The $K_{sp}$ of $PbI_2$ is $1.4 \times 10^{-8}$. The molar mass of $PbI_2$ is $461 \ g \ mol^{-1}$. The molar mass of $Pb(NO_3)_2$ is $331.9 \ g \ mol^{-1}$. Find the solubility of $PbI_2$ in: $(a)$ $500 \ mL$ of water,$(b)$ $500 \ mL$ of $0.10 \ M \ KI$ solution,$(c)$ What is the weight of $PbI_2$ soluble in $500 \ mL$ of solution containing $1.33 \ g$ of $Pb(NO_3)_2$?
(A) For $PbI_2 \rightleftharpoons Pb^{2+} + 2I^-$,$K_{sp} = 4s^3 = 1.4 \times 10^{-8}$. Solving for $s$,$s = (0.35 \times 10^{-8})^{1/3} \approx 1.518 \times 10^{-3} \ M$. Solubility in $500 \ mL = 1.518 \times 10^{-3} \ mol \ L^{-1} \times 0.5 \ L \times 461 \ g \ mol^{-1} \approx 0.35 \ g$.
$(b)$ In $0.10 \ M \ KI$,$[I^-] = 0.10 \ M$. $K_{sp} = [Pb^{2+}][I^-]^2 \implies 1.4 \times 10^{-8} = [Pb^{2+}](0.1)^2$. $[Pb^{2+}] = 1.4 \times 10^{-6} \ M$. Solubility in $500 \ mL = 1.4 \times 10^{-6} \ mol \ L^{-1} \times 0.5 \ L \times 461 \ g \ mol^{-1} \approx 0.322 \times 10^{-3} \ g$.
$(c)$ $1.33 \ g \ Pb(NO_3)_2 = 1.33 / 331.9 \approx 0.004 \ mol$. In $0.5 \ L$,$[Pb^{2+}] = 0.004 / 0.5 = 0.008 \ M$. $K_{sp} = [Pb^{2+}][I^-]^2 \implies 1.4 \times 10^{-8} = (0.008)[I^-]^2$. $[I^-]^2 = 1.75 \times 10^{-6} \implies [I^-] \approx 1.323 \times 10^{-3} \ M$. Solubility of $PbI_2 = [I^-]/2 = 0.6615 \times 10^{-3} \ M$. Weight $= 0.6615 \times 10^{-3} \ mol \ L^{-1} \times 0.5 \ L \times 461 \ g \ mol^{-1} \approx 0.152 \ g$.
$(b)$ In $0.10 \ M \ KI$,$[I^-] = 0.10 \ M$. $K_{sp} = [Pb^{2+}][I^-]^2 \implies 1.4 \times 10^{-8} = [Pb^{2+}](0.1)^2$. $[Pb^{2+}] = 1.4 \times 10^{-6} \ M$. Solubility in $500 \ mL = 1.4 \times 10^{-6} \ mol \ L^{-1} \times 0.5 \ L \times 461 \ g \ mol^{-1} \approx 0.322 \times 10^{-3} \ g$.
$(c)$ $1.33 \ g \ Pb(NO_3)_2 = 1.33 / 331.9 \approx 0.004 \ mol$. In $0.5 \ L$,$[Pb^{2+}] = 0.004 / 0.5 = 0.008 \ M$. $K_{sp} = [Pb^{2+}][I^-]^2 \implies 1.4 \times 10^{-8} = (0.008)[I^-]^2$. $[I^-]^2 = 1.75 \times 10^{-6} \implies [I^-] \approx 1.323 \times 10^{-3} \ M$. Solubility of $PbI_2 = [I^-]/2 = 0.6615 \times 10^{-3} \ M$. Weight $= 0.6615 \times 10^{-3} \ mol \ L^{-1} \times 0.5 \ L \times 461 \ g \ mol^{-1} \approx 0.152 \ g$.
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