Question

One millimole of Ni(NO3)2 dissolves in 270.0 mL of a solution that is 0.500 M in ammonia. The formation constant of Ni(NH3)62+ is 5.5×108 What is the initial concentration of Ni(NO3)2 in the solution? _____M What is the equilibrium concentration of Ni2+(aq ) in the solution? ______M

Answer 1

part a

One millimole of Ni(NO3)2 dissolves in 270 mL :

Initial concentration of Ni(NO3)2 = moles / volume

= 0.001 moles Ni(NO3)2 / 0.270 litres

= 0.0037 M Ni(NO3)2

Part b

Given

Kf for [Ni(NH3)6]2+ = 5.5×10^8

The balanced chemical equation is

Ni+2 + 6 NH3 --> [Ni(NH3)6]2+

Formation constant

Kf = [Ni(NH3)6]2+ / [Ni+2] [NH3]^6

5.5 x 10^8 = [0.0037] / [Ni+2] [0.500]^6

5.5 x 10^8 = [0.0037] / [Ni+2] (0.0156)

[Ni+2] = [0.0037] / (5.5 x 10^8) (0.0156)

Ni+2 = 4.31 x 10^-10 M