Consider a saturated aqueous solution of for which the solubility product is at room temperature. To — Ionic Equilibrium Chemistry Question
Question
Consider a $1 \text{ L}$ saturated aqueous solution of $AgCl$ for which the solubility product $K_{sp}$ is $1.6 \times 10^{-10}$ at room temperature. To this exact solution, exactly $0.1 \text{ moles}$ of solid $CuCl$ ($K_{sp} = 1.0 \times 10^{-6}$) is added and allowed to fully equilibrate. The resultant concentration of $Ag^+$ in the mixed solution massively drops to $1.6 \times 10^{-x} \text{ M}$. What is the exact integer value of $x$?
💡 Solution & Explanation
The $CuCl$ is significantly more soluble than $AgCl$. Its dissolution ($CuCl \rightleftharpoons Cu^+ + Cl^-$) forcefully dictates the overwhelming majority of the chloride ion concentration. Let its solubility be $y$. Since $K_{sp}(CuCl) = 1.0 \times 10^{-6}$, $y^2 = 10^{-6} \Rightarrow y = 10^{-3} \text{ M}$. Because $0.1 \text{ mol}$ of $CuCl$ was added to $1 \text{ L}$, it will only dissolve until $[Cl^-] = 10^{-3} \text{ M}$, creating a saturated $CuCl$ solution with excess solid resting at the bottom. The $Ag^+$ concentration is rigorously determined by this new common $[Cl^-]$: $[Ag^+] = \frac{K_{sp}(AgCl)}{[Cl^-]} = \frac{1.6 \times 10^{-10}}{10^{-3}} = 1.6 \times 10^{-7} \text{ M}$. Thus, $x = 7$.