The solubility product of AgCl is 1.8×10^-10 and the formation constant of [Ag(NH3)2]^+ is 1.6×10^7 (both at 298 K). The molar solubility of AgCl in…

Dissolution in ammonia is the sum of two steps: $\mathrm{AgCl(s)\rightleftharpoons Ag^++Cl^-}$ ($K_{sp}$) and $\mathrm{Ag^++2NH_3\rightleftharpoons[Ag(NH_3)_2]^+}$ ($K_f$). The overall reaction $\mathrm{AgCl+2NH_3\rightleftharpoons[Ag(NH_3)_2]^++Cl^-}$ has $K=K_{sp}K_f=1.8\times10^{-10}\times1.6\times10^7=2.88\times10^{-3}$. If $s$ mol/L dissolves, $[\mathrm{Cl^-}]=[\mathrm{Ag(NH_3)_2^+}]=s$ and $[\mathrm{NH_3}]=0.100-2s$. Thus $s/(0.100-2s)=\sqrt{K}=0.05367$, so $s=0.1(0.05367)/(1+2\times0.05367)=0.004846$ M. The solubility is therefore $4.85\times10^{-3}$ M, about $360$ times larger than the solubility in pure water ($1.34\times10^{-5}$ M). Neglecting the $-2s$ term in the denominator would give $5.37\times10^{-3}$ M, the typical wrong answer, since ammonia consumed by complexation is not negligible here: it uses up nearly 10 percent of the ammonia, which is why the quadratic-free but corrected expression must be used.