The simultaneous solubilities of AgSCN and AgBr are, respectively (), — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

Let solubility of AgSCN be $S_1$ and AgBr be $S_2$. Total $[\text{Ag}^+] = S_1 + S_2$. $K_{sp}(\text{AgSCN}) = S_1(S_1 + S_2) = 1 \times 10^{-12}$ and $K_{sp}(\text{AgBr}) = S_2(S_1 + S_2) = 0.21 \times 10^{-12}$. Adding them gives $(S_1 + S_2)^2 = 1.21 \times 10^{-12}$, so $[\text{Ag}^+] = S_1 + S_2 = 1.1 \times 10^{-6} \text{ M}$. Substituting back, $S_1 = \frac{10^{-12}}{1.1 \times 10^{-6}} = 9.09 \times 10^{-7} \text{ M}$, and $S_2 = \frac{2.1 \times 10^{-13}}{1.1 \times 10^{-6}} = 1.909 \times 10^{-7} \text{ M}$.