How many exact moles of sparingly soluble () will dissolve in exactly of a aqueous solution? If the — Ionic Equilibrium Chemistry Question
Question
How many exact moles of sparingly soluble $CuI$ ($K_{sp} = 5 \times 10^{-12}$) will dissolve in exactly $1.0 \text{ L}$ of a $0.10 \text{ M }$ aqueous $NaI$ solution? If the answer is $y \times 10^{-11} \text{ moles}$, what is the value of $y$?
💡 Solution & Explanation
The $NaI$ solution provides a common iodide ion concentration of $[I^-] = 0.10 \text{ M}$. The dissolution of $CuI$ is $CuI \rightleftharpoons Cu^+ + I^-$. Let the solubility be $s$. The equilibrium expression is $K_{sp} = [Cu^+][I^-]$. Because $s \ll 0.10$, total $[I^-] \approx 0.10 \text{ M}$. Therefore, $5 \times 10^{-12} = (s)(0.10)$, which means $s = \frac{5 \times 10^{-12}}{0.10} = 5 \times 10^{-11} \text{ M}$. In $1.0 \text{ L}$, this is $5 \times 10^{-11} \text{ moles}$. Thus, $y = 5$.