Chemistry › Ionic Equilibria › Solubility
If the ionic product of $\mathrm{Ni}(\mathrm{OH})_2$ is $1.9 \times 10^{-15}$, the molar solubility of…
If the ionic product of $\mathrm{Ni}(\mathrm{OH})_2$ is $1.9 \times 10^{-15}$, the molar solubility of $\mathrm{Ni}(\mathrm{OH})_2$ in $1.0 \mathrm{M} \mathrm{NaOH}$ is
$1.9 \times 10^{-18} \mathrm{M}$ $1.9 \times 10^{-13} \mathrm{M}$ $1.9 \times 10^{-15} \mathrm{M}$ $1.9 \times 10^{-14} \mathrm{M}$ Solution
$\begin{aligned} & \mathrm{NaOH} \rightleftharpoons \underset{\mathrm{CM}}{\mathrm{Na}^{+}}+\underset{\mathrm{CM}}{\mathrm{OH}^{-}} \\ & \mathrm{Ni}(\mathrm{OH})_2 \rightleftharpoons \mathrm{Ni}_x^{2+}+\underset{x}{2 \mathrm{OH}^{-}} \\ & \therefore \quad K_{\mathrm{sp}}=\left[\mathrm{Ni}^{2+}\right]\left[\mathrm{OH}^{-}\right]^2 \\ & =x(x+C)^2 \\ & \end{aligned}$
$
=x\left(x^2+2 C x+C^2\right)
$
or
$
\begin{aligned}
K_{\mathrm{sp}} & \left.=x C^2 \text { (neglecting higher powers of } x\right) \\
x & =\frac{K_{\mathrm{sp}}}{\mathrm{c}^2}=\frac{1.9 \times 10^{-15}}{(1)^2} \\
& =1.9 \times 10^{-15} \mathrm{M}
\end{aligned}
$
Asked in: AP EAMCET 2014
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