Question

what is the solubility of AgCl in 1.0M of NH3

Answer 1

Let's say the solubility of AgCl is S, so the equilibrium concentration of Cl- is S.

what about the conc. of Ag^+? it's not going to be S, because most of the Ag^+ is sequestered away by NH3. however the equilibrium conc. of Ag+ plus the equilibrium conc. of Ag(NH3)2^+ is equal to S.

so we will say the Ag^+ conc is x, and the Ag(NH3)2^+ conc is S-x.

because the question is mum about it, i will assume that the ammonia concentration that we were given is the equilibrium concentration.

there are two equilbria here, so we have two equations:
\[K _{f}=\frac{[[Ag(NH3)_{2}]^+] }{ [Ag^{+}][NH3]^2 }\]\[ K _{sp}=[Ag^{+}][Cl^{-}]\]
These are all equilibrium concs of course. Now plug in S, S-x, and the equilib NH3 conc.
\[K _{f}=\frac{S-x}{(x)(1.0)^2 }=1.7*10^7\]\[ K _{sp}=xS=1.8*10^{-10}\]
i am not very consistent with the units but my back is hurting so you will forgive me right?

Now Plug x=Ksp/S into the first equation, and find S in your favorite way. that's the solubility of AgCl in 1.0 M NH3. if we did everything right, it should be much higher than the solubility of in pure water.

why? lechatlier says that the solubility reaction will shift to the right when we remove the silver ions, one of the products in the AgCl solubility equilibrium. where did the silver ions go? they became the Ag(NH3)2 ions. they are not the same thing!