The solubility product of Zn(OH) is at 25°C. What would be the concentration Zn ion in 0.1 M – NHOH — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

The concentration of $\text{OH}^-$ provided by the 50% ionized $\text{NH}_4\text{OH}$ is $[\text{OH}^-] = 0.50 \times 0.1\text{ M} = 0.05\text{ M}$. Using the solubility product expression for $\text{Zn(OH)}_2$: $K_{sp} = [\text{Zn}^{2+}][\text{OH}^-]^2$. Substituting the values: $10^{-14} = [\text{Zn}^{2+}](0.05)^2 \implies 10^{-14} = [\text{Zn}^{2+}](0.0025)$. Solving for $[\text{Zn}^{2+}]$ gives $\frac{10^{-14}}{2.5 \times 10^{-3}} = 4 \times 10^{-12}\text{ M}$.