A-Level Chemistry May/June 2025 Q3(b)(iii): 5.00 cm³ of 0.100moldm-3 NaOH is added to 10.00cm³ of 0.100 moldm-3 CH3CH2CH2COOH. Calc…

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  (The value for Ka for butanoic acid, $1.5 \times 10^{-5} \text{ mol dm}^{-3}$, is assumed to be provided in the question data.)

  Step 1: Calculate initial moles of acid and base.

  Moles of CH₃CH₂CH₂COOH = concentration × volume $n(\text{acid}) = 0.100 \text{ mol dm}^{-3} \times \frac{10.00}{1000} \text{ dm}^3 = 0.00100 \text{ mol}$

  Moles of NaOH = concentration × volume $n(\text{base}) = 0.100 \text{ mol dm}^{-3} \times \frac{5.00}{1000} \text{ dm}^3 = 0.000500 \text{ mol}$

  Step 2: Calculate moles after reaction.

  CH₃CH₂CH₂COOH + NaOH → CH₃CH₂CH₂COO⁻Na⁺ + H₂O

  The reaction is 1:1. NaOH is the limiting reactant.

  Moles of CH₃CH₂CH₂COOH remaining = $0.00100 - 0.000500 = 0.000500 \text{ mol}$ Moles of CH₃CH₂CH₂COO⁻ formed = $0.000500 \text{ mol}$

  Step 3: Calculate pH.

  Since the moles of the weak acid (CH₃CH₂CH₂COOH) and its conjugate base (CH₃CH₂CH₂COO⁻) are equal, their concentrations in the solution are also equal.

  $[\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}] = [\text{CH}_3\text{CH}_2\text{CH}_2\text{COO}^-]$ (M1)

  For a buffer solution, the Henderson-Hasselbalch equation is: $\text{pH} = \text{p}K_a + \log_{10} \frac{[\text{CH}_3\text{CH}_2\text{CH}_2\text{COO}^-]}{[\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}]}$

  Since the concentrations are equal, the fraction is 1, and $\log_{10}(1) = 0$. Therefore, at the half-equivalence point, $\text{pH} = \text{p}K_a$.

  $\text{p}K_a = -\log_{10}(K_a)$ $\text{p}K_a = -\log_{10}(1.5 \times 10^{-5}) = 4.8239...$

  $\text{pH} = 4.82$ (to 2 d.p.) (M2)

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  How the marks are awarded

  • M1 — Awarded for correctly identifying, either explicitly or through calculation, that the moles (and therefore concentration) of the remaining butanoic acid are equal to the moles (and concentration) of the butanoate salt formed.
  • M2 — Awarded for the correct final answer of pH = 4.82, derived from the pKa of butanoic acid.

  Common mistakes

  • Calculating the pH of the remaining weak acid only, ignoring the buffering effect of the conjugate base (salt) that has been formed.
  • Making a mistake when calculating the initial moles, for example by forgetting to convert the volume from cm³ to dm³.
  • Not recognising the simplification that at half-equivalence pH = pKa, and instead performing a longer, more error-prone calculation involving the total volume and the full Henderson-Hasselbalch equation.
  • Using the wrong Ka value or making a calculator error when finding the pKa from Ka (e.g., forgetting the negative sign in -log(Ka)).

  Examiner tip: Recognise that when a weak acid is exactly half-neutralised by a strong base, the resulting buffer solution has a pH equal to the pKa of the acid, which greatly simplifies the calculation.

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