Find the solubility of AgBr in 3.9 M NH3 [Ksp of AgBr = 5.0 10-13 and Kf of Ag(NH3)2+ = 1.7 107]

Answer 1

AgBr==> Ag^+ + Br^-
Ag^+ + 2NH3 ==> Ag(NH3^+

eqn 1..Ksp = (Ag^+)(Br^-)
Kf = [Ag(NH3)2^+]/(Ag^+)(NH3)^2
Let S = solubility, then
eqn 2..S = (Br^-) = (Ag^+) + [Ag(NH3)2^+]
eqn 3..3.9 = (NH3) + (NH4^+) + 2[Ag(MH3)2^+]

Three equations as above.
1..as is
2..(Ag^+)<< [Ag(NH3)2^+]
3..3.9 = (NH3) as (NH4^+) is small and 2[Ag(NH3)2^+] is much smaller than (NH3)
Three equation and three unknowns.