A-Level Chemistry May/June 2025 Q3(a)(ii): Calculate the solubility, in gdm³, of Cr(OH)3 in pure water at 298 K. Show your working.

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  The dissociation equilibrium for chromium(III) hydroxide is:

  $Cr(OH)_3(s) \rightleftharpoons Cr^{3+}(aq) + 3OH^-(aq)$

  The solubility product constant expression is:

  $K_{sp} = [Cr^{3+}][OH^-]^3$

  Let the molar solubility of $Cr(OH)_3$ be 's' in mol dm⁻³. At equilibrium: $[Cr^{3+}] = s$ $[OH^-] = 3s$

  Substitute these into the $K_{sp}$ expression (assuming $K_{sp} = 1.0 \times 10^{-33} \text{ mol}^4 \text{dm}^{-12}$ from question data):

  $K_{sp} = (s)(3s)^3$ $K_{sp} = (s)(27s^3)$ $K_{sp} = 27s^4$

  Rearrange to solve for s:

  $s = \sqrt[4]{\frac{K_{sp}}{27}}$ $s = \sqrt[4]{\frac{1.0 \times 10^{-33}}{27}}$ $s = 2.466... \times 10^{-9} \text{ mol dm}^{-3}$ $s \approx 2.47 \times 10^{-9} \text{ mol dm}^{-3}$

  To find the solubility in g dm⁻³, multiply the molar solubility by the molar mass ($M_r$) of $Cr(OH)_3$.

  $M_r(Cr(OH)_3) = 52.0 + 3 \times (16.0 + 1.0) = 103.0 \text{ g mol}^{-1}$

  Solubility in g dm⁻³ = $s \times M_r$ Solubility = $(2.466... \times 10^{-9} \text{ mol dm}^{-3}) \times (103.0 \text{ g mol}^{-1})$ Solubility = $2.540... \times 10^{-7} \text{ g dm}^{-3}$

  Solubility = $2.54 \times 10^{-7} \text{ g dm}^{-3}$ (to 3 s.f.)

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  How the marks are awarded

  • M1 — Correctly establishing the stoichiometric relationship between the hydroxide and chromium(III) ion concentrations, shown in the working as $[OH^-] = 3s$ when $[Cr^{3+}] = s$.
  • M2 — Correctly calculating the molar solubility, $s$, by substituting the concentration terms into the $K_{sp}$ expression ($K_{sp} = 27s^4$) and solving to get $2.47 \times 10^{-9} \text{ mol dm}^{-3}$.
  • M3 — Correctly converting the molar solubility into solubility in g dm⁻³ by multiplying by the molar mass of Cr(OH)₃ ($103.0 \text{ g mol}^{-1}$) to obtain the final answer of $2.54 \times 10^{-7} \text{ g dm}^{-3}$.

  Common mistakes

  • Incorrectly setting up the $K_{sp}$ expression, e.g., writing $K_{sp} = (s)(3s)$ instead of $K_{sp} = (s)(3s)^3$.
  • Making a mathematical error when expanding $(3s)^3$, often writing $3s^3$ or $9s^3$ instead of the correct $27s^3$, which leads to an incorrect final expression for $K_{sp}$.
  • Forgetting to convert the final answer from molar solubility (mol dm⁻³) to solubility in g dm⁻³, thereby only completing the first two steps of the calculation.
  • Using an incorrect molar mass for Cr(OH)₃, for example by miscounting the number of oxygen or hydrogen atoms.

  Examiner tip: Always write the dissolution equilibrium equation first to correctly determine the stoichiometric relationships between the ions, which is crucial for setting up the $K_{sp}$ expression.

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