Question

Calculate the solubility of silver chloride in 10.0 M ammonia given the following information: K_sp (AgCI) = 1.6 times 10^-10 Ag^+ (aq) + NH_3(aq) = AgNH_3^+ (aq) K_f1 = 2.2 times 10^3 AgNH_3^+ (aq) NH_3(aq) = Ag(NH_3)_2^+ K_f2 = 8.2 times 10^3

Answer 1

adding two equations we get

Ag+ (aq) + 2NH3 (aq) <---> Ag(NH3)2+ (aq) , Kf = Kf1 x Kf2 = 18.04 x 10^6

we have eq AgCl (s) <---> Ag+ (aq) + Cl- (aq) , Ksp = 1.6 x 10^-10

Now adding above two eq we get

AgCl(s) + 2NH3 (aq) <--> Ag(NH3)2+ (aq) + Cl- (aq) , K = Kfx Ksp = 0.0028864

at equilibrium [NH3] = 10-2X , [Ag(NH3)2+] = X = [Cl-]

K = [Ag(NH3)2+] [Cl-] / [NH3]^2

0.028864 = ( X)(X) / ( 10-2X)^2

0.028864 = ( X/ 10-2X)^2

X/(10-2X) = 0.053725

X = 0.53725-0.10745X

X = 0.485 = [Cl-]

Thus solubility of AgCl = 0.485 moles / liter