Question

You are titrating 50ml of a 0.02M solution of Zn^+2 at pH 7 with 0.05M solution of EDTA. What is the [Zn+2] after 20mL of EDTA has been added? Kzny=3x10^16

Answer 1

I assume that Kzny is Kf for ZnEDTA = 3E16. millimols Zn = 50 x 0.2 = 1.00. millimols EDTA = 20 x 0.05 = 1.00
.......Zn^2+ + EDTA ==> ZnEDTA
I......1.00....1.00......0
C.....-1.00...-1.00......1.00
E......0........0........1.00
With such a large Kf of 3E16, it is reasonable to assume that almost every tiny bit of Zn has formed the complex; that won't be exactly right but with that large value of Kf there won't be many Zn^2+ left by themselves. Then we take the problem in reverse and do this.
(ZnEDTA) = 1 millimol/70 mL = 0.015 but that's and ESTIMATE and you need a more accurate number.
........ZnEDTA ==> Zn^2+ + EDTA
I.......0.015 ......0........0
C..........-x.......x........x
E........0.015-x....x........x

Then substitute the E line into Kf for EDTA and solve for x = (Zn^2+) in mols/L.