One millimole of Ni(NO3)2 dissolves in 260.0 mL of a solution that is 0.300 M in ammonia. The formation constant of Ni(NH3)6^2+ is 5.5×10^8.

Question

What is the initial concentration of Ni(NO3)2 in the solution?

What is the equilibrium concentration of Ni2+(aq) in the solution?

Answer 1

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 [Ni^2+]:

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⁺]