A solution is made that is 1.3×10−3M in Zn(NO3)2 and 0.160M in NH3.
After the solution reaches equilibrium, what concentration of Zn2+(aq) remains?
(I need someone who KNOWS how to do this. The last yahoo answers was wrong.)
2 Answers

Zn2+ + 4NH3 =======> Zn(NH3)4^2+
0.0013 + 4(0.0013) =====> 0.0013
NH3 remaining = 0.160 – 0.0052 = 0.1548 M
Kf = [Zn(NH3)4^2+]/[Zn2+][NH3]^4 = 4 x 10^8
Substituting in the above relationship : [0.0013]/[Zn2+][0.1548]^4 and solving for Zn2+
[Zn2+] = 5.69 x 10^9

the answer should be 8.7 x 10^10 M. I had this problem in my textbook which has the answers in the back.
Zn (aq) + 4NH3 = Zn(NH3)4 2+
kf Zn(NH3)4 2+ = 2.8 x 10^9
2.8 x 10^9 = (1.1 x 10^3) / [ (0.1456)^4 x (x)]
x = 8.7 x 10^10