What concentration of free must be maintained in a solution that is and to prevent AgCl from precipi — Ionic Equilibrium Chemistry Question
Question
What concentration of free $\text{CN}^-$ must be maintained in a solution that is $1.8 \text{ M} - \text{AgNO}_3$ and $0.16 \text{ M} - \text{NaCl}$ to prevent AgCl from precipitating? $K_f$ for $\text{Ag(CN)}_2^- = 6.4 \times 10^{17}$ and $K_{sp}$ for $\text{AgCl} = 1.8 \times 10^{-10}$.
💡 Solution & Explanation
To prevent precipitation of AgCl, we need $[\text{Ag}^+] \le \frac{K_{sp}}{[\text{Cl}^-]} = \frac{1.8 \times 10^{-10}}{0.16} = 1.125 \times 10^{-9} \text{ M}$. The total silver ($1.8 \text{ M}$) will essentially be completely complexed, making $[\text{Ag(CN)}_2^-] \approx 1.8 \text{ M}$. Using the formation constant: $K_f = \frac{[\text{Ag(CN)}_2^-]}{[\text{Ag}^+][\text{CN}^-]^2} \implies 6.4 \times 10^{17} = \frac{1.8}{(1.125 \times 10^{-9})[\text{CN}^-]^2}$. Solving for $[\text{CN}^-]^2$ gives $\frac{1.8}{7.2 \times 10^8} = 25 \times 10^{-10}$. Thus, free $[\text{CN}^-] = 5 \times 10^{-5} \text{ M}$. Therefore, correct answer is B.