Question

What is the minimum concentration of ammonia that should dissolve AgBr to produce a 0.01 M Ag(NH3)2+ solution? [Kf for Ag(NH3)2+ = 1.7x107]

Answer 1

For a solution with 0.01M Ag(NH3)2+, you must have 0.01 M Br- too (electrically neutral). Plug that into your equation for Ks and solve for conc. of free Ag:
Ks = [Ag][Br]

7.7*10^-13 = [Ag] * 0.01 M

[Ag] = 7.7*10^-11 M

Used the computed [Ag] in your equation for Ke to work out how much free NH3 is needed:

Ke = (0.01M Ag(NH3)2+)/([Ag][NH3]^2)

1.7*10^7 = (0.01 M Ag(NH3)2+)/(7.7*10^-11 M *[NH3]^2)

[NH3]^2 = 0.01 M /( 1.7*10^7 * 7.7*10^-11 M)

[NH3]^2 = 7.6394

[NH3] = 2.76 moles/liter

This is *free* NH3, so you've got to add "2*0.01 M" that is bound to the Ag to get total NH3.

Total [NH3] = 2.76 + 2*0.01

= 2.76 +0.02 = 2.78 moles/liter.