For the formation of a complex ion: Ni(NH3)4 2+ the Kf=5.6*10^8?

Question

what is the concentration of free Ni2+ ion in a sol. that initially contains .0090 M Ni2+ and .38 M NH3.

It's for homework and I can't figure it out for the life of me and It's due by midnight, please help mucho points, I'm lost.

Answer 1

Ni2+ + 4 NH3 >> Ni(NH3)4 2+

initial concentration

0.0090. . 0.38

change

-0.0090....0.036 ( => 0.0090 x4)

at equilibrium

0 . . . . . .0.344 . . . .+ 0.0090

Ni2+ + 4 NH3 <-----> Ni(NH3)4+

initial concentration

0 . . . . 0.344 .. . .. . 0.0090

change

+x . . . .+4x . . . . . . 0.0090-4x

5.6 x 10^8 = 0.0090-4x / [Ni2+] ( 0.344-4x)^4 =

= 0.0090 / [Ni2+] ( 0.344)^4 =  0.642 / [Ni2+]

[Ni2+] = 1.1 x 10^-9 M