Question

Find out the solubility of $Ni(OH{)}_2$ in 0.1 M NaOH. Given that the ionic product of $Ni(OH{)}_2$ is 2×10${}^{–15}$ .

• A. $2\times {10}^{-8}M$
• B. $2\times {10}^{-13}M$
• C. $1\times {10}^{-13}M$
• D. $1\times {10}^{8}M$

Answer 1

The correct option is B $2\times {10}^{-13}M$

$Ni(OH{)}_2\rightleftharpoons N{i}^{2+}+2O{H}^{-}$
$\therefore K_{sp}=\left[N{i}^{2+}\right][O{H}^{-}{]}^2$
$\Rightarrow K_{sp}=s\times (0.1{)}^2$
$\Rightarrow s=\frac{2\times {10}^{-15}}{(0.1{)}^2}$
$\Rightarrow s=2\times {10}^{-13}M$