Calculate the solubility of in a 0.1 M NaOH solution. Given that the ionic product () of is . — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

In 0.1 M NaOH, the common $[OH^-]$ is approximately 0.1 M. The equilibrium is $Ni(OH)_2 \rightleftharpoons Ni^{2+} + 2OH^-$. $K_{sp} = [Ni^{2+}][OH^-]^2 = S \times (0.1)^2 = 2 \times 10^{-15}$. Therefore, $S = 2 \times 10^{-13} \text{ M}$.