Q. Calculate the molar solubility of $Ni\left(OH\right)_{2}$ in $0.1MNaOH$ . The solubility product of $Ni\left(OH\right)_{2}$ is $2\times 10^{- 15}$ . (Given: $\sqrt[3]{0 . 5}=0.79$ )

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A

$2\times 10^{- 13}M$

B

$2\times 10^{- 14}M$

C

$7.9\times 10^{- 6}M$

D

$7.9\times 10^{- 4}M$

Solution:

$Ni\left(OH\right)_{2}\left(s\right) & \rightleftharpoons & \left(Ni\right)^{2 +} & + & 2\left(OH\right)^{-}$
$s$ $2s+0.1$
$s\left(2 s + 0 . 1\right)^{2}=2\times \left(10\right)^{- 15}$ $2s+0.1\approx0.1$
$s=2\times 10^{- 13}$

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