Question

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

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

Answer 1

Answer: (A)

$Ni(OH)_2 \rightleftharpoons \underset{(S)}{Ni^{2+}} + \underset{(2S)}{2OH^-} S =$ molar conc. of $Ni(OH)_2$
$NaOH \rightarrow Na^+ +\underset{(0.1\,M}{OH^-}$
$K_{SP} = [Ni^{2+}][OH^-]^2$
$ = (S)(2S + 0.1)^2$
$=(S)(4S^2 + 0.01 + 0.4S)$
$= 4S^3 + 0.01S + 0.4 S^2$
Neglecting higher power of $S$,
$2 \times 10^{-15} = 0.01S$
$S = 2 \times 10^{-13}M$