Question

A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of a 0.03742 M EDTA solution. The solution is then back titrated with 0.02190 M Zn2 solution at a pH of 5. A volume of 20.48 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 0.03742 M EDTA. This solution required 21.41 mL of 0.02190 M Zn2 for back titration. The Ni2 extracted from the column was treated witn 25.00 mL of 0.03742 M EDTA. How many milliliters of 0.02190 M Zn2 is required for the back titration of the Ni2 solution?

How many millimiters of 0.02190 M Zn2+ is required for the back titration of the Ni2+ solution?

Answer 1

Write down the balanced chemical equations as below.

Cu²⁺ (aq) + EDTA^4- (aq) -----> Cu-EDTA^2- (aq) ……(1)

Ni²⁺ (aq) + EDTA^4- (aq) -----> Ni-EDTA^2- (aq) ……(2)

Zn²⁺ (aq) + EDTA^4- (aq) -----> Zn-EDTA^2- (aq) ……(3)

As per the stoichiometric equations,

1 mole EDTA = 1 mole Cu²⁺ = 1 mole Ni²⁺ = 1 mole Zn²⁺

Part 1:

Millimoles of EDTA added to the Cu²⁺/Ni²⁺ mixture = (25.00 mL)*(0.03742 M) = 0.9355 mmole.

Millimoles of Zn²⁺ added = millimoles of EDTA neutralized = (20.48 mL)*(0.02190 M) = 0.448512 mmole.

Millimoles of (Cu²⁺ + Ni²⁺) in 1.00 mL of the mixture = (0.9355 – 0.448512) mmole = 0.486988 mmole.

Part 2:

Millimoles of EDTA added is as before, 0.9355 mmole.

Millimoles of Zn²⁺ added = millimoles of EDTA neutralized = (21.41 mL)*(0.02190 M) = 0.468879 mmole.

Note that the Ni²⁺ was arrested in the column; hence, only Cu²⁺ reacted.

Millimoles of Cu²⁺ reacted = (0.9355 – 0.468879) = 0.466621 mmole.

Remember that we took 2.00 mL aliquot of the sample mixture; hence the millimoles of Cu⁺ in 1.00 mL aliquot of the sample mixture = (0.466621 mmole)*(1.00 mL/2.00 mL) = 0.2333105 mmole.

Part 3:

Millimoles of Ni²⁺ in the mixture = millimoles of EDTA reacted with EDTA = (0.486988 – 0.2333105) = 0.2536775 mmole.

Again, note that we found out the millimoles of Ni²⁺ in 1.00 mL of the sample mixture; therefore, millimoles of Ni²⁺ in 2.00 mL aliquot = millimoles of EDTA reacted = (0.2536775 mmole)*(2.00 mL/1.00 mL) = 0.507355 mmole.

Millimoles of EDTA added = 0.9355 mmole.

Millimoles of EDTA available for reaction with Zn²⁺ = millimoles of Zn²⁺ neutralized = (0.9355 – 0.507355) = 0.428145 mmole.

Milliliters of 0.02190 M Zn²⁺ required for the titration = (0.428145 mmole)/(0.02190 M) = 19.55 mL (ans).