Asked by k
calculate the molar concentration of uncomplexed zn2+(aq) in a solution that contains .22M Zn(NH3)4^2+ and .4978M NH3 at equilibrium. Kf for Zn(NH3)4^2+ is 2.9 x 10^9
Answers
Answered by
DrBob222
.....................Zn^2+ + 4NH3 ==> Zn(NH3)4^2+
Equil...............x..........-.497.............0.22
Kf = 2.9E9 = [Zn(NH3)4]^2+/(Zn^2+)(NH3)^4
Substitute the E line into Kf expression and solve for (Zn^2+)
Equil...............x..........-.497.............0.22
Kf = 2.9E9 = [Zn(NH3)4]^2+/(Zn^2+)(NH3)^4
Substitute the E line into Kf expression and solve for (Zn^2+)
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