The questions are on the acids pH buffers test. Take it under real conditions before reading the mark scheme.
Quick answer: A Brønsted–Lowry acid donates a proton and a base accepts one. This mark scheme gives the model answers and the marks for the 8 questions of the Chemistry 11-12 test on acids pH buffers, from acids and bases to strong against weak acid, with the equivalent answers to accept.
Use these model answers and the marking notes to score yourself out of 20. Award method marks even when the final numerical value is slightly different because of rounding.
Level: Chemistry grades 11-12 (A Level · IB · Class 11-12) · Curriculum: IB R3.1
Question 1 Acids and bases (3 marks)
- A proton (H⁺) donor.
- Pair 1: HSO₄⁻ (acid) / SO₄²⁻ (conjugate base). Pair 2: NH₃ (base) / NH₄⁺ (conjugate acid).
Marking: (a) 1 mark. (b) 1 mark for each correctly identified pair.
Question 2 pH of nitric acid (2 marks)
HNO₃ ionises completely, so [H⁺] = 0.0075 mol/dm³. pH = −log(0.0075) = 2.12.Marking: 1 mark for [H⁺] = 0.0075 mol/dm³; 1 mark for 2.12.
Question 3 pH of propanoic acid (3 marks)
- Ka = [H⁺][CH₃CH₂COO⁻] / [CH₃CH₂COOH].
- [H⁺] = √(1.35 × 10⁻⁵ × 0.200) = √(2.70 × 10⁻⁶) = 1.64 × 10⁻³ mol/dm³; pH = 2.78.
Marking: (a) 1 mark. (b) 1 mark for [H⁺]; 1 mark for pH. Allow 2.8.
Question 4 Hydrocyanic acid buffer (2 marks)
pH = 9.31 + log(0.10 ÷ 0.40) = 9.31 − 0.60 = 8.71.Marking: 1 mark for the correct ratio [A⁻]/[HA] = 0.25; 1 mark for 8.71.
Question 5 Buffer response to alkali (3 marks)
The added OH⁻ reacts with the weak acid: CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O. The acid is plentiful so the OH⁻ is removed almost completely. [CH₃COOH] falls and [CH₃COO⁻] rises only slightly, so the ratio [A⁻]/[HA] and therefore the pH change very little.Marking: 1 mark for the equation; 1 mark for the acid removing OH⁻; 1 mark for the ratio changing only slightly.
Question 6 Titration of lactic acid (3 marks)
- Equivalence at 20.0 cm³: n(NaOH) = 0.0200 × 0.100 = 2.00 × 10⁻³ mol, so c = 2.00 × 10⁻³ ÷ 0.0200 = 0.100 mol/dm³.
- At 10.0 cm³ pH = 3.85, so pKa = 3.85 and Ka = 10⁻³·⁸⁵ = 1.4 × 10⁻⁴ mol/dm³.
Marking: (a) 1 mark. (b) 1 mark for pKa; 1 mark for Ka.
Question 7 pH at equivalence (2 marks)
At equivalence all the lactic acid has been converted to lactate ions. Lactate is a weak base: it accepts H⁺ from water, forming OH⁻, so the solution is slightly alkaline (pH about 8).Marking: 1 mark for lactate (conjugate base) present; 1 mark for reaction with water giving OH⁻.
Question 8 Strong against weak acid (2 marks)
HCl is fully ionised so [H⁺] = 0.10 mol/dm³. Ethanoic acid is only slightly ionised (about 1 %), so [H⁺] is much smaller (1.3 × 10⁻³ mol/dm³) and its pH is higher.Marking: 1 mark for complete against partial ionisation; 1 mark for the linked [H⁺] and pH comparison.
