In: Chemistry
One millimole of Ni(NO3)2 dissolves in 260.0 mL of a solution
that is 0.400 M in ammonia.
The formation constant of Ni(NH3)62+ is 5.5×10^8.
What is the equilibrium concentration of Ni2+(aq ) in
the solution?
Ni^2+(aq) + 6 NH3(aq) <---> Ni(NH3)6^2+(aq)
initial 1/260 = 0.00385 M 0.4 M 0 M
change x M 6x M x M
equilibrium 0.00385-x 0.4-6x x
Kf = [Ni(NH3)6^2+]/[Ni^2+][NH3]^6
at equilibrium, [NH3] = 0.4-6X = 0.4 M
(5.5*10^8) = x/((0.00385-x)(0.4)^6)
x = 0.0038499
at equilibrium, [Ni^2+] = 0.00385-0.0038499 = 1*10^-7 M