Given the solubility product constant K_{sp} for PbBr_2 is 6.6 imes 10^{-6}, what is the molar solubility of PbBr_2 in a 0.500 M KBr solution?

• A. 1.3 imes 10^{-5} M
• B. 0.500 M
• C. 6.6 imes 10^{-6} M
• D. 2.6 imes 10^{-5} M

Answer: (A)

Verified step by step guidance

1. Write the dissociation equation for lead(II) bromide: \(\mathrm{PbBr_2 (s) \rightleftharpoons Pb^{2+} (aq) + 2Br^- (aq)}\).
2. Express the solubility product constant \(K_{sp}\) in terms of the molar solubility \(s\) and bromide ion concentration: \(K_{sp} = [Pb^{2+}][Br^-]^2\).
3. Since the solution already contains 0.500 M KBr, which fully dissociates to provide \([Br^-] = 0.500\) M, account for this common ion effect by setting \([Br^-] = 0.500 + 2s\), but because \(s\) is small, approximate \([Br^-] \approx 0.500\) M.
4. Set \([Pb^{2+}] = s\) (the molar solubility of PbBr\(_2\)) and substitute into the \(K_{sp}\) expression: \(K_{sp} = s \times (0.500)^2\).
5. Solve for \(s\) by rearranging the equation: \(s = \frac{K_{sp}}{(0.500)^2}\) to find the molar solubility of PbBr\(_2\) in the 0.500 M KBr solution.