The Kf for the complex ion Ag(NH3)2+ is 1.7 x 107. The Ksp for AgCl is 1.6 x 10

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Explanation / Answer

AgCl--->Ag+ Cl- Ksp= 1.6x10^-10 Ag + 2NH3 ---> Ag(NH3)2 Kf= 1.7x10^7 net reaction: AgCl + 2NH3 ----> Cl- + Ag(NH3)2 so Ksp * Kf = .00272 = 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 = 0.00272 = x^2 / 1 - 2x x^2 + 0.00544x - 0.00272 = 0 x = [AgNH3)2+]] = molar solubility of Ag(NH3)2+ = 0.0495M = 5x10^-2M.

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