A solution contains a mixture of and , which are to be separated by selective precipitation. Calcula — Ionic Equilibrium Chemistry Question
Question
A solution contains a mixture of $\text{Ag}^+ \ (0.10\text{ M})$ and $\text{Hg}_2^{2+} \ (0.10\text{ M})$, which are to be separated by selective precipitation. Calculate the maximum concentration of iodide ion at which one of them gets precipitated almost completely. What per cent of that metal ion is precipitated, before the start of precipitation of second metal ion? $K_{sp}(\text{AgI}) = 8.5 \times 10^{-17}$ and $K_{sp}(\text{Hg}_2\text{I}_2) = 2.5 \times 10^{-26}$.
💡 Solution & Explanation
Precipitation thresholds: For AgI, $[\text{I}^-] = \frac{8.5 \times 10^{-17}}{0.1} = 8.5 \times 10^{-16}\text{ M}$. For $\text{Hg}_2\text{I}_2$, $[\text{I}^-] = \sqrt{\frac{2.5 \times 10^{-26}}{0.1}} = 5 \times 10^{-13}\text{ M}$. AgI requires less $[\text{I}^-]$ and precipitates first. It continues precipitating until $[\text{I}^-]$ reaches $5 \times 10^{-13}\text{ M}$ (the maximum concentration before $\text{Hg}_2\text{I}_2$ starts). At this point, the remaining $[\text{Ag}^+] = \frac{8.5 \times 10^{-17}}{5 \times 10^{-13}} = 1.7 \times 10^{-4}\text{ M}$. The percentage left $= \frac{1.7 \times 10^{-4}}{0.1} \times 100\% = 0.17\%$. The percentage precipitated $= 100\% - 0.17\% = 99.83\%$.