A-Level Chemistry May/June 2024 Q1(b): Barium hydroxide, Ba(OH)2, is a strong base. A 250.0 cm³ solution of Ba(OH)2 with a pH…

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  Step 1: Calculate pOH and [OH⁻]

  Given pH = 12.2. The relationship between pH and pOH in aqueous solutions at 298 K is $pH + pOH = 14.0$.

  $pOH = 14.0 - 12.2 = 1.8$

  The concentration of hydroxide ions, [OH⁻], can be calculated from the pOH.

  $[OH⁻] = 10^{-pOH} = 10^{-1.8} = 0.015848...$ mol dm⁻³ $[OH⁻] \approx 0.0158$ mol dm⁻³

  (Alternatively, calculate [H⁺] first: $[H⁺] = 10^{-12.2} = 6.3095... \times 10^{-13}$ mol dm⁻³. Then use $K_w = [H⁺][OH⁻]$ to find $[OH⁻] = \frac{1.0 \times 10^{-14}}{6.3095... \times 10^{-13}} = 0.015848...$ mol dm⁻³)

  Step 2: Calculate the concentration of Ba(OH)₂

  Barium hydroxide is a strong base that dissociates fully in water according to the equation: $Ba(OH)_2(aq) \rightarrow Ba^{2+}(aq) + 2OH^-(aq)$

  The stoichiometric ratio of Ba(OH)₂ to OH⁻ is 1:2. Therefore, the concentration of Ba(OH)₂ is half the concentration of OH⁻.

  $[Ba(OH)_2] = \frac{[OH⁻]}{2} = \frac{0.015848...}{2} = 7.924... \times 10^{-3}$ mol dm⁻³ $[Ba(OH)_2] \approx 7.92 \times 10^{-3}$ mol dm⁻³

  Step 3: Calculate the moles of Ba(OH)₂

  The volume of the solution is 250.0 cm³, which must be converted to dm³.

  Volume = $\frac{250.0}{1000} = 0.2500$ dm³

  Moles = Concentration × Volume Moles of Ba(OH)₂ = $(7.924... \times 10^{-3}) \times 0.2500 = 1.981... \times 10^{-3}$ mol Moles of Ba(OH)₂ $\approx 1.98 \times 10^{-3}$ mol

  Step 4: Calculate the mass of Ba(OH)₂

  The molar mass (Mr) of Ba(OH)₂ is given as 171.3 g mol⁻¹.

  Mass = Moles × Mr Mass of Ba(OH)₂ = $(1.981... \times 10^{-3}) \times 171.3 = 0.3394...$ g

  Rounding to 3 significant figures: Mass of Ba(OH)₂ = 0.339 g

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  How the marks are awarded

  • M1 — Correctly calculating the hydrogen ion concentration as $[H^+] = 10^{-12.2}$ (or $6.31 \times 10^{-13}$ mol dm⁻³). This is the first step in relating pH to concentration.
  • M2 — Correctly calculating the hydroxide ion concentration as 0.0158 mol dm⁻³. This can be found from [H⁺] using Kw, or more directly by first calculating pOH = 1.8 and then using $[OH⁻] = 10^{-1.8}$.
  • M3 — Correctly calculating the concentration of barium hydroxide by applying the 1:2 stoichiometric ratio from the dissociation equation, i.e., $[Ba(OH)_2] = [OH⁻] / 2 = 7.92 \times 10^{-3}$ mol dm⁻³.
  • M4 — Correctly calculating the final mass by first finding the moles of Ba(OH)₂ in the 250.0 cm³ solution (moles = concentration × 0.2500 dm³) and then multiplying by the Mr (171.3) to get the final answer of 0.339 g.

  Common mistakes

  • Forgetting the 1:2 stoichiometry and incorrectly assuming $[Ba(OH)_2] = [OH⁻]$, which leads to a final mass that is double the correct answer (0.678 g).
  • Making a volume conversion error, such as multiplying the concentration by 250.0 instead of 0.2500, resulting in a mass 1000 times too large.
  • Confusing pH and pOH, for example by calculating $[Ba(OH)_2] = (10^{-12.2}) / 2$, which incorrectly uses the [H⁺] value to find the concentration of the base.
  • Stopping after calculating the concentration of Ba(OH)₂, failing to proceed to calculate the moles and mass required for the specific volume.

  Examiner tip: For multi-step calculations involving pH, always map out the path from the given information (pH) to the required quantity (mass), paying close attention to stoichiometry and unit conversions at each stage.

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