I think it helps to draw a picture.
......Zn^2+ + 4NH3 ==> Zn(NH3)4 +
E......x......0.3109...0.22
The problem gives the equilibrium conditions; therefore, you need not worry about a reaction.
Kf = [Zn(NH3)4}^2+/(Zn^2+)(NH3)^4
(Zn^2+) = x
[Zn(NH3)4]^2+ = 0.22M
(NH3) = 0.3109
Solve for (Zn^2+) which is the only unknown in the equation.
Calculate the molar concentration of uncomplexed Zn2+ (aq) in a solution that contains 0.22 mol of Zn(NH3)4 2+ per liter and 0.3109 M NH3 at equilibrium. Kf for Zn(NH3)4 2+ is
2.9 X 10^9
I started this way...
(0.3109 + 4)2/0.22-x)= 2.9 X 10^9
1 answer
1 answer