Chemistry - Ionic Equilibrium Question with Solution | TestHub

Given: ${\left[{\mathrm{K}}_{\mathrm{sp}}\right]}_{{\mathrm{PbI}}_2}=8\times {10}^{-9}$

To calculate : solubility of ${\mathrm{PbI}}_2$ in $0.1\mathrm{M}$ solution of $\mathrm{Pb}{\left({\mathrm{NO}}_2\right)}_2$

$\left(\mathrm{I}\right) \mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2\to {\mathrm{Pb}}_{\left(\mathrm{aq}\right)}^{+2}+2{\mathrm{NO}}_3^{-}\left(\mathrm{aq}\right)$

$\begin{array}{ccc}0.1 \mathrm{M}& -& -\\ -& 0.1 \mathrm{M} & 0.2 \mathrm{M}\end{array}$

$\left(\mathrm{II}\right) {\mathrm{PbI}}_2\left(\mathrm{s}\right)\rightleftharpoons {\mathrm{Pb}}^{+2}\left(\mathrm{aq}\right)+2{\mathrm{I}}^{-}\left(\mathrm{aq}\right)$

$\begin{array}{cc}\mathrm{s} & 2\mathrm{s}\end{array}$

$\left[{\mathrm{Pb}}^{+2}\right]=\mathrm{s}+0.1$

$\simeq 0.1$

Now: ${\mathrm{K}}_{\mathrm{sp}}=8\times {10}^{-9}=\left[{\mathrm{Pb}}^{+2}\right]{\left[{\mathrm{I}}^{-}\right]}^2$

$\Rightarrow 8\times {10}^{-9}=0.1\times {\left(2 \mathrm{s}\right)}^2$

$\Rightarrow 8\times {10}^{-8}=4 {\mathrm{s}}^2\Rightarrow \mathrm{s}=\sqrt{2}\times {10}^{-4}$

$\Rightarrow \mathrm{S}=141\times {10}^{-6}\mathrm{M}$

$\Rightarrow \mathrm{x}=141$