Question

The solubility product of $Pb{I}_2$ is $8.0\times {10}^{-9}$. The solubility of lead iodide in 0.1 molar solution of lead nitrate is $x\times {10}^{-6}mol/L$. Given: $\sqrt{2}=1.41$. The value of $x$ is

Answer 1

$Pb{I}_2\left(s\right)\rightleftharpoons P{b}^{2+}\left(aq\right)+2I^{-}\left(aq\right)$
$S+0.12S$
$K_{sp}\left(Pb{I}_2\right)=8\times {10}^9$
$K_{sp}=\left[P{b}^{2+}\right][I^{-}{]}^2$
$8\times {10}^{-9}=(S+0.1)(2S{)}^2$
$\Rightarrow S^2=2\times {10}^{-8}$
$S=1.414\times {10}^{-4}mol/L$
$=x\times {10}^{-6}mol/L$
$\therefore x=141.4\approx 141$