons, not all hydrogen atoms; counting the C–H hydrogen in methanoic acid gives the wrong answer.

3. (a) Fe(s) + 2H⁺(aq) → Fe²⁺(aq) + H₂(g) [1] (b) CO₃²⁻(aq) + 2H⁺(aq) → CO₂(g) + H₂O(l) [1] (c) MgO(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂O(l) [1] (d) H⁺(aq) + OH⁻(aq) → H₂O(l) [1] Examiner insight: leave out spectator ions (K⁺, Li⁺, Cl⁻) and keep MgO(s) as a solid; an equation whose charges do not balance should not be expected to gain credit.

4. New concentration = 0.240 × 20.0 / 500 = 0.00960 mol dm⁻³ [1] HNO₃ is a strong acid, so [H⁺] = 0.00960 mol dm⁻³ [1] pH = –log(0.00960) = 2.02 [1] Examiner insight: a pH worked out from the undiluted 0.240 mol dm⁻³ cannot be right.

5. (a) [OH⁻] = 2 × 0.0315 = 0.0630 mol dm⁻³ [1] [H⁺] = (1.00 × 10⁻¹⁴) / 0.0630 = 1.59 × 10⁻¹³ mol dm⁻³ [1] pH = 12.80 [1] (b) [H⁺] = √(2.92 × 10⁻¹⁴) = 1.71 × 10⁻⁷ mol dm⁻³ [1] pH = 6.77 [1] Neutral, because [H⁺] = [OH⁻] in pure water [1] Examiner insight: in (b), “acidic because pH is below 7” is wrong; link neutrality to [H⁺] = [OH⁻].

6. (a) Ka = [H⁺]² / [C₆H₅COOH] [1] [H⁺] = √(6.46 × 10⁻⁵ × 0.0380) = 1.57 × 10⁻³ mol dm⁻³ [1] pH = 2.80 [1] (b) [H⁺] = [C₆H₅COO⁻], because the dissociation of water is negligible [1] [C₆H₅COOH]equilibrium ≈ 0.0380 mol dm⁻³, because dissociation is so small [1] Examiner insight: “the acid is weak” is not an approximation; say what is assumed equal to what.

7. [H⁺] = 10⁻²·⁸⁴ = 1.45 × 10⁻³ mol dm⁻³ [1] Ka = (1.45 × 10⁻³)² / 0.160 [1] Ka = 1.31 × 10⁻⁵ mol dm⁻³ [1] pKa = –log(1.31 × 10⁻⁵) = 4.88 [1] Examiner insight: give Ka its units, mol dm⁻³, and give pKa none.

8. (a) [H⁺] = √(3.3 × 10⁻³ × 0.0200) = 8.12 × 10⁻³ mol dm⁻³ [1] pH = 2.09 [1] (b) Percentage = (8.12 × 10⁻³ / 0.0200) × 100 = 41% [1] The approximation assumes [HW]equilibrium ≈ 0.0200 mol dm⁻³, i.e. very little dissociation [1] But 41% has dissociated, so [HW]equilibrium is much less than 0.0200 mol dm⁻³ and the assumption fails [1] (c) Higher (the approximation overestimates [H⁺]) [1] Examiner insight: name the specific approximation that fails; “approximations are inaccurate” earns little.

9. (a) n(propanoic acid) = 0.0600 × 0.150 = 0.00900 mol; n(OH⁻) = 0.0350 × 0.100 = 0.00350 mol [1] n(propanoate) = 0.00350 mol; n(acid) left = 0.00900 – 0.00350 = 0.00550 mol [1] [H⁺] = 1.35 × 10⁻⁵ × 0.00550 / 0.00350 = 2.12 × 10⁻⁵ mol dm⁻³ [1] pH = 4.67 [1] (b) Added H⁺ reacts with the conjugate base: CH₃CH₂COO⁻ + H⁺ → CH₃CH₂COOH [1] The equilibrium CH₃CH₂COOH ⇌ H⁺ + CH₃CH₂COO⁻ shifts left, removing most of the added H⁺, so [H⁺] and pH change very little [1] Examiner insight: in (a), subtract the OH⁻ from the acid; in (b), name the species that removes H⁺ and the direction of shift.

