Answer to Question #26133 in Inorganic Chemistry for emily
.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
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
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