What is the ratio of $\left[M{g}^{2+}\right]$  and  $\left[A{l}^{3+}\right]$ in 1 litre saturated solution of  $Mg{\left(OH\right)}_2$ and $Al{\left(OH\right)}_3$  by making proper approximation ?  Given  $K_{sp}\text{of }Mg{\left(OH\right)}_2=4\times {10}^{–12}$, $K_{sp}\text{\hspace{0.17em}of }Al{\left(OH\right)}_3=1\times {10}^{–33}$
 Report your answer in multiples of  ${10}^{17}.$ (Round off to nearest integer)
 

 $Mg{\left(OH\right)}_2\underset{ }{\overset{ }{\rightleftharpoons }}M{g}^{2+}+2\overline{O}H$
                              x        2x + 3y
 $Al{\left(OH\right)}_3\underset{ }{\overset{ }{\rightleftharpoons }}A{l}^{3+}+3\overline{O}H$
                                      y       2x + 3y
 $\text{As }{K}_{sp}ofMg{\left(OH\right)}_2>>K_{sp}ofAl{\left(OH\right)}_3$
 $2x+3y\approx 2x$
Now
 $$\begin{gathered} \left[M{g}^{2+}\right]{\left[\overline{O}H\right]}^2=4\times {10}^{-12} \\ x\times {\left(2x\right)}^2=4\times {10}^{-12} \\ x={10}^{-4} \\ \left[A{l}^{3+}\right]{\left[\overline{O}H\right]}^3=1\times {10}^{-33} \\ y\times {\left(2x\right)}^3=1\times {10}^{-33} \\ y\times 8\times {10}^{-12}=1\times {10}^{-33} \\ y=\frac{{10}^{-21}}{8}. \\ \therefore \frac{x}{y}=\frac{{10}^{-4}}{\frac{{10}^{-21}}{8}}=8\times {10}^{17} \end{gathered}$$