Keq = Kf*Ksp = [Ag(NH3)2]*(Cl^-)/(NH3)^2
From my viewpoint, the 2 for NH3 does play a role.
Combine the Ksp and Kf equilibria for AgCl and Ag(NH3)2+ respectively and demonstrate Hess's law to determine the equilibrium constant for the dissolution of AgCl in NH3.
AgCl + 2NH3 --> Ag(NH3)2 + Cl-
K= (Ksp)(Kf)
If there is only one mole of both AgCl and Ag(NH3)2, both the Ksp and Kf remain with an exponent of one, correct? The coefficient in front of the NH3 doesn't play a role?
2 answers
Dr Bob is 100% right. You must take the NH3 into account in the formation of the silver-amine complex. It is easier to see if simultaneous equations are written.
(1) AgCl <=> Ag^+ + Cl^-
(2) Ag+ + 2NH3 <=> Ag[NH3]2^+
_______________________
(3) AgCl + 2NH3 <=> Ag[NH2]2^+ + Cl-
Ceq:--- (0.10M-2x) x x
(1) Ksp = 1.8E-10
(2) Kf = 1.7E+7
(3) Knet = Ksp x Kf
= (1.8E-10)(1.7E+7)
= 0.0031
Using net equation for analysis of solubility and assuming 0.10M in NH3:
AgCl + 2NH3 <=> (Ag[NH3]2^+)+Cl^-
Ceq ---(0.10M-2x)* x x
__________
The Knet (0.0031) of this equation is too large to allow dropping 2x and avoiding significant error in calculations. However, the Kcplx expression can be reduced to a non-quadratic form by taking square root of both sides of Knet expression.
(Knet)^2
=([Ag(NH2)2^+][Cl^-])/([NH3]^2)
=[(x)(x)/(0.10-2x)^2]=(x^2)/(0.1-2x)^2
[(0.0031)^1/2]=[(x^2)/(0.1-2x)^2]^1/2
0.055 = (x)/(0.10-2x)
0.055(0.10-2x) = (x)
0.0055-0.11x = x
Add 0.11 to both sides & solve for x
1.11x = 0.0055
x = solubility of AgCl in presence of 0.01M NH3 = (0.0055/1.11)M = 0.005M
[AgCl](in HOH) = 1.34e-5M
[AgCl](With Common-ion) ~ 9.0E-9M
[AgCl](with complex form'n) ~ .005M
(1) AgCl <=> Ag^+ + Cl^-
(2) Ag+ + 2NH3 <=> Ag[NH3]2^+
_______________________
(3) AgCl + 2NH3 <=> Ag[NH2]2^+ + Cl-
Ceq:--- (0.10M-2x) x x
(1) Ksp = 1.8E-10
(2) Kf = 1.7E+7
(3) Knet = Ksp x Kf
= (1.8E-10)(1.7E+7)
= 0.0031
Using net equation for analysis of solubility and assuming 0.10M in NH3:
AgCl + 2NH3 <=> (Ag[NH3]2^+)+Cl^-
Ceq ---(0.10M-2x)* x x
__________
The Knet (0.0031) of this equation is too large to allow dropping 2x and avoiding significant error in calculations. However, the Kcplx expression can be reduced to a non-quadratic form by taking square root of both sides of Knet expression.
(Knet)^2
=([Ag(NH2)2^+][Cl^-])/([NH3]^2)
=[(x)(x)/(0.10-2x)^2]=(x^2)/(0.1-2x)^2
[(0.0031)^1/2]=[(x^2)/(0.1-2x)^2]^1/2
0.055 = (x)/(0.10-2x)
0.055(0.10-2x) = (x)
0.0055-0.11x = x
Add 0.11 to both sides & solve for x
1.11x = 0.0055
x = solubility of AgCl in presence of 0.01M NH3 = (0.0055/1.11)M = 0.005M
[AgCl](in HOH) = 1.34e-5M
[AgCl](With Common-ion) ~ 9.0E-9M
[AgCl](with complex form'n) ~ .005M
1 answer