QCE Chemistry Paper 2 free practice questions with answers

Question 1 (6 marks) [Solubility]

The solubility product of CaF₂ is K_sp = 3.9 × 10−¹¹ at 25 °C: CaF₂(s)

Ca²+ + 2F−.

(a) Calculate the solubility of CaF₂ in water, in mol L−¹.

(a) Answer: 2.14 × 10−⁴ mol L−¹

• Let the solubility be s. Then [cation] = s and [anion] = 2s.
• K_sp = s × (2s)² = 4s³
• s = ³√(3.9 × 10−¹¹ ÷ 4) = 2.14 × 10−⁴ mol L−¹

(b) Equal volumes of a 8.2 × 10−⁴ mol L−¹ solution of the cation and a 7.2 × 10−³ mol L−¹ solution of the anion (ratio 1 : 2) are mixed. Calculate the ion product Q and state whether a precipitate forms.

(b) Answer: Q = 5.31 × 10−⁹; Q > K_sp so a precipitate forms

• Mixing equal volumes halves each concentration: [cation] = 4.1 × 10−⁴ mol L−¹, [anion] = 3.6 × 10−³ mol L−¹.
• Q = [cation] × [anion]² = 5.31 × 10−⁹
• Compare with K_sp = 3.9 × 10−¹¹: Q is greater than K_sp, so a precipitate forms.

K_sp = [cation]^a [anion]^b: write the ion concentrations in terms of the solubility s (remembering multiples like 2s). A precipitate forms when the ion product Q exceeds K_sp.