AQA A-Level Physical Chemistry 3.1.12 Acids and Bases Exam Practice
3.1.12

Acids and Bases Exam Practice

Exam-style practice questions with detailed mark schemes and examiner tips.

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Structured Questions

Answer the questions below in the spaces provided. Check your answers with the mark schemes and examiner tips.

Question 1: Standard pH Calculations

9 marks

Calculate the pH of each of the following solutions at 298 K, giving all answers to 2 decimal places. (Kw = 1.00 x 10^-14 mol^2 dm^-6 at 298 K).

(a) Calculate the pH of a 0.035 mol dm^-3 solution of nitric acid (HNO3). [2]

(b) Calculate the pH of a 0.025 mol dm^-3 solution of potassium hydroxide (KOH). [3]

(c) Propanoic acid (CH3CH2COOH) is a weak acid with Ka = 1.35 x 10^-5 mol dm^-3 at 298 K. Calculate the pH of a 0.120 mol dm^-3 solution of propanoic acid, stating any assumptions made. [4]

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(a)

  • HNO3 is a strong acid, so [H^+] = 0.035 mol dm^-3 [1]
  • pH = -log10(0.035) = 1.46 [1]

(b)

  • KOH is a strong base, so [OH^-] = 0.025 mol dm^-3 [1]
  • [H^+] = Kw / [OH^-] = 1.00 × 10^-14 / 0.025 = 4.00 × 10^-13 mol dm^-3 [1]
  • pH = -log10(4.00 × 10^-13) = 12.40 [1]

(c)

  • [H^+] = √(Ka × [HA]) = √(1.35 × 10^-5 × 0.120) [1]
  • [H^+] = √(1.62 × 10^-6) = 1.27 × 10^-3 mol dm^-3 [1]
  • pH = -log10(1.27 × 10^-3) = 2.90 (accept 2.89) [1]
  • Assumptions (any one from):
    • Assume [H^+] = [A^-] at equilibrium (i.e. negligible H^+ contribution from the autoionisation of water)
    • Assume [HA] at equilibrium equals the initial [HA] (i.e. dissociation of the weak acid is negligible) [1]
Examiner Tip: pH values must always be quoted to 2 decimal places in A-Level Chemistry. The number of decimal places in a logarithmic value corresponds to the number of significant figures in the original concentration.

Question 2: pH Curves & Indicators

6 marks

A titration is carried out by adding 0.100 mol dm^-3 sodium hydroxide solution to 25.0 cm^3 of 0.100 mol dm^-3 ethanoic acid solution.

(a) Outline the key characteristics of the pH titration curve for this reaction. Specify the initial pH, the equivalence point pH, and the pH range of the vertical section. [3]

(b) Explain how you would select a suitable indicator for this titration using the pH curve. [3]

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(a)

  • Initial pH should start in the weak acid range (typically pH 2.5 to 3.0) [1]
  • The equivalence point pH is basic / greater than 7 (typically pH 8.5 to 9.0) because the salt formed (sodium ethanoate) hydrolyses slightly [1]
  • The vertical section of the pH curve lies entirely in the basic region, typically spanning from pH 7.0 to 10.0 [1]

(b)

  • The chosen indicator must change colour completely within the steep / vertical section of the pH titration curve [1]
  • The pH range of the indicator's transition / colour change must fall entirely within this vertical section [1]
  • Phenolphthalein (pH range 8.3 to 10.0) is suitable because its transition range matches the vertical section of a weak acid-strong base titration curve [1]
Examiner Tip: Make sure your sketches reflect key features: starting pH (strong acid is ~1, weak acid is ~3), equivalence point pH (strong-strong is 7, weak-strong is >7, strong-weak is <7), and the pH at half-neutralisation (where pH = pKa).

Question 3: Buffer Solution Calculations & Action

8 marks

A buffer solution is prepared by mixing 50.0 cm^3 of 0.200 mol dm^-3 ethanoic acid (CH3COOH) with 50.0 cm^3 of 0.150 mol dm^-3 sodium ethanoate (CH3COONa). For ethanoic acid, Ka = 1.76 x 10^-5 mol dm^-3 at 298 K.

(a) Calculate the pH of this buffer solution at 298 K. [4]

(b) Explain, with the help of a chemical equation, how this buffer solution minimises pH change when a small volume of hydrochloric acid is added. [4]

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(a)

  • Moles of CH3COOH = 0.200 × (50.0 / 1000) = 0.0100 mol [1]
  • Moles of CH3COO^- (from salt) = 0.150 × (50.0 / 1000) = 0.00750 mol [1]
  • [H^+] = Ka × (Moles of HA / Moles of A^-) = 1.76 × 10^-5 × (0.0100 / 0.00750) = 2.35 × 10^-5 mol dm^-3 [1]
  • pH = -log10(2.35 × 10^-5) = 4.63 [1]

(b)

  • Hydrochloric acid dissociates to release H^+ ions into the solution [1]
  • The conjugate base (ethanoate ions, CH3COO^-) in the buffer reacts with the added H^+ ions to form ethanoic acid:
    CH3COO^-(aq) + H^+(aq) → CH3COOH(aq) [2] (1 mark for correct equation; 1 mark for state symbols/explanation)
  • Because the added H^+ ions are converted into the weak acid, the concentration of hydrogen ions remains almost constant, minimising the change in pH [1]
Examiner Tip: Since the acid and conjugate base are in the same total volume, you can use moles directly in the buffer expression instead of calculating concentrations, as the volume term cancels out.
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