Answer to Question #26133 in Inorganic Chemistry for emily

Question

.160-mole quantity of NiCl2 is added to a liter of 1.20M NH3 solution. what is the concentration of Ni2+ ions at equilibrium? assume the formation constant of Ni(NH3)6^2+ is 5.5*10^8

Answer 1

The starting concentration of NH₃ solutionis 1.20 M.
The starting concentration of NiCl₂ is 0.160 mole in 1 L = 0.160 M
The starting concentration of Ni(NH₃)₆²⁺ is 0

Formation constant is: K = [Ni(NH₃)₆²⁺] / [Ni²⁺][NH₃]⁶

Equilibrium concentrations:
If equilibrium concentrations of Ni(NH₃)₆²⁺ is x, equilibriumconcentrations of Ni²⁺ is 0.160-x, and equilibrium concentrations of NH₃ is 1.20^-6x.

K = [x]/[0.160 - x][1.20 - 6x]⁶
x = 0.159 M

So concentration of Ni²⁺ is 0.160 - 0.159 = 0.001 M