Question

Given the following equilibria at 25oC:
AgCl(s) <----> Ag+(aq) + Cl-(aq); Ksp = 1.6 x 10-10
Ag+(aq) + 2NH3(aq) <----> Ag(NH3)2(aq); Kf = 1.7 x 107
What are the equilibrium concentration of Ag+ in the saturated solution?

Answer 1

Q1.

AgCl(s) <----> Ag+(aq) + Cl-(aq); Ksp = 1.6 x 10-10

if saturated

Ksp = [Ag+][Cl-]

Assume, stoichiometriccaly --> [Ag+]= [Cl-] = S

1.6*10^-10 = (S)(S)

S = sqrt(1.6*10^-10)

S = 1.2649*10^-5

[Ag+] = S = 1.2649*10^-5 M

b)

K = Ksp × Kf = (1.6*10^-10)*(1.7*10^7) =0.00272

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

0.00272 = (1.2649*10^-5)^2 / ((M - 2*(1.2649*10^-5 ))^2

((M - 2*(1.2649*10^-5 ))^2 = ((1.2649*10^-5)^2 )/0.00272

((M - 2*(1.2649*10^-5 ))^2 = 5.882*10^-8

M - 2*(1.2649*10^-5 = sqrt( 5.882*10^-8)

M - 2*(1.2649*10^-5) = 0.000242

M = 0.000242 + 2*(1.2649*10^-5)

M = 0.000267298