[NYM]

Could somebody please check my results?
That would make me very happy :p

a) Find the solubility of AgCl in 0,1 M NH₃ (aq).
b) Using the result from a), deduce a general expression for the solubility of AgCl in NH₃ (aq)

This is what I did:
a)
AgCl + 2 NH₃(aq) ->Ag(NH₃)₂⁺ + Cl^-
K = ([Ag(NH₃)⁺][Cl^-])/[NH₃]²

At equilibrium the actual concentrations will be:
[NH₃] = c - 2x
Ag(NH₃)₂⁺ = x
Cl^- = x

Since I don't have K for the reaction, I'll split it in to different reactions where I do have the K-values:
AgCl -> Ag⁺ + Cl^-
K_o=[Ag⁺][Cl^-]

and

Ag⁺ + 2 NH₃(aq) -> Ag(NH₃)₂⁺
K_k=[Ag(NH₃)₂⁺]/([Ag⁺][(NH₃)²

Then I'll multiply them to eliminate [Ag⁺]:
K_o*K_k = ([Ag⁺][Cl^-])*([Ag(NH₃)₂⁺]/([Ag⁺][(NH₃)² = ([Ag(NH₃)⁺][Cl^-])/[NH₃]²

That is the K-value from the original reaction.
I'll insert the actual concentrations:
K_o*K_k = x²/(c-2x)² .
I get x = 5,0*1^0^-3 M.

b) Basically, I just have to isolate x, right?
I get:
x = (c*(K_o*K_k)^(1/2))/(1+2*(*(K_o*K_k)^(1/2))

Thanks!

[Borek]

http://www.chemicalforums.com/index.php?topic=10045.msg47696#msg47696

[NYM]

Thanks! I should have search it