객관식
A solution containing Ca(NO3)2 is mixed with a solution of Li2S to form a solution that is 2.1 × 10^-5 M in calcium ion and 4.75 × 10^-5 M in sulfide ion. What will happen once these solutions are mixed? Ksp (CaS) = 2.3 × 10^-9.
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18. Aqueous Equilibrium
Solubility Product Constant: Ksp
Multiple Choice
Given that the solubility product constant (Ksp) for PbBr2 is 6.6 × 10^{-6}, what is the molar solubility of PbBr2 in a 0.500 M Pb(NO3)2 solution?
A
1.1 × 10^{-6} M
B
1.8 × 10^{-5} M
C
6.6 × 10^{-6} M
D
3.3 × 10^{-3} M
검증된 단계별 안내1
Write the dissociation equation for PbBr\_2 in water: \(\mathrm{PbBr_2 (s) \rightleftharpoons Pb^{2+} (aq) + 2Br^- (aq)}\).
Express the solubility product constant \(K_{sp}\) in terms of the molar solubility \(s\) and the initial concentration of \(\mathrm{Pb^{2+}}\) from \(\mathrm{Pb(NO_3)_2}\): \(K_{sp} = [\mathrm{Pb^{2+}}][\mathrm{Br^-}]^2\).
Since the solution already contains \(0.500\,M\) \(\mathrm{Pb^{2+}}\) from \(\mathrm{Pb(NO_3)_2}\), let the molar solubility of \(\mathrm{PbBr_2}\) be \(s\). Then, \([\mathrm{Pb^{2+}}] = 0.500 + s\) and \([\mathrm{Br^-}] = 2s\).
Substitute these concentrations into the \(K_{sp}\) expression: \(6.6 \times 10^{-6} = (0.500 + s)(2s)^2 = (0.500 + s)4s^2\).
Assuming \(s\) is very small compared to \(0.500\), approximate \(0.500 + s \approx 0.500\) and solve for \(s\) by rearranging: \(6.6 \times 10^{-6} = 0.500 \times 4s^2\), then solve for \(s\).
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