The silver ion concentration in a 0.2 M solution of Ag(NH)NO is (, ) — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

The complex dissociation is $\text{Ag(NH}_3)_2^+ \rightleftharpoons \text{Ag}^+ + 2\text{NH}_3$. Let $[\text{Ag}^+] = x$, then $[\text{NH}_3] = 2x$. The initial concentration of the complex is $0.2 \text{ M}$. $K_{\text{diss}} = \frac{[\text{Ag}^+][\text{NH}_3]^2}{[\text{Ag(NH}_3)_2^+]} \implies 6.8 \times 10^{-8} = \frac{x(2x)^2}{0.2} = 20x^3$. Solving for $x^3$: $x^3 = \frac{6.8 \times 10^{-8}}{20} = 3.4 \times 10^{-9}$. We are given that $1.5^3 = 3.4$, so $x^3 = 1.5^3 \times 10^{-9}$, giving $x = 1.5 \times 10^{-3} \text{ M}$.