Multiple Choice
Given that the solubility product constant (Ksp) of PbF2 is 3.6 × 10^{-8}, what is the molar solubility of PbF2 in pure water?
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18. Aqueous Equilibrium
Solubility Product Constant: Ksp
Multiple Choice
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
Verified step by step guidance1
Write the dissociation equation for calcium fluoride (CaF2): \(\mathrm{CaF_2 (s) \rightleftharpoons Ca^{2+} (aq) + 2F^- (aq)}\).
Express the solubility product constant (\(K_{sp}\)) in terms of the molar solubility \(s\) and fluoride ion concentration: \(K_{sp} = [Ca^{2+}][F^-]^2\).
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.
Substitute \([Ca^{2+}] = s\) and \([F^-] \approx 0.030\) into the \(K_{sp}\) expression: \(K_{sp} = s \times (0.030)^2\).
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.
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