Question

Ionic product of $Ni{\left(OH\right)}_2$ is $2.0\times {10}^{-15}$. Molar solubility of $Ni{\left(OH\right)}_2$ in $0.10M$ $NaOH$ will be ________.

• A. $1.0\times {10}^{-13}M$
• B. $2.0\times {10}^{-13}M$
• C. $4.0\times {10}^{-13}M$
• D. $8.0\times {10}^{-13}M$

Answer 1

Answer: (B)

$2.0\times {10}^{-13}M$

Let the solubility of $Ni{\left(OH\right)}_2=S$

$Ni{\left(OH\right)}_2+H_{2}O\rightleftharpoons {Ni}^{2+}\left(aq\right)+2{OH}^{-}\left(aq\right)$

$S$ $S$ $2S$

$K_{sp}=2.0\times {10}^{-16}=\left[{Ni}^{2+}\right]{\left[{OH}^{-}\right]}^2$

$=\left[S\right]{[0.10+2S]}^2$

As $K_{sp}$ is small $2S<<0.10$

therefore $(0.10+2S)=0.10$

Hence, $2.0\times {10}^{-15}=S{\left(0.10\right)}^2$

$S=2.0\times {10}^{-13}M=\left[{Ni}^{2+}\right]=\left[Ni\right(OH{)}_2]$

Hence, the correct option is $\text{B}$