IB DP Chemistry Reactivity 3.1: Proton transfer: Kw, Ka, Kb, buffers and titration curves – 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.

These questions cover the higher level half of Reactivity 3.1, “Proton transfer reactions”, numbered to match the Diploma Programme Chemistry guide, which the IB published in February 2023 for courses first assessed in 2025. Every understanding from Reactivity 3.1.9 to 3.1.17 appears in at least one question. The guide classes all of them as additional higher level, so the whole set is HL only, for sessions from 2025 onwards.

If a method is unfamiliar, go back to the proton transfer study guide or the shorter revision notes. Browse other units on the DP Chemistry hub, log what you have finished on the DP Chemistry checklist, and use a free diagnostic quiz to choose your next topic.

Data for every question: all solutions are at 298 K, where Kw = 1.00 × 10⁻¹⁴. Give pH values to 2 decimal places and other answers to 2 or 3 significant figures unless told otherwise.

Questions

1. A cleaning solution has pH 9.65. Calculate its pOH and its [OH⁻]. [2]

2. Use these values: acid P, Ka = 2.5 × 10⁻³; acid Q, pKa = 4.10; acid R, Ka = 3.2 × 10⁻⁶; acid S, pKa = 2.20. List the four acids from strongest to weakest, showing how you compared them. [2]

3. Hydrofluoric acid, HF, is a weak acid with pKa = 3.17.

(a) Write the Kb expression for the fluoride ion. [1] (b) Show that Ka(HF) × Kb(F⁻) = Kw. [1] (c) Calculate Kb for F⁻. [1]

4. Calculate the pH of 0.150 mol dm⁻³ aqueous trimethylamine, (CH₃)₃N, given Kb = 6.3 × 10⁻⁵. State the assumptions you make. [4]

5. A 0.0200 mol dm⁻³ solution of a weak monoprotic acid HZ has pH 3.40. Calculate Ka and pKa for HZ. [3]

6. Three salts are dissolved separately in water: ammonium iodide, potassium carbonate and sodium nitrate.

(a) Predict whether each solution is acidic, neutral or basic. [3] (b) Write equations for the reactions with water that explain your answers for the ammonium ion and the carbonate ion. [2]

7. Hydrocyanic acid has Ka = 6.2 × 10⁻¹⁰. Calculate the pH of 0.0750 mol dm⁻³ sodium cyanide solution. [4]

8. 20.0 cm³ of an aqueous weak base B is titrated with 0.125 mol dm⁻³ hydrochloric acid. The pH curve falls steeply at 16.0 cm³. After 8.0 cm³ of acid has been added, the pH is 10.70.

(a) Calculate the concentration of B. [2] (b) Deduce pKb and Kb for B. [2] (c) Calculate the pH of the base solution before any acid is added. [2] (d) Explain, with an equation, why the pH at the equivalence point is below 7. [2]

9. An indicator is represented as HInd.

(a) Write the equilibrium for HInd in water, and explain which colour you see in a strongly acidic solution. [2] (b) Use these ranges: bromocresol green 3.8–5.4; phenol red 6.8–8.4; thymolphthalein 9.3–10.5. Select an indicator for titrating aqueous methylamine with hydrochloric acid, giving a reason. [1] (c) Explain why bromocresol green is unsuitable for titrating a weak acid with sodium hydroxide. [1]

10. (a) Distinguish between the equivalence point and the end point of a titration. [2] (b) Explain why no indicator gives a sharp end point when ethanoic acid is titrated with aqueous ammonia. [1]

11. (a) State the composition of a basic buffer made from ammonia. [1] (b) Write equations to show how this buffer resists a change in pH when small amounts of acid and of alkali are added. [2]

12. Phenylethanoic acid, C₆H₅CH₂COOH, has Ka = 4.9 × 10⁻⁵. A buffer is made by mixing 50.0 cm³ of 0.160 mol dm⁻³ phenylethanoic acid with 24.0 cm³ of 0.200 mol dm⁻³ sodium hydroxide.

