The concentration of in a saturated solution of () dramatically drops due to the common ion effect w — Ionic Equilibrium Chemistry Question
Question
The concentration of $Ag^+$ in a saturated solution of $AgCl$ ($K_{sp} = 1.6 \times 10^{-10}$) dramatically drops due to the common ion effect when mixed with $0.1 \text{ M } NaCl$. Calculate the exact $[Ag^+]$ in the resulting $0.1 \text{ M } NaCl$ medium. If the concentration is $x \times 10^{-9} \text{ M}$, calculate the value of $x$.
💡 Solution & Explanation
The dissociation of $AgCl$ is $AgCl_{(s)} \rightleftharpoons Ag^+_{(aq)} + Cl^-_{(aq)}$. The $NaCl$ dissociates completely, providing $[Cl^-] = 0.1 \text{ M}$. Because the $K_{sp}$ is very small, the added $Cl^-$ from $AgCl$ is negligible compared to $0.1 \text{ M}$. The solubility product expression is $K_{sp} = [Ag^+][Cl^-]$. Thus, $[Ag^+] = \frac{K_{sp}}{[Cl^-]} = \frac{1.6 \times 10^{-10}}{0.1} = 1.6 \times 10^{-9} \text{ M}$. So, $x = 1.6$.