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Find out the solubility of $Ni(OH)_2$ in 0.1 M NaOH. Given that the ionic product of $Ni(OH)_2$ is $2\times10^{-15}$.
A
$2\times10^{-8}\ M$
B
$1\times10^{-13}\ M$
C
$1\times10^{8}\ M$
D
$2\times10^{-13}\ M$
Detailed Solution
$Ni(OH)_2 \rightleftharpoons Ni^{2+} + 2OH^-$ (solubility s)
In 0.1 M NaOH, $[OH^-] = 2s + 0.1 \approx 0.1$ M (common ion effect)
$K_{sp} = [Ni^{2+}][OH^-]^2 \Rightarrow 2 \times 10^{-15} = s(0.1)^2$
$s = 2 \times 10^{-13}$ M