The questions are on the buffer solutions test. Take it under real conditions before reading the mark scheme.
Quick answer: A buffer is a solution of a weak acid and its conjugate base that resists pH change. This mark scheme gives the model answers and the marks for the 8 questions of the AP Chemistry test on buffer solutions, from defining a buffer to not a buffer, with the equivalent answers to accept.
Each model answer is followed by a marking guide. Follow-through marks apply where a later step correctly uses an earlier wrong value.
Level: AP Chemistry · Curriculum: AP Chem Unit 8
Question 1 Defining a buffer (2 marks)
A buffer resists changes in pH when small amounts of acid or base are added. It consists of a weak acid and its conjugate base, for example CH₃COOH and CH₃COONa.Marking: 1 mark for the definition, 1 mark for a correct pair.
Question 2 Buffer pH (3 marks)
pH = pKa + log₁₀([F⁻]/[HF]) = 3.17 + log₁₀(0.25/0.15) = 3.17 + 0.22 = 3.39.Marking: 1 mark for the equation, 1 for substitution, 1 for the answer.
Question 3 Adding acid (3 marks)
HCl reacts with NH₃: NH₃ = 0.15 − 0.020 = 0.13 mol; NH₄⁺ = 0.20 + 0.020 = 0.22 mol. pH = 9.25 + log₁₀(0.13/0.22) = 9.02.Marking: 1 mark for the stoichiometry, 1 for the new amounts, 1 for the pH.
Question 4 Effective range (2 marks)
The ratio [A⁻]/[HA] is between 0.1 and 10 in this range, so both components remain in comparable amounts. Each can neutralise added base or acid, giving high buffer capacity.Marking: 1 mark for the ratio, 1 for the capacity argument.
Question 5 Reading a titration curve (3 marks)
- The steep rise is at 20 cm³.
- At half-equivalence, 10 cm³, pH = 4.8, so pKa = 4.8.
- Phenolphthalein, whose range includes the equivalence pH of about 8.7.
Marking: 1 mark for each part.
Question 6 Strong acid titration (2 marks)
n(HCl) = 0.0200 × 0.150 = 3.00 × 10⁻³ mol. V = 3.00 × 10⁻³ / 0.100 = 0.0300 dm³ = 30.0 cm³.Marking: 1 mark for the amount, 1 for the volume.
Question 7 Equivalence pH of HCN (3 marks)
Equivalence at 25.0 cm³: [CN⁻] = 5.00 × 10⁻³ mol / 0.0500 dm³ = 0.100 mol/dm³. Kb = 1.0 × 10⁻¹⁴ / 4.9 × 10⁻¹⁰ = 2.04 × 10⁻⁵. [OH⁻] = √(2.04 × 10⁻⁵ × 0.100) = 1.429 × 10⁻³ mol/dm³; pOH = 2.85; pH = 11.15. CN⁻ is a weak base, so it makes the solution basic.Marking: 1 mark for [CN⁻], 1 for the pH, 1 for the explanation.
Question 8 Not a buffer (2 marks)
HCl is a strong acid and Cl⁻ is a negligible base, so there is no basic species to neutralise added acid. A buffer needs a weak acid and its conjugate base.Marking: 1 mark for the point about a strong acid, 1 for the missing conjugate pair.
