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


 Calculate molar solubility of AgCl in 3 M NH3


 


KSP OF AgCl=1.8 *10power -10


 


kf of[Ag(NH3)2]+=1.6 *10POWER7


Grade:11

2 Answers

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

Best Of luck

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Thanks

Aman Bansal

Askiitian Expert

Rishi Sharma
askIITians Faculty 646 Points
9 months 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

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