Question

A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of a 0.04503 M EDTA solution. The solution is then back titrated with 0.02327 M Zn2 solution at a pH of 5. A volume of 22.80 mL of the Zn2 solution was needed to reach the xylenol orange end point. A 2.000-mL aliquot of the Cu2 and Ni2 solution is fed through an ion-exchange column that retains Ni2 . The Cu2 that passed through the column is treated with 25.00 mL 0.04503 M EDTA. This solution required 21.76 mL of 0.02327 M Zn2 for back titration. The Ni2 extracted from the column was treated witn 25.00 mL of 0.04503 M EDTA. How many milliliters of 0.02327 M Zn2 is required for the back titration of the Ni2 solution?

Answer 1

asume EDTA be Y4-; it complexes 1:1 with M2+ ions:

M2+ + Y4- -> [MY]2- (M2+ = Cu2+, Ni2+)
1st Step What was the number of mmol of Y4- added at the beginning:
MV = 25.00 x 0.04503 = 1.12575 mmol
How many mmol not reacted (back-titrated) = 22.8 x 0.02327 = 0.53
Therefore mmol consumed in first titration = 1.12575 - 0.53 = 0.595 mmol = [Cu2+] + [Ni2+] note this is in 1.000 mL

2nd Step Titration of Ni2+; mmol of Y4- added = 25.00 x 0.04503 = 1.12575 mmol
mmol of Y4- not reacted = 21.76 x 0.02327 = 0.5063 mmol
Therefore mmol of Ni2+ = 1.12575 - 0.5063 = 0.6194 mmol
From Step 1 in 2.000 mL of the Cu2+/Ni2+ solution there will be 2 x 0.595 = 1.19 mmol
Combining results from Step 1 and 2 we have the [Cu2+] = 1.19 - 0.6194 = 0.5706 mmol

3rd Step mmol of Y4- added 25.00 x 0.04503* = 0.1.12575 mmol (of course same as Step 2: silly me!
(* First value of conc used: 0.04503 M Step 2, but 0.04503 M in Step 3, but we’re told they are the same: change if it is the other way around)
Therefore Y4- not consumed = 1.12575 - 0.5703 = 0.55545 mmol
This will require 0.55545 / 0.02327 = 23.867 mL for the back titration