Question

A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of a 0.05893 M EDTA solution. The solution is then back titrated with 0.02306 M Zn2 solution at a pH of 5. A volume of 19.89 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 0.05893 M EDTA. This solution required 24.30 mL of 0.02306 M Zn2 for back titration. The Ni2 extracted from the column was treated witn 25.00 mL of 0.05893 M EDTA. How many milliliters of 0.02306 M Zn2 is required for the back titration of the Ni2 solution?

Answer 1

Write down the three equations for the reactions.

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

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

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

As per the stoichiometric equations, there is 1:1 mole ratio of reaction between the metal ion and EDTA.

Divide the problem into three parts.

Part 1:

The millimoles of EDTA added for the mixed titration = (25.00 mL)*(0.05893 M) = 1.47325 mmole.

Millimoles of EDTA reacted with Zn²⁺ = millimoles of Zn²⁺ added = (19.89 mL)*(0.02306 M) = 0.4586634 mmole.

Therefore, millimoles of EDTA that reacted with Cu²⁺ and Ni²⁺ = sum of the millimoles of Cu²⁺ and Ni²⁺ = (1.47325 – 0.4586634) = 1.0145866 mmole.

Part 2:

Millimoles of EDTA added = same as above = 1.47325 mmole.

Millimoles of EDTA that reacted with Zn²⁺ = millimoles of Zn²⁺ = (24.30 mL)*(0.02306 M) = 0.560358 mmole.

Since Ni²⁺ was retained in the column, hence, only Cu²⁺ reacted; therefore, millimoles of Cu²⁺ in the original mixture = (millimoles of EDTA added) – (millimoles of EDTA reacted with Zn²⁺) = (1.47325 – 0.560358) mmole = 0.912892 mmole.

Part 3:

Millimoles of Ni²⁺ in the mixture = (1.0145866 – 0.912892) = 0.1016946 mmole = millimoles of EDTA reacted with Ni²⁺.

Therefore, millimoles of EDTA available for reaction with Zn²⁺ = (1.47325 – 0.1016946) = 1.3715554 mmole = millimoles of Zn²⁺ reacted.

Volume of Zn²⁺ reacted = (1.3715554 mmole)/(0.02306 M) = 59.47768 mL ≈ 59.48 mL (ans).