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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−30.0500 \ mol \ dm^{-3} solution of propanoic acid at 298 K. The KaK_a for propanoic acid is 1.35×10−5 mol dm−31.35 \times 10^{-5} \ mol \ dm^{-3}.

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  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)
    • [CH3CH2COOH]eqm≈[CH3CH2COOH]initial=0.0500 mol dm−3[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+]=6.75×10−7=8.216×10−4 mol dm−3[H^+] = \sqrt{6.75 \times 10^{-7}} = 8.216 \times 10^{-4} \ mol \ dm^{-3}
  5. Calculate pH: pH=−log10(8.216×10−4)=3.09pH = -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−30.800 \ 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 CH3COOH=1.75×10−5 mol dm−3CH_3COOH = 1.75 \times 10^{-5} \ mol \ dm^{-3})

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  1. Calculate moles of conjugate base (ethanoate): Moles of CH3COONa=massMr=12.382.0=0.150 molCH_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 CH3COOH=concentration×volume=0.800×2501000=0.200 molCH_3COOH = \text{concentration} \times \text{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+]=(1.75×10−5)×0.2000.150=2.333×10−5 mol dm−3[H^+] = (1.75 \times 10^{-5}) \times \frac{0.200}{0.150} = 2.333 \times 10^{-5} \ mol \ dm^{-3}
  5. Calculate pH: pH=−log10(2.333×10−5)=4.63pH = -log_{10}(2.333 \times 10^{-5}) = 4.63 (to 2 d.p.)
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