Question

The solubility product of $Ni ( OH )_{2}$ is $2.0 \times 10^{-15} .$ The molar solubility of $Ni ( OH )_{2}$ in $0.10\, M\, NaOH$ is

• A. $3.2 \times 10^{-12}$
• B. $2.0 \times 10^{-13}$
• C. $4.34 \times 10^{-12}$
• D. $0.58 \times 10^{-4}$

Answer 1

Answer: (B)

$Ni ( OH )_{2} \rightleftharpoons \underset{s}{Ni^{2+}}+\underset{2s}{2OH^{-}}$

Total $\left[ OH ^{-}\right]=2 s+0.1$

Solubility product $=(s)(2s+0.1)^{2} $

$=4 s^{3}+0.4 s^{2}+0.01 s=2 \times 10^{-15}$

$s$ is very small so by neglecting $4 s^{3}$ and $0.4 s^{2}$

$0.01 s=2 \times 10^{-15}$

$ \Rightarrow s=2 \times 10^{-13}$