Multiple Choice
Which term describes the ability of a substance to dissolve in water?
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6. Chemical Quantities & Aqueous Reactions
Solubility Rules
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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
Verified step by step guidance1
Write the dissociation equation for lead(II) bromide: $\mathrm{PbBr_2 (s) \rightleftharpoons Pb^{2+} (aq) + 2Br^- (aq)}$.
Express the solubility product constant $K_{sp}$ in terms of the molar solubility $s$ and bromide ion concentration: $K_{sp} = [Pb^{2+}][Br^-]^2$.
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.
Set $[Pb^{2+}] = s$ (the molar solubility of PbBr$_2$) and substitute into the $K_{sp}$ expression: $K_{sp} = s \times (0.500)^2$.
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.
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