IB DP Chemistry Reactivity 3.1: Proton transfer: acids, bases and pH – 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 – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

This set covers Reactivity 3.1, “Proton transfer reactions”, in IB DP Chemistry. Its content list comes from the February 2023 Diploma Programme Chemistry guide, first examined in 2025. The questions span understandings Reactivity 3.1.1 to 3.1.8, all shared by SL and HL, so each one suits either level; the set applies to sessions from 2025 onwards.

Work through the study guide first and recap with the revision notes. The course hub, checklist and free diagnostics show where to go next.

Data for every question: Kw = 1.0 × 10⁻¹⁴ at 298 K. Ar values: C 12.01, Ca 40.08, O 16.00. Assume 298 K unless a question says otherwise.

Questions

1. Hydrogensulfate ions react with methanoate ions:

HSO₄⁻(aq) + HCOO⁻(aq) ⇌ SO₄²⁻(aq) + HCOOH(aq)

For the forward reaction, identify which species acts as the Brønsted–Lowry acid, which acts as the base, and the two conjugate acid–base pairs. [3]

2. Deduce the formula of:

(a) the conjugate base of each of NH₄⁺, HCOOH and H₂PO₄⁻ [2] (b) the conjugate acid of each of CH₃NH₂, SO₄²⁻ and H₂O [2]

3. The hydrogenphosphate ion, HPO₄²⁻, is amphiprotic.

(a) State what amphiprotic means and which feature of HPO₄²⁻ lets it accept a proton. [1] (b) Write equations for HPO₄²⁻ reacting with H₃O⁺(aq) and with OH⁻(aq). [2] (c) Classify Na₂O, Al₂O₃ and P₄O₁₀ as acidic, basic or amphoteric, and write an equation for P₄O₁₀ reacting with water. [3]

4.

(a) A solution contains 4.6 × 10⁻³ mol dm⁻³ of H⁺ ions. Calculate its pH. [1] (b) Calculate [H⁺] in a solution of pH 9.45. [1] (c) A solution of nitric acid, concentration 0.0060 mol dm⁻³, is diluted by a factor of 100. Calculate the pH after dilution. [2]

5.

(a) Calculate the pH of 0.0450 mol dm⁻³ lithium hydroxide. [2] (b) Three solutions at 298 K have: A, [H⁺] = 2.0 × 10⁻⁶ mol dm⁻³; B, [OH⁻] = 1.0 × 10⁻⁷ mol dm⁻³; C, [OH⁻] = 6.3 × 10⁻⁴ mol dm⁻³. Classify each as acidic, neutral or basic, using the relative values of [H⁺] and [OH⁻]. [3]

6. At 313 K, Kw = 2.9 × 10⁻¹⁴ (use this value).

(a) Calculate [H⁺] and the pH of pure water at 313 K. [2] (b) Explain why this water is neutral even though its pH is not 7. [1] (c) Explain why Kw is larger at 313 K than at 298 K. [2]

7. A 0.050 mol dm⁻³ solution of a weak monoprotic acid, HA, has pH 3.15.

(a) Calculate the percentage of HA molecules that are ionized. [2] (b) Calculate the pH of 0.050 mol dm⁻³ hydrochloric acid. [1] (c) Calculate how many times greater [H⁺] is in the hydrochloric acid than in the HA solution. [1] (d) A student describes HA as “a dilute acid” to explain its higher pH. Comment on this. [2]

8.

(a) Nitrous acid, HNO₂, is a stronger acid than ethanoic acid. Predict, with a reason, whether this equilibrium lies to the left or the right: CH₃COOH(aq) + NO₂⁻(aq) ⇌ CH₃COO⁻(aq) + HNO₂(aq) [3] (b) Explain why HBr is a stronger acid than HF. [2]

9.

(a) Write balanced equations for: (i) nitric acid with calcium carbonate; (ii) butanoic acid with potassium hydroxide; (iii) ammonia with hydrobromic acid; (iv) methylamine with nitric acid. [5] (b) Identify the parent acid and base of lithium methanoate and of magnesium sulfate. [2]

10. 20.00 cm³ of 0.0600 mol dm⁻³ sodium hydroxide is placed in a flask and titrated with 0.150 mol dm⁻³ hydrochloric acid, while pH is recorded with a probe.

