The solubility product of AgCl is . The equilibrium constant of the reaction<br>AgCl(s) + Br AgBr(s) — Ionic Equilibrium Chemistry Question
Question
The solubility product of AgCl is $1.0 \times 10^{-10}$. The equilibrium constant of the reaction<br>AgCl(s) + Br$^-$ $\rightleftharpoons$ AgBr(s) + Cl$^-$<br>is 200 and that of the reaction<br>2AgBr(s) + S$^{2-} \rightleftharpoons$ Ag$_2$S(s) + 2Br$^-$<br>is $1.6 \times 10^{24}$. What is the $K_{sp}$ of Ag$_2$S?
💡 Solution & Explanation
For the first substitution: $K_1 = \frac{K_{sp}(\text{AgCl})}{K_{sp}(\text{AgBr})} = 200 \implies K_{sp}(\text{AgBr}) = \frac{1.0 \times 10^{-10}}{200} = 5 \times 10^{-13}$. For the second substitution: $K_2 = \frac{[K_{sp}(\text{AgBr})]^2}{K_{sp}(\text{Ag}_2\text{S})} = 1.6 \times 10^{24}$. Substituting $K_{sp}(\text{AgBr})$: $1.6 \times 10^{24} = \frac{(5 \times 10^{-13})^2}{K_{sp}(\text{Ag}_2\text{S})} = \frac{25 \times 10^{-26}}{K_{sp}(\text{Ag}_2\text{S})}$. Thus, $K_{sp}(\text{Ag}_2\text{S}) = \frac{25 \times 10^{-26}}{1.6 \times 10^{24}} = 15.625 \times 10^{-50} \approx 1.56 \times 10^{-49}$.