 # Calculate molar solubility of AgCl in 3 M NH3KSP OF AgCl=1.8 *10power -10kf of[Ag(NH3)2]+=1.6 *10POWER7

11 years ago

Daer Naitik,

AgCl--->Ag+ Cl- Ksp= 1.8x10^-10
Ag + 2NH3 ---> Ag(NH3)2 Kf= 1.6x10^7
net reaction: AgCl + 2NH3 ----> Cl- + Ag(NH3)2
so Ksp * Kf = .00288 = net equilbrium constant

so, for the formation of the complex, you would use the net equilibrium constant for the values of your ICE chart of the formation of the Ag(NH3)2 and solve for x

Kf = [Ag(NH3)2] / [Ag][NH3]^2
so Knet = [Ag(NH3)2+][Cl-] / [NH3]^2
Knet = 0.00288 = x^2 / 0.1 - 2x
x^2 + 0.00576x - 0.000288 = 0
x =0.0143M = molar solubility of AgCl in NH3

Best Of luck

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Thanks

Aman Bansal

2 years ago
Dear Student,
Please find below the solution to your problem.

AgCl--->Ag+ Cl- Ksp= 1.8x10^-10
Ag + 2NH3 ---> Ag(NH3)2 Kf= 1.6x10^7
net reaction: AgCl + 2NH3 ----> Cl- + Ag(NH3)2
so Ksp * Kf = .00288 = net equilbrium constant
so, for the formation of the complex, you would use the net equilibrium constant for the values of your ICE chart of the formation of the Ag(NH3)2 and solve for x
Kf = [Ag(NH3)2] / [Ag][NH3]^2
so Knet = [Ag(NH3)2+][Cl-] / [NH3]^2
Knet = 0.00288 = x^2 / 0.1 - 2x
x^2 + 0.00576x - 0.000288 = 0
x =0.0143M = molar solubility of AgCl in NH3

Thanks and Regards