Find out the solubility of $Ni(OH)_2$ in 0.1 M $NaOH$. Given that the ionic product of $Ni(OH)_2$ is $2 imes

Question

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

Accepted Answer

$Ni(OH)_2 ightleftharpoons Ni^{2+} + 2OH^-$. $NaOH ightarrow Na^+ + OH^-$. Total $[OH^-] = 2s + 0.1 \approx 0.1$. Ionic product = $[Ni^{2+}][OH^-]^2$. $2 	imes 10^{-15} = s(0.1)^2$. $s = 2 	imes 10^{-13}$. Solubility of $Ni(OH)_2 = 2 	imes 10^{-13}$ M.

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