# Solve this following

Question:

The molar solubility of $\mathrm{Zn}(\mathrm{OH})_{2}$ in $0.1 \mathrm{M} \mathrm{NaOH}$ solution is $\mathrm{x} \times 10^{-18} \mathrm{M}$. The value of $\mathrm{x}$ is

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

Solution:

$\mathrm{Zn}(\mathrm{OH})_{2}(\mathrm{~s}) \rightleftharpoons \mathrm{Zn}^{+2}(\mathrm{aq})+2 \mathrm{OH}$ (aq)

$(0.1+2 \mathrm{~s})=0.1$

$\mathrm{K}_{\mathrm{sp}}=\mathrm{S}(0.1)^{2}$

$2 \times 10^{-20}=\mathrm{s} \times 10^{-2} \Rightarrow \mathrm{s}=2 \times 10^{-18}$

$=x \times 10^{-18}$

$x=2$