Question

Find out the solubility of $Ni(OH{)}_2$ in $0.1MNaOH$. 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)

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