(a) Calculate the pH at the start. [2] (b) Calculate the volume of acid needed to reach the equivalence point. [2] (c) Calculate the pH after 20.00 cm³ of acid has been added. [2] (d) Describe the shape of the pH curve, giving the pH-axis intercept, the equivalence point and the pH the curve approaches. [2]

11. A lake affected by acid rain has pH 4.30. A sample of unpolluted rain from the same area has pH 5.60.

(a) Sulfur trioxide in the air dissolves in rainwater. Write the equation. [1] (b) Calculate how many times greater [H⁺] is in the lake than in the unpolluted rain. [2] (c) Powdered limestone, CaCO₃, is added to the lake. Write an ionic equation for its reaction with H⁺(aq) and identify the Brønsted–Lowry base. [2] (d) The lake holds 2.0 × 10⁶ dm³ of water. Assuming only the free H⁺ must be neutralized, calculate the mass of CaCO₃ needed. [3]

Answers

1. HSO₄⁻ donates a proton, so it is the acid [1]. HCOO⁻ accepts it, so it is the base [1]. Pairs: HSO₄⁻ / SO₄²⁻ and HCOOH / HCOO⁻ [1] Examiner insight: Write each pair acid first and include charges; HSO₄⁻ / HCOOH differs by more than one proton, so it is not a conjugate pair at all.

2. (a) NH₃ and HCOO⁻ [1]; HPO₄²⁻ [1] (b) CH₃NH₃⁺ and HSO₄⁻ [1]; H₃O⁺ [1] Examiner insight: Check every charge: adding H⁺ raises it by one, removing H⁺ lowers it by one. A correct formula with the wrong charge is a wrong species.

3. (a) It can act as a Brønsted–Lowry acid or base; it has lone pairs on its oxygen atoms that can accept a proton [1] (b) HPO₄²⁻(aq) + H₃O⁺(aq) → H₂PO₄⁻(aq) + H₂O(l) [1]; HPO₄²⁻(aq) + OH⁻(aq) → PO₄³⁻(aq) + H₂O(l) [1] (c) Na₂O basic, Al₂O₃ amphoteric, P₄O₁₀ acidic [1] [1]; P₄O₁₀(s) + 6H₂O(l) → 4H₃PO₄(aq) [1] Examiner insight: For (a), name both roles, donor and acceptor; mentioning only one leaves the definition incomplete.

4. (a) pH = −log₁₀(4.6 × 10⁻³) = 2.34 [1] (b) [H⁺] = 10⁻⁹·⁴⁵ = 3.5 × 10⁻¹⁰ mol dm⁻³ [1] (c) After dilution [H⁺] = 0.0060 / 100 = 6.0 × 10⁻⁵ mol dm⁻³ [1]; pH = 4.22 (up 2 units from 2.22) [1] Examiner insight: Show the log expression you used; a bare pH that is slightly off leaves no method for an examiner to reward.

5. (a) [OH⁻] = 0.0450 mol dm⁻³, so [H⁺] = 1.0 × 10⁻¹⁴ / 0.0450 = 2.2 × 10⁻¹³ mol dm⁻³ [1]; pH = 12.65 [1] (b) A: [OH⁻] = 5.0 × 10⁻⁹, so [H⁺] > [OH⁻], acidic [1]. B: [H⁺] = 1.0 × 10⁻⁷ = [OH⁻], neutral [1]. C: [H⁺] = 1.6 × 10⁻¹¹ < [OH⁻], basic [1] Examiner insight: Compare [H⁺] with [OH⁻] as asked: work out the missing concentration each time rather than quoting pH alone.

6. (a) [H⁺] = [OH⁻], so [H⁺] = √(2.9 × 10⁻¹⁴) = 1.7 × 10⁻⁷ mol dm⁻³ [1]; pH = 6.77 [1] (b) [H⁺] still equals [OH⁻], which is what neutral means [1] (c) The ionization of water is endothermic [1]; raising the temperature shifts the equilibrium to the right, increasing [H⁺] and [OH⁻], so Kw rises [1] Examiner insight: “Neutral” is defined by equal concentrations; an answer that just says “pH 7 is neutral” contradicts the calculation and loses the explanation.

7. (a) [H⁺] = 10⁻³·¹⁵ = 7.1 × 10⁻⁴ mol dm⁻³ [1]; percentage ionized = (7.1 × 10⁻⁴ / 0.050) × 100 = 1.4% [1] (b) HCl is fully ionized, [H⁺] = 0.050 mol dm⁻³, pH = 1.30 [1] (c) 0.050 / 7.08 × 10⁻⁴ = 71 times [1] (d) The two acids have the same concentration, so “dilute” does not explain the difference [1]; HA is weak, only partly ionized, so it releases fewer H⁺ ions [1] Examiner insight: Strength and concentration are separate ideas; an answer that swaps “weak” and “dilute” misses the exact point the question is testing.

