Chemistry - Ionic Equilibrium Question with Solution | TestHub

$\mathrm{Mg}(\mathrm{OH}{)}_2\rightleftharpoons {\mathrm{Mg}}^{2+}+2{\mathrm{OH}}^{-}$

$x$ $2x+3y$

$Al(OH{)}_3\rightleftharpoons A{l}^{3+}+3O{H}^{-}$

$y$$3y+2x$

Since ${\mathrm{K}}_{\mathrm{sp}}$ of $\mathrm{Mg}(\mathrm{OH}{)}_2>{\mathrm{K}}_{\mathrm{sp}}$ of $\mathrm{Al}(\mathrm{OH}{)}_3$

$\begin{array}{c}\therefore x\gg y\therefore 2x+3y\simeq 2x\end{array}$

$4\times 10^{-12}=\left[M{g}^{2+}\right]{\left[O{H}^{-}\right]}^2=x\times (2x{)}^2\therefore x=10^{-4}$

Similarly

$\begin{array}{rlr}& 1\times 10^{-33}=\left[A{l}^{3+}\right]{\left[O{H}^{-}\right]}^3& 1\times 10^{-33}=y\times (2x{)}^3\end{array}$

$\therefore y=\frac{10^{-21}}{8}$ Thus ratio will be $\frac{x}{y}=8\times 10^{17}$

∴ Ans. $=8\times 10^{+17}\times 10^{-17}=8$