Calculate the molar solubility (to 4 decimal places) of AgBr in 2.4 M NH3 (Kf for Ag(NH3)2+1 is 1.7x107 and Ksp for AgBr is 5.0 x 10-13).

Answer 1

This is a problem dealing with complex ion formation and solubility. It is improper to solve to 4 decimal places, so I'll leave that part to you. I will solve to 2 decimals and even that isn't valid based on the Ksp and Kf values.

Complex ion is Ag(NH₃)₂⁺

Formation of this ion can be viewed as...

Ag⁺(aq) + 2NH₃(aq) ==> Ag(NH₃)₂⁺(aq) ... Kf = 1.7x10⁷

AgBr(s) <==> Ag+(aq) + Br-(aq) ... Ksp = 5.0x10^-13

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2NH₃(aq) + AgBr(s) ==> Ag(NH₃)₂⁺(aq) + Br^-(aq) ... K = (1.7x10⁷)(5.0x10^-13) = 8.5x10^-6

2.4.....................................0.....................0..........Initial

-2x..................................+x....................+x..........Change

2.4-2x...............................x......................x..........Equilibrium

K = 8.5x10^-6 = [Ag(NH₃)₂⁺][Br^-] / [NH₃]²

8.5x10^-6 = (x)(x) / (2.4 -2x)² = x² / (2.4 -2x)² and taking square root of both sides, we have....

2.92 = x / 2.4 - 2x

7.01 - 5.84x = x

6.84x = 7.01

x = 1.03 M = molar solubility

(be sure to check all the math for possible errors)