[skibum143]

Find the solubility of AgBr in 2.7 M NH3 [Ksp of AgBr = 5.0E-13 and Kf of Ag(NH3)2+ = 1.7E7].

So: AgBr -> Ag+ + Br- (Ksp)
and: Ag+ + 2NH3 -> Ag(NH3)2 (Kf)

Ag+ Cancels out

so AgBr + 2NH3 -> Ag(NH3)2 + Br-

K = [Br-] / [NH3]^2
K = [5.0E-13] / [2.7]^2
k = 6.9E-14

Not sure what I'm doing wrong, but that answer is incorrect. I also tried using the Kf value for the concentration of NH3, but that didn't work either.

[rabolisk]

K = [Br-]/[NH3]^2 is correct. The next step is wrong. Remember that you're not solving for any equilibrium constant. In fact, you know the value of K.

[AWK]

K = [Br-]/[NH3]^2 is incorrect. Nominator also should contain concentration of Ag(NH₃)₂⁺ which is approximately equal to [Br^-]