Given that the solubility product constant (Ksp) for CaF2 is 4.0 × 10^{-11}, what is the molar solubility of CaF2 in a 0.030 M NaF solution?

• A. 1.2 × 10^{-7} M
• B. 2.2 × 10^{-5} M
• C. 7.3 × 10^{-6} M
• D. 4.4 × 10^{-9} M

Answer: (D)

Verified step by step guidance

1. Write the dissociation equation for calcium fluoride (CaF2): \(\mathrm{CaF_2 (s) \rightleftharpoons Ca^{2+} (aq) + 2F^- (aq)}\).
2. Express the solubility product constant (\(K_{sp}\)) in terms of the molar solubility \(s\) and fluoride ion concentration: \(K_{sp} = [Ca^{2+}][F^-]^2\).
3. Since the solution already contains 0.030 M NaF, which fully dissociates to provide fluoride ions, the fluoride ion concentration is \([F^-] = 0.030 + 2s\) (where \(s\) is the molar solubility of CaF2). Because \(s\) is expected to be very small compared to 0.030, approximate \([F^-] \approx 0.030\) M.
4. Substitute \([Ca^{2+}] = s\) and \([F^-] \approx 0.030\) into the \(K_{sp}\) expression: \(K_{sp} = s \times (0.030)^2\).
5. Solve for \(s\) (the molar solubility of CaF2) by rearranging the equation: \(s = \frac{K_{sp}}{(0.030)^2}\). This will give the molar solubility in the presence of 0.030 M NaF.