The solubility product of \(\mathrm{Ni}(\mathrm{OH})_2\) at \(298 \mathrm{~K}\) is \(2 \times 10^{-15}…

The solubility product of \(\mathrm{Ni}(\mathrm{OH})_2\) at \(298 \mathrm{~K}\) is \(2 \times 10^{-15} \mathrm{~mol}^3 \mathrm{dm}^{-9}\). The \(\mathrm{pH}\) value if, its aqueous and saturated solution is
  1. 5
  2. 7.5
  3. 9
  4. 13

Solution

\(\begin{aligned} & \mathrm{Ni}(\mathrm{OH})_2 ightleftharpoons \underset{S \mathrm{~mol} / \mathrm{L}}{\mathrm{Ni}^{2+}}+\underset{2 S \mathrm{~mol} / \mathrm{L}}{2 \mathrm{OH}^{-}} \\ & \begin{aligned} \Rightarrow K_{\mathrm{sp}}=\left[\mathrm{Ni}^{2+}ight]\left[\mathrm{OH}^{-}ight]^2 & =S \times(2 S)^2=4 S^3 \\ & =2 \times 10^{-15}(\mathrm{~mol} / \mathrm{L})^3 \end{aligned} \end{aligned}\) \(\begin{aligned} & =2 \times 10^{-15} \mathrm{~mol}^3 \mathrm{dm}^{-9} \\ \Rightarrow \quad S & =\left(\frac{2}{4} \times 10^{-15}ight)^{1 / 3} \simeq 8 \times 10^{-6} \mathrm{~mol} / \mathrm{L} \\ {\left[\mathrm{OH}^{-}ight] } & =2 S=2 \times 8 \times 10^{-6} \mathrm{~mol} / \mathrm{L} \\ \mathrm{pOH} & =6-\log 16=4.983 \simeq 5 \\ \therefore \quad & \mathrm{pH}=14-\mathrm{pOH}=14-5=9 \text { at } 25^{\circ} \mathrm{C}\end{aligned}\) ,

Asked in: JEE-TOPICTESTS-CHEMISTRY

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