(a) Calculate the amounts, in mol, of the acid and its conjugate base in the buffer. [2] (b) Calculate the pH of the buffer. [2] (c) 8.0 × 10⁻⁴ mol of HCl is added. Calculate the new pH. [2] (d) Predict, with a reason, how the pH of the buffer from (b) changes if you double its volume by adding water. [1]

Answers

1. pOH = 14.00 − 9.65 = 4.35 [1]. [OH⁻] = 10^(−4.35) = 4.5 × 10⁻⁵ mol dm⁻³ [1]. [2] Examiner insight: Label which value is pOH and which is [OH⁻]; a correct number under the wrong label does not answer the question.

2. Convert to one scale: pKa(P) = 2.60 and pKa(R) = 5.49 [1]. Strongest to weakest: S > P > Q > R [1]. [2] Examiner insight: Show the conversions; an order given without comparable values does not show how you decided.

3. (a) Kb = [HF][OH⁻] / [F⁻] [1]. [1] (b) Ka × Kb = ([H⁺][F⁻] / [HF]) × ([HF][OH⁻] / [F⁻]); [F⁻] and [HF] cancel to leave [H⁺][OH⁻] = Kw [1]. [1] (c) pKb = 14.00 − 3.17 = 10.83, so Kb = 10^(−10.83) = 1.5 × 10⁻¹¹ [1]. [1] Examiner insight: In a “show that”, write both expressions in full and state what cancels; jumping straight to Kw shows nothing.

4. Kb = [(CH₃)₃NH⁺][OH⁻] / [(CH₃)₃N], taking [(CH₃)₃NH⁺] = [OH⁻] and [(CH₃)₃N] ≈ 0.150 [1]. [OH⁻] = √(6.3 × 10⁻⁵ × 0.150) = 3.07 × 10⁻³ mol dm⁻³ [1]. pOH = 2.51 [1]. pH = 14.00 − 2.51 = 11.49 [1]. [4] Examiner insight: Stating the two assumptions is part of the method; without them the square root has no justification.

5. [H⁺] = 10^(−3.40) = 3.98 × 10⁻⁴ mol dm⁻³ [1]. Ka = (3.98 × 10⁻⁴)² ÷ 0.0200 = 7.9 × 10⁻⁶ [1]. pKa = 5.10 [1]. [3] Examiner insight: Square [H⁺] because [Z⁻] equals it; using [H⁺] once gives a Ka that is far too large.

6. (a) Ammonium iodide: acidic [1]. Potassium carbonate: basic [1]. Sodium nitrate: neutral [1]. [3] (b) NH₄⁺(aq) + H₂O(l) ⇌ NH₃(aq) + H₃O⁺(aq) [1]. CO₃²⁻(aq) + H₂O(l) ⇌ HCO₃⁻(aq) + OH⁻(aq) [1]. [2] Examiner insight: Show H₃O⁺ or OH⁻ as a product; that species is what links the equation to the pH.

7. Kb(CN⁻) = 1.00 × 10⁻¹⁴ ÷ 6.2 × 10⁻¹⁰ = 1.61 × 10⁻⁵ [1]. [OH⁻] = √(1.61 × 10⁻⁵ × 0.0750) = 1.10 × 10⁻³ mol dm⁻³ [1]. pOH = 2.96 [1]. pH = 11.04 [1]. [4] Examiner insight: The data give Ka of the parent acid, so convert to Kb first; using Ka directly gives an acidic pH for a basic salt.

