What is the ratio of moles of $$ Mg ( OH)_2 $$ and $$ Al(OH)_3 $$ present in 1 lit saturated aqueous solution of $$ Mg (OH)_2 $$ & $$Al(OH)_3 (K_{sp} \ of \ Mg (OH)_2 = 4 \times 10^{-12} $$ and $$ K_{sp} \ of \ Al(OH)_3 = 1 \times 10^{-33} $$). Give answer by multiplying by $$ 10^{-16} . $$
Correct option is A. 80
$$ Mg(OH)_2 \leftrightharpoons \begin{matrix} Mg^{+2} \quad + \quad 2OH^- \\ x \quad \quad 2x + 3y \end{matrix} \quad \quad \quad K_{sp} of Mg(OH)_ 2of Al(OH)_3 $$
$$ Al(OH)_{ 3 }\leftrightharpoons \quad \begin{matrix} Al^{ 3+ }\quad +\quad 3OH^{ - } \\ y\quad \quad 3y+2x \end{matrix}\quad \quad \quad \quad \begin{matrix} x>>y \\ 2x+3y \end{matrix} \simeq 2x $$
$$ \because 4 \times 10^{-12} = [Mg^{2+} ] [OH^-]^2 $$
$$ = x \times (2x)^2 $$
$$ x = 10^{-4} $$
$$ 1 \times 10^{-33} = [Al^{3+}] [ OH^-]^3 $$
$$ 1 \times 10^{-33} = (y) (2x)^3 $$
$$ 1 \times 10^{-33} = y \times ( 10^{-4} \times 2)^3 $$
$$ y = \dfrac {10^{-21}}{8} \quad so \quad \dfrac {x}{y} = 8 \times 10^{17} \quad \quad 8 \times 10^{17} \times 10^{-16} = 80 $$