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

Aman Bansal
592 Points
9 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

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Aman Bansal

Rishi Sharma
9 months ago
Dear Student,

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