Question

The molar solubility of $Zn{\left(OH\right)}_2$ in 0.1 M NaOH solution is $x\times {10}^{-18}M$. The value of x is .......(Nearest integer)
(Given: The solubility product of $Zn\left(OH\right){}_2\text{is}2\times {10}^{-20}$)

Answer 1

$Zn(OH{)}_2$ is dissolved in $0.1M$ of $NaOH$

$Zn\left(OH\right){}_2\left(s\right)\rightleftharpoons Z{n}^{+2}\left(aq\right)+2OH\left(aq\right)$
$S(0.1+2S)$

$Zn(OH{)}_2$ is less soluble, so 2S is negligible compared to 0.1.
$\therefore (0.1+2S)\simeq 0.1$

$K_{sp}=S{\left(0.1\right)}^2$

$2\times {10}^{-20}=S\times {10}^{-2}$

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

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

Thus, x=2.