Question

Calculate solubility of $Pb{I}_2$ $(K_{sp}=1.4\times 10{}^{-8})$ in water at ${25}^{o}C$, which is $90\%$ dissociated.

Answer 1

Given:-

$K_{sp}\left(Pb{I}_2\right)=1.4\times {10}^{-8}$

$\alpha =90$%

$Pb{I}_2\left(s\right)\rightleftharpoons P{b}^{+2}\left(aq\right)+2I^{⊝}\left(aq\right)$

$0.9S$ $2\times 0.9S$

$S⟶$ Solubility

$K_{sp}=0.9S(2\times 0.9S{)}^2$

$\Rightarrow 1.4\times {10}^{-8}=4(0.9{)}^3{S}^3$

$\Rightarrow S^3=\frac{1.4\times {10}^{-8}}{4\times (0.9{)}^3}$

$\Rightarrow S=0.78\times {10}^{-2}$ Moles/litre