Let's look at the equilibrium equation for the reaction taking place:
Ni(NO3)2 + 6NH3 <==> [Ni(NH3)6]2+ + 2NO3-
More simply, as a net ionic equation without spectator ions, we have...
Ni2+ + 6NH3 <==> [Ni(NH3)6]2+ ... Kf = 5.5x108
Kf = 5.5x108 = [Ni(NH3)6]2+ / [Ni2+][NH3]6
Initial [Ni2+]:
1 mmol Ni(NO3)2 / 0.260 L = 3.85 mM Ni(NO3)2 = 3.85 mM Ni2+ = 0.00385 M NI2+
[Ni2+] at equilibrium:
Ni2+ + 6NH3 <==> [Ni(NH3)6]2+
Because the value of Kf is so large, we can assume that almost all of the reactants are consumed to the maximum amount (i.e. dictated by limiting supply). Almost all Ni2+(0.00385 M) will be consumed, and 6x that amount (0.0231 M) of the NH3 will be consumed. We will call the final [Ni2+] x since almost all will be consumed but in reality there will be a small amount in solution.
So, we can set up an ICE table as follows:
Ni2+ + 6NH3 <==> [Ni(NH3)6]2+
0.00385...0.300...............0...........Initial
-0.00385...-0.0231.....+0.00385.....Change
...x.............0.277............0.00385......Equilibrium
Kf = 5.5x108 = [Ni(NH3)6]2+ / [Ni2+][NH3]6
5.5x108 = 0.00385 / (x)(0.277)6
5.5x108 = 0.00385 / 0.000452x
x = 1.5x10-8 M = [Ni2+]