Question

The solubility product of lead iodide is $1.4\times {10}^{-8}$. Calculate its molar solubility in $0.1M$ $KI$ solution.

Answer 1

Let the solubility of $Pb{I}_2$ be $S$. Then
$Pb{I}_2\rightleftharpoons {Pb}^{2+}+2I^{-}$
$S$ $S$ $2S$
Potassium iodide is a strong electrolyte and is completely ionised. It shall provide $I^{-}$ ion concentration $=0.1M$
$\left[{Pb}^{2+}\right]=S$
$\left[I^{-}\right]=(2S+0.1)M$
$K_{sp}=\left[{Pb}^{2+}\right]{\left[I^{-}\right]}^2=S\times {(2S+0.1)}^2=4S^3+0.01S+0.4S^2$
Neglecting $S^3$ and $S^2$
$1.4\times {10}^{-8}=0.01S$
or $S=\frac{1.4\times {10}^{-8}}{0.01}=1.4\times {10}^{-6}mol{L}^{-1}$