A-Level Chemistry May/June 2025 Q1(b): A sample of 0.333 g of strontium oxide, SrO, is completely dissolved in distilled water…

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  The reaction of strontium oxide with water forms strontium hydroxide:

  $SrO(s) + H_2O(l) \rightarrow Sr(OH)_2(aq)$

  Step 1: Calculate moles of SrO and Sr(OH)₂

  Molar mass of SrO, $M_r(SrO) = 87.6 + 16.0 = 103.6 \text{ g mol}^{-1}$

  Moles of SrO = $\frac{\text{mass}}{M_r} = \frac{0.333}{103.6} = 3.214 \times 10^{-3} \text{ mol}$

  From the 1:1 stoichiometry of the reaction, moles of Sr(OH)₂ produced is also $3.214 \times 10^{-3} \text{ mol}$.

  Step 2: Calculate the concentration of hydroxide ions, [OH⁻]

  Strontium hydroxide is a strong base and dissociates fully in water:

  $Sr(OH)_2(aq) \rightarrow Sr^{2+}(aq) + 2OH^-(aq)$

  Each mole of Sr(OH)₂ produces two moles of OH⁻ ions. Moles of OH⁻ = $2 \times 3.214 \times 10^{-3} = 6.429 \times 10^{-3} \text{ mol}$

  The solution volume is $250.0 \text{ cm}^3 = 0.2500 \text{ dm}^3$.

  Concentration of OH⁻, $[OH^-] = \frac{\text{moles}}{\text{volume}} = \frac{6.429 \times 10^{-3}}{0.2500} = 0.0257 \text{ mol dm}^{-3}$

  Step 3: Calculate pOH

  pOH = $-\log_{10}[OH^-]$ pOH = $-\log_{10}(0.0257) = 1.59$

  Step 4: Calculate pH

  At 298 K, pH + pOH = 14.00 pH = $14.00 - pOH$ pH = $14.00 - 1.59 = 12.41$

  pH = 12.41

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  How the marks are awarded

  • M1 — Correctly calculating the moles of SrO as 0.333 ÷ 103.6, leading to 3.214 × 10⁻³ mol, and stating that this is also the number of moles of Sr(OH)₂.
  • M2 — Correctly calculating the concentration of hydroxide ions, [OH⁻]. This requires doubling the moles of Sr(OH)₂ to get moles of OH⁻ (6.429 × 10⁻³ mol) and then dividing by the volume in dm³ (0.2500) to get 0.0257 mol dm⁻³.
  • M3 — Correctly calculating an intermediate value from [OH⁻]. In this method, it is for calculating the pOH: -log(0.0257) = 1.59.
  • M4 — Correctly calculating the final pH by subtracting the pOH from 14.00, and giving the answer to the required two decimal places (12.41).

  Common mistakes

  • Forgetting the 1:2 stoichiometry between Sr(OH)₂ and OH⁻, leading to an [OH⁻] that is half the correct value and an incorrect final pH of 12.11.
  • Calculating pOH correctly (1.59) but then stating this as the final pH, forgetting to subtract from 14.
  • Making a volume conversion error, such as dividing by 250 instead of 0.250, resulting in a vastly incorrect concentration.
  • Failing to give the final answer to two decimal places as instructed in the question, which would lose the final accuracy mark.

  Examiner tip: For pH calculations involving bases, always determine the [OH⁻] first by considering the stoichiometry of dissociation before using the pOH or Kₗ relationships.

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