Edexcel International A Level Chemistry: Acid-base equilibria: pH, Ka and Kw (YCH11) – 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.

Outcomes 14.1 to 14.14 are tested here, all from Topic 14 (Acid-base Equilibria). The source document is the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017. Unit 4, Rates, Equilibria and Further Organic Chemistry, examines this IA2 topic. New to a method? Read the study guide or revision notes first. See also the checklist, hub and free diagnostic.

Use these values throughout: Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K; Ar(H) = 1.0, Ar(C) = 12.0, Ar(O) = 16.0. Give pH values to two decimal places.

Questions

1. Define a Brønsted–Lowry acid and a Brønsted–Lowry base. [2]

2. Dihydrogenphosphate ions react with ammonia:

H₂PO₄⁻(aq) + NH₃(aq) ⇌ HPO₄²⁻(aq) + NH₄⁺(aq)

(a) Identify the two conjugate acid-base pairs. [2] (b) Write an equation to show H₂PO₄⁻ acting as a base when it reacts with H₃O⁺. [1]

3.

(a) Define the term pH. [1] (b) Calculate the pH of a solution with [H⁺] = 6.3 × 10⁻³ mol dm⁻³. [1] (c) Calculate [H⁺] in a cleaning solution of pH 9.62. [1]

4. A bottle contains 0.0290 mol dm⁻³ nitric acid.

(a) Calculate its pH. [1] (b) Water is added to 10.0 cm³ of this acid until the total volume is 500 cm³. Calculate the new pH. [2]

5.

(a) Using HCl and HCOOH as examples, with an equation for each, explain how a strong acid differs from a weak acid in terms of dissociation. [2] (b) A student writes: “0.0010 mol dm⁻³ hydrochloric acid is a weak acid because it is so dilute.” Explain why the student is wrong. [1]

6. 2-methylpropanoic acid, (CH₃)₂CHCOOH, is a weak monoprotic acid with pKa = 4.86 at 298 K.

(a) Write the expression for Ka of 2-methylpropanoic acid. [1] (b) Calculate the pH of a 0.0840 mol dm⁻³ solution of the acid. [3] (c) State the two assumptions you made in part (b). [2] (d) Calculate the percentage of the acid molecules that are dissociated, and comment on whether your assumption about the equilibrium concentration of the acid is justified. [1]

7.

(a) Define the ionic product of water, Kw. [1] (b) Define pKw. [1] (c) Calculate the pH of 0.0420 mol dm⁻³ sodium hydroxide at 298 K. [2] (d) Calculate the pH of 0.00875 mol dm⁻³ strontium hydroxide, Sr(OH)₂, at 298 K. [2]

8. Three weak acids have these values at 298 K: acid P, Ka = 1.3 × 10⁻³ mol dm⁻³; acid R, pKa = 4.20; acid S, Ka = 2.0 × 10⁻⁵ mol dm⁻³. Place the acids in order of strength, strongest first, showing how you compared them. [2]

9. Seven solutions, each 0.0900 mol dm⁻³, were tested with a pH meter at 298 K. The solutions were hydrochloric acid, ethanoic acid, ammonium chloride, sodium chloride, sodium ethanoate, aqueous ammonia and sodium hydroxide.

Solution	J	K	L	M	N	O	P
pH	8.9	1.0	11.1	5.1	13.0	7.0	2.9

(a) Identify each solution J to P. [3] (b) Explain why the ammonium chloride solution has a pH below 7. [2]

10. A student compared two acids, X and Y, both 0.0500 mol dm⁻³ at 298 K. Each was diluted 10, 100 and 1000 times and the pH measured.

Dilution factor	none	10	100	1000
Acid X	1.30	2.30	3.30	4.30
Acid Y	2.76	3.27	3.83	4.48

(a) Identify which acid is strong. Explain your answer using the data. [2] (b) Explain why the pH of Y rises by about 0.5 for each of the first two tenfold dilutions. [2] (c) Suggest why the rise in pH for Y between 100 and 1000 times dilution is larger than 0.5. [1] (d) Use the undiluted reading for Y to calculate Ka for acid Y. [2]

11. To find Ka for glycolic acid, HOCH₂COOH (a monoprotic acid), a technician weighed out 1.14 g, dissolved it and made the volume up to 250 cm³ in a volumetric flask. The pH meter read 2.53.

