Question

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: $\sqrt[3]{0 . 5}=0.79$ )

• A. $2\times 10^{- 13}M$
• B. $2\times 10^{- 14}M$
• C. $7.9\times 10^{- 6}M$
• D. $7.9\times 10^{- 4}M$

Answer 1

Answer: (A)

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