The molar solubility of $Zn(OH)_2 $ in 0.1 M NaOH solution is $x \times 10^{-18} M$. The value of x is _ _ _ _ . (Nearest integer)

(Given: The solubility product of $Zn(OH)_2 $ is $2 \times 10^{-20} $)

*Solution*

$Zn{(OH)_2}\rightleftharpoons \mathop {Z{n^{2 + }}}\limits_s + 2\mathop {O{H^ - }}\limits_{2s + 0.1} $

${K_{sp}} = [Z{n^{2 + }}]{[O{H^ - }]^2} = s.{(2s + 0.1)^2}$

$\therefore 2 \times {10^{ - 20}} = s.{(2s + 0.1)^2} \approx {0.1^2} \times s$

$ \Rightarrow s = 2 \times {10^{ - 18}}$

$\therefore x = 2 $