A Level Chemistry: Acids, Bases, Buffers and Partition Coefficients — Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Acids, Bases, Buffers and Partition Coefficients study guide

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Section A

1. Define pKa. [1]

2. State the expression for Kw and its value at 298 K, including units. [2]

3. State two conditions needed for a solution to act as an effective buffer. [2]

4. Write the Ksp expression for lead(II) chloride, PbCl₂. [2]

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Section B

5. Calculate the pH of 0.0800 mol dm⁻³ HNO₃ (a strong acid). [2]

6. Calculate the pH of 0.0400 mol dm⁻³ KOH. [3]

7. Propanoic acid, CH₃CH₂COOH, has Ka = 1.35 × 10⁻⁵ mol dm⁻³.

(a) Calculate the pH of 0.200 mol dm⁻³ propanoic acid. [3]

(b) A buffer is made from 0.150 mol dm⁻³ propanoic acid and 0.300 mol dm⁻³ sodium propanoate. Calculate its pH. [3]

(c) State and explain what happens to the pH of the buffer in (b) if a small amount of dilute HCl is added. [2]

8. Ksp of calcium fluoride, CaF₂, is 3.90 × 10⁻¹¹ mol³ dm⁻⁹.

(a) Write the Ksp expression for CaF₂ and calculate its molar solubility in pure water. [3]

(b) Calculate the solubility of CaF₂ in 0.100 mol dm⁻³ NaF(aq). [3]

9. A solute Y has a partition coefficient Kpc (hexane/water) = 6.00. 8.00 g of Y is shaken with 40 cm³ of hexane and 40 cm³ of water until equilibrium is reached. Calculate the mass of Y in each layer. [3]

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Section C

10. The HCO₃⁻/H₂CO₃ (hydrogencarbonate/carbonic acid) system buffers blood pH close to 7.4.

(a) Write an equation showing how this buffer removes excess H⁺. [1]

(b) Explain, referring to the reservoir of H₂CO₃, how excess OH⁻ would be neutralised instead. [2]

11. Explain, in terms of Le Chatelier’s principle, why adding excess NaCl(aq) to a saturated solution of AgCl reduces the concentration of dissolved Ag⁺. [2]

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Answers

1. pKa = −log₁₀ Ka [1].

2. Kw = [H⁺][OH⁻] [1] = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K [1].

3. Any two of: contains a weak acid (or weak base) and its conjugate base (or conjugate acid); both species present in significant, comparable concentrations; concentrations are large enough that the added acid/base is small in comparison [2 — one mark each, max 2].

4. Ksp = [Pb²⁺][Cl⁻]² [2 — 1 mark for correct ions, 1 mark for correct power on Cl⁻].

5. [H⁺] = 0.0800 mol dm⁻³ (complete dissociation) [1]. pH = −log₁₀(0.0800) = 1.10 [1].

6. [OH⁻] = 0.0400 mol dm⁻³ [1]. [H⁺] = Kw/[OH⁻] = 1.00 × 10⁻¹⁴/0.0400 = 2.50 × 10⁻¹³ mol dm⁻³ [1]. pH = −log₁₀(2.50 × 10⁻¹³) = 12.60 [1].

7. (a) [H⁺] = √(Ka × C) = √(1.35 × 10⁻⁵ × 0.200) [1] = √(2.70 × 10⁻⁶) = 1.64 × 10⁻³ mol dm⁻³ [1]. pH = −log₁₀(1.64 × 10⁻³) = 2.78 [1].

(b) [H⁺] = Ka × [HA]/[A⁻] = 1.35 × 10⁻⁵ × (0.150/0.300) [1] = 6.75 × 10⁻⁶ mol dm⁻³ [1]. pH = −log₁₀(6.75 × 10⁻⁶) = 5.17 [1].

(c) The pH stays almost unchanged (falls only very slightly) [1]. The added H⁺ is removed by the propanoate ion, CH₃CH₂COO⁻ + H⁺ → CH₃CH₂COOH, shifting only a small amount of the buffer’s composition rather than allowing free H⁺ to accumulate [1].

8. (a) Ksp = [Ca²⁺][F⁻]² [1]. In pure water, [F⁻] = 2s, so Ksp = s(2s)² = 4s³ [1]. s = ∛(Ksp/4) = ∛(3.90 × 10⁻¹¹/4) = ∛(9.75 × 10⁻¹²) = 2.14 × 10⁻⁴ mol dm⁻³ [1].

(b) [F⁻] ≈ 0.100 mol dm⁻³ (from the NaF, since so little CaF₂ dissolves in comparison) [1]. [Ca²⁺] = Ksp/[F⁻]² = 3.90 × 10⁻¹¹/(0.100)² = 3.90 × 10⁻⁹ mol dm⁻³ [1]. This is far less soluble than in pure water — the common F⁻ ion suppresses CaF₂’s solubility [1].

9. Let mass in hexane = m; mass in water = (8.00 − m). Equal volumes, so Kpc = m/(8.00 − m) = 6.00 [1]. m = 6.00(8.00 − m) = 48.0 − 6.00m → 7.00m = 48.0 → m = 6.86 g in hexane [1], and 1.14 g in water [1].

10. (a) HCO₃⁻ + H⁺ → H₂CO₃ [1].

(b) The H₂CO₃ reservoir dissociates to release H⁺ (H₂CO₃ ⇌ H⁺ + HCO₃⁻), which then neutralises the added OH⁻ (H⁺ + OH⁻ → H₂O) [1], so — exactly as with the acetic acid/acetate buffer — a small addition shifts the equilibrium only slightly and blood pH barely changes [1].

11. Adding NaCl increases [Cl⁻] [1]; by Le Chatelier’s principle, the equilibrium AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq) shifts left to oppose the increase, removing Ag⁺ from solution as more AgCl precipitates, until Ksp is satisfied again [1].

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Where marks are usually lost

• Applying the weak-acid approximation ([H⁺] = √(Ka × C)) to a strong acid, or using the strong-acid method for a weak one.
• Forgetting the buffer equation uses the ratio [HA]/[A⁻], not either concentration alone.
• Missing the stoichiometric factor when writing Ksp expressions for compounds like PbCl₂ or CaF₂ (the power on the ion, and the corresponding multiplier when solving for s in pure water).
• Using the common-ion concentration as if it came partly from the sparingly soluble salt, rather than treating it as approximately equal to the concentration of the more soluble common-ion source.
• Describing a biological buffer (such as the blood’s HCO₃⁻/H₂CO₃ system) as working by a different mechanism from a laboratory buffer — the underlying chemistry (a weak acid/conjugate base pair resisting pH change) is identical.