A of is dissolved in of . If of is added to this solution, will precipitate? and . — Ionic Equilibrium Chemistry Question
Question
A $0.1\text{ mole}$ of $\text{AgNO}_3$ is dissolved in $1\text{ L}$ of $1\text{ M} – \text{NH}_3$. If $0.01\text{ mole}$ of $\text{NaCl}$ is added to this solution, will $\text{AgCl(s)}$ precipitate? $K_{sp} \text{ for AgCl} = 1.8 \times 10^{-10}$ and $K_{stab} \text{ for Ag(NH}_3)_2^+ = 1.6 \times 10^7$.
💡 Solution & Explanation
$\text{Ag}^+$ reacts with $\text{NH}_3$ to form a complex. Initial $[\text{Ag}^+] = 0.1\text{ M}$, which uses $0.2\text{ M}$ of $\text{NH}_3$. Free $[\text{NH}_3] = 1.0 - 0.2 = 0.8\text{ M}$. $K_{stab} = \frac{[\text{Ag(NH}_3)_2^+]}{[\text{Ag}^+][\text{NH}_3]^2} \implies 1.6 \times 10^7 = \frac{0.1}{[\text{Ag}^+](0.8)^2}$. Solving for free $[\text{Ag}^+] = 9.7 \times 10^{-9}\text{ M}$. Ionic product $= [\text{Ag}^+][\text{Cl}^-] = (9.7 \times 10^{-9})(0.01) = 9.7 \times 10^{-11}$. Since this is less than the $K_{sp} (1.8 \times 10^{-10})$, AgCl will not precipitate.