The solubility of $PbS{O}_4$ in $0.01M$ $N{a}_{2}S{O}_4$ solution is $\left(K_{sp}\text{of}PbS{O}_4\text{is}1.25\times {10}^{-9}\right)$

$\left[P{b}^{2+}\right]\left[S{O}_4^{2-}\right]=K_{sp}=1.25\times {10}^{-9}$

$\left[P{b}^{2+}\right]=\text{solubility=}\frac{\text{1}\text{.25}\times {\text{10}}^{-9}}{\left[S{O}_4^{2-}\right]}$

                                     $=\frac{1.25\times {10}^{-9}}{0.01}$

                                     $=1.25\times {10}^{-7}M$