I think the easiest way to solve this type problem is to assume that the reaction goes to completion and forms the complex, as follows:
............Cu + 4NH3 ==> Cu(NH3)4^2+
I...........0.1...6.4.......0
C..........-0.1..-0.4......+0.1
E............0....6.0......0.1
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and considering the large Kf for Cu(NH3)4^2+ it isn't unreasonable that Cu(NH3)4 will be essentially 0.1M. THEN we reverse the whole thing and calculate the small component of the Cu that ionizes. We make the E line the new I line.
-------------------------------------
I........0.......6.0........0.1
C.......+x......+4x........0.1-x
E.......+x......6.0-4x....0.1-x
Then Kf = 1.1E13 = [Cu(NH3)2]/(Cu^2+)((NH3)^4
Substitute the E line and solve for x = (Cu^2+). You will have some complicated math if you go with (0.1-x)/(x)(6+x)^4.
I would make the simplifying assumptions that 0.1-x = 0.1 and 6.0+x = 6.0 which makes the final equation look like this.
1.1E13 = 0.1/(x)(6)^4 and solve for x.
Calculate [Cu2+] in a 0.10 M CuSO4 (aq) solution that is also 6.4 M in free NH3 .
Cu2+(aq) + 4NH3(aq) --->[Cu(NH3)4]2+ (aq)
kf = 1.1 x 10^13
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