A-Level Chemistry May/June 2025 Q5(a)(ii): The value for the acid dissociation constant, Ka, of HClO(aq) is 3.70 x 10-8. Calculate…

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  Calculation of [HClO]

  Step 1: Calculate the hydrogen ion concentration, [H⁺]

  The relationship between pH and [H⁺] is given by: $pH = -log_{10}[H^+]$

  Therefore, $[H^+] = 10^{-pH}$

  Given $pH = 4.51$: $[H^+] = 10^{-4.51}$ $[H^+] = 3.090 imes 10^{-5} \, mol\,dm^{-3}$

  Step 2: Calculate the concentration of HClO, [HClO]

  The acid dissociation constant, $K_a$, for the weak acid HClO is given by: $K_a = \frac{[H^+][ClO^-]}{[HClO]}$

  For a weak acid, we assume that $[H^+] = [ClO^-]$. Therefore, the expression simplifies to: $K_a = \frac{[H^+]^2}{[HClO]}$

  Rearranging to solve for [HClO]: $[HClO] = \frac{[H^+]^2}{K_a}$

  Substitute the known values ($[H^+] = 3.090 imes 10^{-5} \, mol\,dm^{-3}$ and $K_a = 3.70 imes 10^{-8}$): $[HClO] = \frac{(3.090 imes 10^{-5})^2}{3.70 imes 10^{-8}}$ $[HClO] = \frac{9.548 imes 10^{-10}}{3.70 imes 10^{-8}}$ $[HClO] = 0.02580 \, mol\,dm^{-3}$

  Final Answer: $[HClO] = 0.0258 \, mol\,dm^{-3}$ (to 3 s.f.)

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  How the marks are awarded

  • M1 — Correctly calculating the hydrogen ion concentration, [H+], from the given pH value using the formula [H+] = 10^-pH.
  • M2 — Correctly rearranging the Ka expression to make [HClO] the subject and substituting the calculated [H+] and given Ka values to find the final concentration, given to at least 2 significant figures.

  Common mistakes

  • Incorrectly rearranging the Ka expression, for example calculating [HClO] = Ka / [H+]².
  • Forgetting to square the [H+] concentration in the numerator, incorrectly using [HClO] = [H+] / Ka.
  • Making a calculator error when computing 10⁻⁴.⁵¹, for example by omitting the negative sign.
  • Providing the final answer to only one significant figure, which is below the minimum requirement of two.

  Examiner tip: Master the algebraic rearrangement of the weak acid dissociation expression, Ka = [H+]²/[HA], as you may be required to solve for any of its variables.

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