Question

The next 7 questions are related to the titration of 60.0 mL of a 0.0250 M Zn2+ solution with 0.0600 M EDTA in a solution buffered at pH 11. Assume that the temperature is 25 oC and that the formation constant for Zn2+ is 3.13 x 1016 at this temperature.

How many mmols of Zn2+ are present in the solution before the titration begins?

What volume of the EDTA solution is needed to reach the equivalence point?

What is the conditional formation constant for Zn at this pH?

What is the pZn of the analyte solution before the titration begins?

What is the pZn of the solution after 15 mL of titrant have been added?

What is the pZn at the equivalence point of the titration?

What is the pZn of the solution after 30 mL of titrant have been added?

Answer 1

[alphaY4-] at pH 9 = 5.4 x 10^-2

Kf = 3.13 x 10^16

Kf' = [alpha[Y4-].Kf = 1.69 x 10^15

(a) mmol of Zn2+ before titration = 0.0250 M x 60 ml = 1.5 mmol

(b) Volume of EDTA required to reach equivalence point = 1.5 mmol/0.06 M = 25 ml

(c) Kf' at equivalence point = 1.69 x 10^15

(d) pZn before titration begin = -log(0.025) = 1.60

(e) pZn after 15 ml of titrant have been added

mmols of Zn2 = 1.5 mmol

mmols of EDTA = 0.06 x 15 = 0.9 mmol

remaining Zn2 = 0.6 mmol

[Zn2] = 0.6/75 = 8 x 10^-3 M

pZn2 = -log(8 x 10^-3) = 2.10

(f) pZn at euivalence point

[ZnY2-] = 1.5/85 = 0.018 M

ZnY2- <===> Zn2+ + EDTA

let x amount has gone under change

Kf' = 1.69 x 10^15 = 0.018/x^2

x = 3.26 x 10^-9 M

pZn = -log(3.26 x 10^-9) = 8.49

(g) pH after 30 ml of titrant added

excess mmols of EDTA = 0.06 M x 5 ml = 0.3 mmol

[EDTA] = 0.3/90 = 3.33 x 10^-3 M

[ZnY2-] = 1.5/90 = 0.017 M

Kf' = 1.69 x 10^15 = (0.017)/[Zn2+](3.33 x 10^-3)

[Zn2+] = 3.02 x 10^-15 M

pZn = 14.52