10. [H⁺] = 10⁻⁵·¹⁵ = 7.08 × 10⁻⁶ mol dm⁻³ [1] [C₃H₇COO⁻] = Ka × [HA] / [H⁺] = 1.48 × 10⁻⁵ × 0.120 / (7.08 × 10⁻⁶) = 0.251 mol dm⁻³ [1] n = 0.251 × 0.250 = 0.0627 mol [1] mass = 0.0627 × 110.0 = 6.90 g [1] Examiner insight: write the rearranged Ka expression before substituting; forgetting to convert 250 cm³ to 0.250 dm³ gives a mass 1000 times too large.

11. (a) n(acid) = 0.0300 × 0.0960 = 2.88 × 10⁻³ mol; V = 2.88 × 10⁻³ / 0.160 = 18.0 cm³ [1] (b) [H⁺] = √(1.48 × 10⁻⁵ × 0.0960) = 1.19 × 10⁻³ mol dm⁻³ [1] pH = 2.92 [1] (c) 9.00 cm³ is half the equivalence volume, so [HA] = [A⁻] and [H⁺] = Ka [1] pH = pKa = –log(1.48 × 10⁻⁵) = 4.83 [1] (d) Phenolphthalein [1] Its range, 8.3–10.0, lies inside the steep section (about pH 7–11); the other two change colour before equivalence [1] (e) HIn ⇌ H⁺ + In⁻; HIn is colourless and In⁻ is pink [1] Added OH⁻ removes H⁺, so the equilibrium shifts right and the pink In⁻ form appears at the end point [1] (f) After equivalence the pH rises only slowly and levels off lower, so there is no sharp steep section [1] An indicator needs a sharp pH change across its range at equivalence, so none is suitable for a weak acid–weak base titration [1] Examiner insight: in (d), a choice without a reason linking its range to the steep section is unlikely to gain the second mark.

12. (a) [H⁺] = 10⁻⁷·⁴⁰ = 3.98 × 10⁻⁸ mol dm⁻³ [1] (b) Ka = [H⁺][HCO₃⁻] / [H₂CO₃], so [HCO₃⁻] / [H₂CO₃] = Ka / [H⁺] [1] = (7.9 × 10⁻⁷) / (3.98 × 10⁻⁸) = 20 (to 2 s.f.) [1] (c) OH⁻ from the base reacts with H⁺ to form water, lowering [H⁺] [1] H₂CO₃ dissociates, so H₂CO₃ ⇌ H⁺ + HCO₃⁻ shifts right, replacing the H⁺ and keeping blood pH within 7.35–7.45 [1] (d) Calibrate with buffer solutions of known pH, e.g. pH 4, 7 and 10 [1] Rinse the electrode with deionised water between solutions and blot dry [1] Immerse the bulb, stir gently and record the value only when the reading is steady, at constant temperature [1] Examiner insight: in (d), each mark needs a separate practical point.

Where marks are usually lost

• Using –log[OH⁻] as the pH of a strong base instead of going through Kw.
• Not doubling [OH⁻] for Ba(OH)₂ and other group 2 hydroxides.
• Stating “the acid is weak” instead of the two specific approximations.
• Describing buffer action without naming the reacting species and the direction of shift.
• Picking an indicator that changes colour before or after the steep part of the curve.
• Giving units for pH or pKa, or leaving them off Ka.

Next steps

• Go over the revision notes again.
• Return to the study guide for any outcome you found hard.
• Browse the course hub for the next topic, and record what you have finished on the printable statement-by-statement checklist.
• Try a free 10-minute diagnostic on your weakest topics.
• Book a free trial class and work through buffers or titration curves with one of our chemistry teachers.

Official syllabus

Source document: OCR A Level Chemistry A (code H432), specification Version 3.1 dated May 2026; first assessment 2017. These questions follow section 5.1.3 Acids, bases and buffers, in Module 5: Physical chemistry and transition elements.