calculate Cu2+ in a 0.15M CuSO2 (aq) solution that is also 6.5M in free NH3.

Question

Cu^2+(aq)+ 4NH3(aq)<=> [Cu(NH3)4]^2+ kf=1.1*10^13

Answer 1

Do you mean CuSO4?
Kf = 1.1E13 = [Cu(NH3)4]^2+/(Cu^2+)(NH3)^4.

Do this ICE chart in two steps. Here is the first one. With Kf so large the reaction goes to completion (almost) like this.
...Cu^2+ + NH3 ==>Cu(NH3)²⁺₄
I..0.15.....6.5.......0
C..-0.15...-0.6.....+0.15
E....0......5.9......0.15

Now redo the ICE chart BUT showing the small amount of reverse reaction like this.
....Cu^2+ + 4NH3 ==> Cucomplex^2+
I...0.......5.9.......0.15
C...+x......+4x........-x
E...x......5.9+4x.....0.15-x

Now substitute this last E line into the kf expression and solve for x = (Cu^2+)
Post your work if you get stuck.