chem help please!!?

Question

Find the solubility of AgI in 2.8 M NH3 [Ksp of AgI = 8.3 *10-17 and Kf of Ag(NH3)2+ = 1.7 * 107].

Answer 1

At equilibrium both the conditions of the solubility product {Ag+}{I-} = Ksp and of the complex formation:
{complex]/[Ag+][NH3]^2= Kf
have to be met. Note that although AgI goes into solution as ions, the I- stays in solution , but a good part of the Ag+ is converted to the complex. This you have to figure out.
Without going through the gory math, [NH3] will be about what it is above, since the amount of Ag+ added is expected to be much less in comparison. You have 3 equations to solve, the third of which is {I-} = [Ag+]+[complex]