The solubility of ${\mathrm{PbCl}}_2$ in water is 0.01 M 25° C. Its maximum concentration in 0.1 M NaCl will be:

${\mathrm{K}}_{\mathrm{sp}}{\mathrm{PbCl}}_2=4{\mathrm{s}}^3=4\times (0.01{)}^3=4\times {10}^{-6}$

In NaCl solution for ${\mathrm{PbCl}}_2$

${\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Pb}}^2\right]{\left[{\mathrm{Cl}}^{-}\right]}^2$

$$\begin{gathered} {\mathrm{K}}_{0=}=4\times {\left({10}^{-2}\right)}^3=4\times {10}^{-6}; \\ 4\times {10}^{-6}={\left(\mathrm{Pb}\right)}^2\times (0.1{)}^2 \end{gathered}$$

$\therefore \left[{\mathrm{Pb}}^{2+}\right]=4\times {10}^{-4}\mathrm{M}$