8. (a) n(HCl) = 0.125 × 0.0160 = 2.00 × 10⁻³ mol, and B reacts 1 : 1 [1]. [B] = 2.00 × 10⁻³ ÷ 0.0200 = 0.100 mol dm⁻³ [1]. [2] (b) 8.0 cm³ is half-equivalence, so pOH = pKb [1]. pKb = 14.00 − 10.70 = 3.30, Kb = 5.0 × 10⁻⁴ [1]. [2] (c) [OH⁻] = √(5.0 × 10⁻⁴ × 0.100) = 7.08 × 10⁻³ mol dm⁻³ [1]. pOH = 2.15, pH = 11.85 [1]. [2] (d) The salt contains BH⁺, which donates a proton to water: BH⁺(aq) + H₂O(l) ⇌ B(aq) + H₃O⁺(aq) [1]. Cl⁻ comes from a strong acid and does not react with water, so the solution is acidic [1]. [2] Examiner insight: At half-equivalence a weak base gives pOH = pKb, not pH = pKb; convert 10.70 before you write pKb.

9. (a) HInd(aq) ⇌ H⁺(aq) + Ind⁻(aq) [1]. High [H⁺] shifts the equilibrium to the left, so the colour of HInd is seen [1]. [2] (b) Bromocresol green: the salt formed is acidic, so the equivalence pH is below 7 and its range 3.8–5.4 lies in the steep section [1]. [1] (c) It changes colour between pH 3.8 and 5.4, in the buffer region, long before the equivalence point, which is above 7 [1]. [1] Examiner insight: Justify an indicator choice by linking the salt to the equivalence pH; naming the indicator alone does not show the reasoning.

10. (a) Equivalence point: acid and base have reacted in the stoichiometric ratio [1]. End point: the indicator changes colour [1]. [2] (b) The curve has no steep vertical section, so the pH, and the indicator colour, change gradually over a large volume [1]. [1] Examiner insight: Define each term on its own; “the end point is when the reaction finishes” describes the equivalence point.

11. (a) Aqueous ammonia with an ammonium salt such as NH₄Cl [1]. [1] (b) Added acid: NH₃ + H⁺ → NH₄⁺ [1]. Added alkali: NH₄⁺ + OH⁻ → NH₃ + H₂O [1]. [2] Examiner insight: Give one equation for added acid and one for added alkali; two equations for the same addition only cover half the question.

12. (a) n(acid) at start = 0.0500 × 0.160 = 0.00800 mol; n(OH⁻) = 0.0240 × 0.200 = 0.00480 mol [1]. After reaction: n(acid) = 0.00320 mol, n(conjugate base) = 0.00480 mol [1]. [2] (b) [H⁺] = Ka × n(acid) / n(base) = 4.9 × 10⁻⁵ × 0.00320 ÷ 0.00480 = 3.27 × 10⁻⁵ mol dm⁻³ [1]. pH = 4.49 [1]. [2] (c) The base converts H⁺ to acid: n(base) = 0.00400 mol and n(acid) = 0.00400 mol [1]. Equal amounts, so pH = pKa = −log₁₀(4.9 × 10⁻⁵) = 4.31 [1]. [2] (d) Acid and conjugate base are diluted equally, leaving their ratio fixed, so the pH stays almost unchanged [1]. [1] Examiner insight: Subtract the added OH⁻ from the weak acid before substituting; using the starting amount of acid gives the wrong ratio.

Where marks are usually lost

• Stopping at pOH when the question asks for pH.
• Writing [H₂O] inside a Kb expression.
• Converting Ka to Kb by dividing the wrong way round; Kb = Kw ÷ Ka.
• Using √(Ka × c) for a buffer after partial neutralisation.
• Writing pH = pKb at the half-equivalence point of a weak base titration.
• Predicting pH 7 at equivalence for every titration.
• Giving hydrolysis equations for spectator ions such as Na⁺, K⁺, I⁻ or NO₃⁻.
• Explaining buffer action with only one equation.
• Claiming that dilution changes buffer pH; it reduces capacity instead.

Next steps

• Proton transfer revision notes
• Proton transfer study guide
• All DP Chemistry units on the hub
• DP Chemistry topic checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

These questions are based on the Diploma Programme Chemistry guide (International Baccalaureate Organization), first assessment 2025, published February 2023, sub-topic Reactivity 3.1, understandings 3.1.9 to 3.1.17 (additional higher level).