If the solubility of ${\mathrm{Ag}}_2{\mathrm{SO}}_4$ in ${10}^{-2} \mathrm{M} {\mathrm{Na}}_2{\mathrm{SO}}_4$ solution be $2 \mathrm{x}{10}^{-8} \mathrm{M}$ then ${\mathrm{K}}_{\mathrm{sp}}$ of ${\mathrm{Ag}}_2{\mathrm{SO}}_4$ will be:

$\left[{\mathrm{Na}}_2{\mathrm{SO}}_4\right]={10}^{-2}\mathrm{m}$

Solubility of  ${\mathrm{Ag}}_2{\mathrm{SO}}_4$ $=2\times {10}^{-8}\mathrm{m}=\mathrm{s}$

${\mathrm{K}}_{\mathrm{sp}}$ of ${\mathrm{Ag}}_2{\mathrm{SO}}_4$ = ?

${\mathrm{Na}}_2{\mathrm{SO}}_4⟶2{\mathrm{Na}}^{+}+{\mathrm{SO}}^{2-}$

${\mathrm{Ag}}_2{\mathrm{SO}}_4\rightleftharpoons {\mathrm{ag}}^{+}+{\mathrm{SO}}_4^{2-}$

$\begin{array}{l}S=2\times {10}^{-8}\\ \left[{\mathrm{Ag}}^{+}\right]=4\times {10}^{-8}\end{array}$

${\mathrm{K}}_{\mathrm{sp}}=[\mathrm{Ag}{]}^2\left[{\mathrm{SO}}_2^{4-}\right]$

${\mathrm{K}}_{\mathrm{sp}}=16\times {10}^{-16}\times {10}^{-2}$

$\begin{array}{l}={\left(2\times 2\times {10}^{-8}\right)}^2\left({10}^{-2}\right)\\ =16\times {10}^{-16-2}\\ =16\times {10}^{-18}\end{array}$