1. 2×10⁻¹³ M 2. 2×10⁻⁸ M 3. 1×10⁻¹³ M 4. 1×10⁸ M

Answer: 1. 2×10⁻¹³ M In 0.1 M NaOH the common ion OH^- is already present, so the OH^- from Ni(OH)₂ can be ignored. mathrmNi(OH)₂ ⇌ Ni²⁺ + 2OH^- , K_sp=[mathrmNi²⁺][OH^-]² . Let solubility be s : s=[mathrmNi²⁺] , [OH^-]≈0.1 M. s=fracK_sp(0.1)²=frac2×10⁻¹⁵10⁻²=2×10⁻¹³\ M Option (1) is correct.

Topic: Solubility product· Equilibrium

Find out the solubility of Ni(OH)_2 in 0.1 M NaOH. Given that the ionic product of Ni(OH)_2 is 2×10^-15.

1. $2\times10^{-13}$ M

2. $2\times10^{-8}$ M

3. $1\times10^{-13}$ M

4. $1\times10^{8}$ M

Doubtly Asked 30 Oct 2026 NEET UG 2020 Practice test neet Save WhatsApp Report

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Answer: 1. $2\times10^{-13}$ M

In 0.1 M NaOH the common ion $\mathrm{OH^-}$ is already present, so the $\mathrm{OH^-}$ from $\mathrm{Ni(OH)_2}$ can be ignored.

$\mathrm{Ni(OH)_2 \rightleftharpoons Ni^{2+} + 2OH^-}$, $K_{sp}=[\mathrm{Ni^{2+}}][\mathrm{OH^-}]^2$.

Let solubility be $s$: $s=[\mathrm{Ni^{2+}}]$, $[\mathrm{OH^-}]\approx0.1$ M.

$$s=\frac{K_{sp}}{(0.1)^2}=\frac{2\times10^{-15}}{10^{-2}}=2\times10^{-13}\ \mathrm{M}$$

Option (1) is correct.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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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 2×10^-15.

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