8. (a) HNO₂ is the stronger acid, so CH₃COOH is the weaker acid [1]; CH₃COO⁻ is then the stronger base and NO₂⁻ the weaker base [1]; the equilibrium lies towards the weaker acid and base, so to the left [1] (b) The H–Br bond is longer and weaker than the H–F bond [1], so HBr loses its proton (ionizes) more readily in water [1] Examiner insight: For (b), link bond strength to ease of losing a proton; quoting “Br is less electronegative” alone does not explain the trend.

9. (a) (i) 2HNO₃(aq) + CaCO₃(s) → Ca(NO₃)₂(aq) + H₂O(l) + CO₂(g) [1] [1] (ii) CH₃CH₂CH₂COOH(aq) + KOH(aq) → CH₃CH₂CH₂COOK(aq) + H₂O(l) [1] (iii) NH₃(aq) + HBr(aq) → NH₄Br(aq) [1] (iv) CH₃NH₂(aq) + HNO₃(aq) → CH₃NH₃NO₃(aq) [1] (b) Lithium methanoate: methanoic acid and lithium hydroxide [1]. Magnesium sulfate: sulfuric acid and magnesium oxide or hydroxide (or carbonate) [1] Examiner insight: In (i) check all three products, including CO₂, and that the nitrate ions balance; in (iii) and (iv) a water product makes the equation wrong.

10. (a) [OH⁻] = 0.0600 mol dm⁻³; [H⁺] = 1.0 × 10⁻¹⁴ / 0.0600 = 1.7 × 10⁻¹³ mol dm⁻³ [1]; pH = 12.78 [1] (b) n(NaOH) = 0.0600 × 0.02000 = 1.20 × 10⁻³ mol [1]; V(HCl) = 1.20 × 10⁻³ / 0.150 = 8.00 × 10⁻³ dm³ = 8.00 cm³ [1] (c) n(HCl) = 0.150 × 0.02000 = 3.00 × 10⁻³ mol; excess H⁺ = 1.80 × 10⁻³ mol in 40.00 cm³, [H⁺] = 0.045 mol dm⁻³ [1]; pH = 1.35 [1] (d) Starts at pH 12.78 and falls slowly, then drops almost vertically around 8.00 cm³, passing pH 7 at the equivalence point [1]; then levels off, approaching but staying above pH 0.82 (the pH of 0.150 mol dm⁻³ HCl) [1] Examiner insight: In (c) the total volume is 40.00 cm³, not 20.00 cm³; leaving out the added acid volume overstates [H⁺] and gives the wrong pH.

11. (a) SO₃(g) + H₂O(l) → H₂SO₄(aq), giving sulfuric acid [1] (b) Ratio = 10⁻⁴·³⁰ / 10⁻⁵·⁶⁰ = 10¹·³⁰ [1] = 20 times [1] (c) CaCO₃(s) + 2H⁺(aq) → Ca²⁺(aq) + H₂O(l) + CO₂(g) [1]; the base is the carbonate ion, CO₃²⁻ (in CaCO₃) [1] (d) [H⁺] = 10⁻⁴·³⁰ = 5.0 × 10⁻⁵ mol dm⁻³; n(H⁺) = 5.0 × 10⁻⁵ × 2.0 × 10⁶ = 1.0 × 10² mol [1]; n(CaCO₃) = 50 mol [1]; mass = 50 × 100.09 = 5.0 × 10³ g (5.0 kg) [1] Examiner insight: A pH difference of 1.30 is not “1.3 times”; convert both pH values to [H⁺] or use 10 to the power of the difference.

Where marks are usually lost

• Pairing a reactant with a product that differs by more than one proton, or listing the base first in a pair.
• Writing a conjugate with the wrong charge.
• Using “amphoteric” and “amphiprotic” as if they mean the same.
• For alkalis, taking −log₁₀[OH⁻] and calling it pH.
• Treating a pH difference as a simple ratio rather than a power of 10.
• Forgetting CO₂ with carbonates, or adding water with ammonia and amines.
• Ignoring the extra volume of titrant when finding a concentration after mixing.
• Sketching a pH curve whose steep section is not vertical, or whose end plateau reaches the pH of the titrant.
• Saying neutral water must have pH 7 at all temperatures.

Next steps

• Revision notes for this unit
• Study guide for this unit
• IB DP Chemistry course hub
• Printable syllabus checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

This unit is drawn from Reactivity 3.1 (understandings 3.1.1 to 3.1.8) of the Diploma Programme Chemistry guide, first examined in 2025 and issued by the International Baccalaureate Organization in February 2023.