NEET 2020 Chemistry Question 92 Solution

Sodium hydroxide (NaOH) is a strong base that dissociates completely in aqueous solution: $$\text{NaOH (aq)} \to \text{Na}^+ \text{(aq)} + \text{OH}^- \text{(aq)}$$ Given $$\text{[NaOH]} = 0.1\text{ M}$$, the contribution of hydroxide ions from NaOH is $$[\text{OH}^-]_{\text{NaOH}} = 0.1\text{ M} = 10^{-1}\text{ M}$$.

Nickel hydroxide is a sparingly soluble salt that establishes the following dissolution equilibrium: $$\text{Ni(OH)}_2 \text{ (s)} \rightleftharpoons \text{Ni}^{2+} \text{(aq)} + 2\text{OH}^- \text{(aq)}$$ If $$s$$ is the molar solubility of $$\text{Ni(OH)}_2$$ in this solution: $$[\text{Ni}^{2+}] = s$$ $$[\text{OH}^-] = 0.1 + 2s$$

Because $$K_{sp} = 2 \times 10^{-15}$$ is extremely small, $$s \ll 0.1$$, and hence $$2s$$ is negligible compared to $$0.1$$: $$[\text{OH}^-] \approx 0.1\text{ M} = 10^{-1}\text{ M}$$

Substitute into the solubility product expression: $$K_{sp} = [\text{Ni}^{2+}][\text{OH}^-]^2$$ $$2 \times 10^{-15} = s \times (0.1)^2 = s \times 10^{-2}$$ $$s = \frac{2 \times 10^{-15}}{10^{-2}} = 2 \times 10^{-13}\text{ M}$$ This confirms the approximation ($$2 \times (2 \times 10^{-13}) = 4 \times 10^{-13} \ll 0.1$$).