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9701 · 25.1

Acids and bases — practice questions

Practice and worked examples for 9701 Acids and bases. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

Calculate the pH of a $0.0500 \ mol \ dm^{-3}$ solution of propanoic acid at 298 K. The KaK_a for propanoic acid is $1.35 \times 10^{-5} \ mol \ dm^{-3}$.

Show solution outline
  1. Write the Ka expression: CH3CH2COOH⇌H++CH3CH2COO−CH_3CH_2COOH \rightleftharpoons H^+ + CH_3CH_2COO^- Ka=[H+][CH3CH2COO−][CH3CH2COOH]K_a = \frac{[H^+][CH_3CH_2COO^-]}{[CH_3CH_2COOH]}
  2. State assumptions: For a weak acid, we assume the dissociation is small, so:
    • [H+]=[CH3CH2COO−][H^+] = [CH_3CH_2COO^-] (from stoichiometry)
    • $[CH_3CH_2COOH]{eqm} \approx [CH_3CH_2COOH]{initial} = 0.0500 \ mol \ dm^{-3}$
  3. Substitute into Ka expression: 1.35×10−5=[H+]20.05001.35 \times 10^{-5} = \frac{[H^+]^2}{0.0500}
  4. Solve for [H⁺]: [H+]2=(1.35×10−5)×0.0500=6.75×10−7[H^+]^2 = (1.35 \times 10^{-5}) \times 0.0500 = 6.75 \times 10^{-7} $[H^+] = \sqrt{6.75 \times 10^{-7}} = 8.216 \times 10^{-4} \ mol \ dm^{-3}$
  5. Calculate pH: $pH = -log_{10}(8.216 \times 10^{-4}) = 3.09$ (to 2 d.p.)

Worked example 2

A buffer solution is made by dissolving 12.3 g of sodium ethanoate (CH3COONaCH_3COONa) in 250 cm³ of $0.800 \ mol \ dm^{-3}$ ethanoic acid. Calculate the pH of the resulting buffer solution. (MrM_r of CH3COONa=82.0CH_3COONa = 82.0; KaK_a for $CH_3COOH = 1.75 \times 10^{-5} \ mol \ dm^{-3}$)

Show solution outline
  1. Calculate moles of conjugate base (ethanoate): Moles of $CH_3COONa = \frac{mass}{M_r} = \frac{12.3}{82.0} = 0.150 \ mol$. This is moles of A−A^-.
  2. Calculate moles of weak acid (ethanoic acid): Moles of $CH_3COOH = concentration \times volume = 0.800 \times \frac{250}{1000} = 0.200 \ mol$. This is moles of HAHA.
  3. Use the Ka expression (or Henderson-Hasselbalch): We can use the mole ratio directly since they are in the same volume. Ka=[H+][A−][HA]K_a = \frac{[H^+][A^-]}{[HA]} Rearranging for [H+][H^+]: [H+]=Ka×[HA][A−]=Ka×moles(HA)moles(A−)[H^+] = K_a \times \frac{[HA]}{[A^-]} = K_a \times \frac{moles(HA)}{moles(A^-)}
  4. Substitute values and calculate [H⁺]: $[H^+] = (1.75 \times 10^{-5}) \times \frac{0.200}{0.150} = 2.333 \times 10^{-5} \ mol \ dm^{-3}$
  5. Calculate pH: $pH = -log_{10}(2.333 \times 10^{-5}) = 4.63$ (to 2 d.p.)
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