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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 $2 x 10^{- 15}$

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$2 x 10^{- 8} M$

$1 x 10^{- 13} M$

$1 x 10^{8} M$

$2 x 10^{- 13} M$

answer is D.

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Detailed Solution

$\begin{aligned} 0 .1 M NaOH = 0 .1 M {OH}^{-} ions \\ Ni ({OH}_{2}) \to {Ni}^{2 +} + 2 {OH}^{-} \\ K_{SP} = S \times {(10^{- 1})}^{2} \\ S = \frac{2 \times 10^{- 15}}{10^{- 2}} = 2 \times 10^{- 13} M \end{aligned}$

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