Given that the solubility product constant (Ksp) of PbF2 is 3.6 × 10^{-8}, what is the molar solubility of PbF2 in pure water?

• A. 2.1 × 10^{-3} mol/L
• B. 3.6 × 10^{-8} mol/L
• C. 1.9 × 10^{-3} mol/L
• D. 1.1 × 10^{-3} mol/L

Answer: (D)

Verified step by step guidance

1. Write the dissociation equation for PbF\_2 in water: \(\mathrm{PbF_2 (s) \rightleftharpoons Pb^{2+} (aq) + 2F^- (aq)}\).
2. Define the molar solubility of PbF\_2 as \(s\), which represents the concentration of \(\mathrm{Pb^{2+}}\) ions in solution at equilibrium.
3. Express the equilibrium concentrations of the ions in terms of \(s\): \([\mathrm{Pb^{2+}}] = s\) and \([\mathrm{F^-}] = 2s\) because two fluoride ions are produced per formula unit dissolved.
4. Write the expression for the solubility product constant \(K_{sp}\) in terms of \(s\): \(K_{sp} = [\mathrm{Pb^{2+}}][\mathrm{F^-}]^2 = s \times (2s)^2 = 4s^3\).
5. Solve for \(s\) by rearranging the equation: \(s = \sqrt[3]{\frac{K_{sp}}{4}}\). Substitute the given \(K_{sp} = 3.6 \times 10^{-8}\) to find the molar solubility.