(a) Calculate Ka for glycolic acid. [4] (b) Calculate pKa. [1]

12. A student mixes 20.0 cm³ of 0.150 mol dm⁻³ hydrochloric acid with 30.0 cm³ of 0.120 mol dm⁻³ sodium hydroxide at 298 K.

(a) Calculate the pH of the mixture. [4] (b) At 60 °C, Kw = 9.6 × 10⁻¹⁴ mol² dm⁻⁶ (use this value). Calculate the pH of pure water at 60 °C. [2] (c) Explain why pure water at 60 °C is neutral even though its pH is not 7.00. [1] (d) Use the values of Kw at 298 K and 60 °C to deduce whether the ionisation of water is exothermic or endothermic. Explain your answer. [2]

Answers

1. Acid: a species that donates a proton, H⁺ [1]. Base: a species that accepts a proton, H⁺ [1]. [2] Examiner insight: Write “proton” or “H⁺”; writing “hydrogen” alone is too vague to describe what is transferred.

2. (a) H₂PO₄⁻ and HPO₄²⁻ [1]; NH₄⁺ and NH₃ [1]. (b) H₂PO₄⁻ + H₃O⁺ → H₃PO₄ + H₂O [1] Examiner insight: Check that each pair differs by one H⁺ and that the charges in your equation balance.

3. (a) pH = −log₁₀[H⁺], with [H⁺] in mol dm⁻³ [1]. (b) pH = −log₁₀(6.3 × 10⁻³) = 2.20 [1] (c) [H⁺] = 10⁻⁹·⁶² = 2.40 × 10⁻¹⁰ mol dm⁻³ [1] Examiner insight: A definition of pH needs the minus sign and the base-10 log; “the negative log of hydrogen” is incomplete.

4. (a) Nitric acid is fully dissociated, so [H⁺] = 0.0290 mol dm⁻³; pH = 1.54 [1]. (b) New concentration = 0.0290 × 10.0/500 = 5.80 × 10⁻⁴ mol dm⁻³ [1]; pH = −log₁₀(5.80 × 10⁻⁴) = 3.24 [1]. Examiner insight: Show the dilution step before taking the log.

5. (a) HCl is a strong acid: it is fully dissociated, HCl(aq) → H⁺(aq) + Cl⁻(aq) [1]. HCOOH is a weak acid: it is only partly dissociated, HCOOH(aq) ⇌ H⁺(aq) + HCOO⁻(aq), with most molecules undissociated [1]. (b) Strength depends on the degree of dissociation, not concentration; HCl is still fully dissociated however dilute it is, so it is a dilute strong acid [1]. Examiner insight: Use a one-way arrow for the strong acid and an equilibrium arrow for the weak acid; the arrows carry the meaning.

6. (a) Ka = [H⁺][(CH₃)₂CHCOO⁻] / [(CH₃)₂CHCOOH] [1] (b) Ka = 10⁻⁴·⁸⁶ = 1.38 × 10⁻⁵ mol dm⁻³ [1]; [H⁺] = √(1.38 × 10⁻⁵ × 0.0840) = 1.08 × 10⁻³ mol dm⁻³ [1]; pH = 2.97 [1]. (c) [H⁺] = [(CH₃)₂CHCOO⁻], because the ionisation of water is negligible [1]. The equilibrium concentration of the acid equals 0.0840 mol dm⁻³, because very little of it dissociates [1]. (d) 1.08 × 10⁻³/0.0840 × 100 = 1.3%, which is small, so the assumption is justified [1]. Examiner insight: State each assumption as an equality between named concentrations; “the acid is weak” is not an assumption.

7. (a) Kw = [H⁺][OH⁻] (the ionic product of water) [1]. (b) pKw = −log₁₀Kw [1]. (c) [H⁺] = 1.00 × 10⁻¹⁴/0.0420 = 2.38 × 10⁻¹³ mol dm⁻³ [1]; pH = 12.62 [1]. (d) [OH⁻] = 2 × 0.00875 = 0.0175 mol dm⁻³ [1]; [H⁺] = 1.00 × 10⁻¹⁴/0.0175 = 5.71 × 10⁻¹³, so pH = 12.24 [1]. Examiner insight: Strontium hydroxide releases two OH⁻ per formula; using 0.00875 directly gives 11.94, a common slip.

8. Convert to one scale: P has pKa 2.89 and S has pKa 4.70 (or R has Ka 6.3 × 10⁻⁵) [1]. Order: P > R > S [1]. Examiner insight: Show the conversion to one scale; an unsupported order is a guess.

