if $K_{\mathrm{s}}\left({\mathrm{PbF}}_2\right)=7.2\times {10}^{-8}{\mathrm{M}}^3$ and $K_{\mathrm{s}}\left({\mathrm{MgF}}_2\right)=1.44\times {10}^{-7}{\mathrm{M}}^3$ then the saturation solubilities of

${\mathrm{PbF}}_2\text{ and }{\mathrm{MgF}}_2$ present in the same solution, respectively, are

Let x and y be the respective solubilities of ${\mathrm{PbF}}_2\text{ and }{\mathrm{MgF}}_2$. Then $\left(\left[{\mathrm{Pb}}^{2+}\right]=x,\left[{\mathrm{Mg}}^{2+}\right]=y\text{ and }\left[{\mathrm{F}}^{-}\right]=2(x+y)\right.$

For ${\mathrm{PbF}}_2$     $K_{\mathrm{s}}=\left[{\mathrm{Pb}}^{2+}\right]{\left[{\mathrm{F}}^{-}\right]}^2=4x(x+y{)}^2=7.2\times {10}^{-8}\mathrm{M}$

For  ${\mathrm{MgF}}_2:$      $K_{\mathrm{s}}=\left[{\mathrm{Mg}}^{2+}\right]\left[{\mathrm{F}}^{-}\right]=4y(x+y{)}^2=14.4\times {10}^{-8}\mathrm{M}$

Solving for x and y, we get x $=1.26\times {10}^{-3}\mathrm{M}$ and $y=2.52\times {10}^{-3}\mathrm{M}$