SATHEE: Equilibrium - Result Question 64

Equilibrium - Result Question 64

####67. Find out the solubility of $N i (O H)_{2}$ in $0.1 M$ $N a O H$ . Given that the ionic product of $N i (O H)_{2}$ is $2 \times 10^{- 15}$

[2020]

(a) $2 \times 10^{- 8} M$

(b) $1 \times 10^{- 13} M$

(d) $2 \times 10^{- 13} M$

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Solution:

(d) $N i (O H)^{2} \to \underset{s}{N i^{2 +}} + \underset{2 s}{2 O H^{-}}$

$N a O H ⟶ N a + O H^{-}$

Total $[O H^{-}] = 2 s + 0.1 \approx 0.1$

Ionic product $= [N i]^{2 +} [O H]^{2}$

$2 \times 10^{- 15} = s (0.1)^{2}$

$s = 2 \times 10^{- 13}$

Solubility of $N i (O H)_{2} = 2 \times 10^{- 13} M$

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Equilibrium - Result Question 64

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