9. (a) Acids: K = hydrochloric acid, P = ethanoic acid [1]. Bases: N = sodium hydroxide, L = ammonia [1]. Salts: M = ammonium chloride, O = sodium chloride, J = sodium ethanoate [1]. (b) NH₄⁺ is a weak (Brønsted–Lowry) acid [1]; it donates H⁺ to water, NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺, so [H⁺] > [OH⁻] [1]. Examiner insight: Explain the salt’s pH through the ion that donates or accepts protons; “made from a strong acid” alone is not enough.

10. (a) X is strong [1]; its pH rises by exactly 1.00 for each tenfold dilution, so [H⁺] falls tenfold, matching full dissociation with [H⁺] = c [1]. (b) For a weak acid [H⁺] = √(Ka × c) [1]; a tenfold fall in c lowers [H⁺] by √10, which raises pH by log₁₀√10 = 0.5 [1]. (c) At high dilution a larger fraction of Y dissociates, so [HA] is no longer close to c and the square-root approximation fails [1]. (d) [H⁺] = 10⁻²·⁷⁶ = 1.74 × 10⁻³ mol dm⁻³ [1]; Ka = (1.74 × 10⁻³)²/0.0500 = 6.04 × 10⁻⁵ mol dm⁻³, using the unrounded [H⁺] (6.0 × 10⁻⁵ to two significant figures) [1]. Examiner insight: Quote numbers from the table to support “strong” or “weak”; a description without data ignores the evidence given.

11. (a) Mr = 76.0, so n = 1.14/76.0 = 0.0150 mol [1]; c = 0.0150/0.250 = 0.0600 mol dm⁻³ [1]; [H⁺] = 10⁻²·⁵³ = 2.95 × 10⁻³ mol dm⁻³ [1]; Ka = (2.95 × 10⁻³)²/0.0600 = 1.45 × 10⁻⁴ mol dm⁻³, using the unrounded [H⁺] [1]. (b) pKa = −log₁₀(1.45 × 10⁻⁴) = 3.84 [1]. Examiner insight: Divide by 0.250 dm³, not 250; the cm³ figure makes the concentration a thousand times too small.

12. (a) n(H⁺) = 0.0200 × 0.150 = 3.00 × 10⁻³ mol; n(OH⁻) = 0.0300 × 0.120 = 3.60 × 10⁻³ mol [1]. Excess OH⁻ = 6.0 × 10⁻⁴ mol [1]; [OH⁻] = 6.0 × 10⁻⁴/0.0500 = 0.0120 mol dm⁻³ [1]; [H⁺] = 1.00 × 10⁻¹⁴/0.0120 = 8.33 × 10⁻¹³, so pH = 12.08 [1]. (b) In pure water [H⁺] = [OH⁻], so [H⁺] = √(9.6 × 10⁻¹⁴) = 3.10 × 10⁻⁷ mol dm⁻³ [1]; pH = 6.51 [1]. (c) [H⁺] still equals [OH⁻], which is the meaning of neutral [1]. (d) Kw is larger at the higher temperature, so the equilibrium H₂O ⇌ H⁺ + OH⁻ shifts right on heating [1]; an equilibrium constant that increases with temperature means the forward reaction is endothermic [1]. Examiner insight: Divide the excess moles by the total volume of the mixture (50.0 cm³), not by the volume of one solution.

Where marks are usually lost

• Naming a conjugate pair whose members differ by two protons.
• Confusing “weak” (dissociation) with “dilute” (concentration).
• Taking log₁₀ of a mass or a moles value instead of a concentration in mol dm⁻³.
• Forgetting to double [OH⁻] for group 2 hydroxides.
• Rounding [H⁺] to two significant figures before squaring it for Ka.
• Leaving Ka without units, or giving pH or pKa units.
• Using pH 7 as the test for neutral at temperatures other than 298 K.

Next steps

• pH, Ka and Kw revision notes
• pH, Ka and Kw study guide
• Edexcel IAL Chemistry hub
• Printable Edexcel IAL Chemistry checklist
• All free 10-minute diagnostics
• Book a free trial class
• Titration curves and buffers (14.15 onwards)
• Introduction to kinetics and equilibria (Unit 2 groundwork)

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017, Pearson Education Limited. Pages in this set teach Topic 14: Acid-base Equilibria (Unit 4), outcomes 14.1 to 14.14.