Question

A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of a 0.04328 M EDTA solution. The solution is then back titrated with 0.02246 M Zn2 solution at a pH of 5. A volume of 21.60 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 0.04328 M EDTA. This solution required 21.32 mL of 0.02246 M Zn2 for back titration. The Ni2 extracted from the column was treated witn 25.00 mL of 0.04328 M EDTA. How many milliliters of 0.02246 M Zn2 is required for the back titration of the Ni2 solution.

Answer 1

The balanced chemical equations are:

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⁺

The millimoles of EDTA added to the Cu²⁺/Ni²⁺ mixture = (25.00 mL)*(0.04328 M)

= 1.082 millimole.

The millimoles of Zn²⁺ added = millimoles of EDTA neutralized

= (21.60 mL)*(0.02246 M) = 0.48514 millimole

The millimoles of (Cu²⁺ + Ni²⁺) in 1.00 mL of the mixture = (1.082 – 0.48514) milli moles

= 0.5969 milli moles

The millimoles of EDTA added is as before, 1.082 mmole.

The millimoles of Zn²⁺ added = milli moles of EDTA neutralized = (21.32 mL)*(0.02246 M)

= 0.47885 milli moles

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

The millimoles of Cu²⁺ reacted = (1.082 – 0.47885) = 0.60315 mmole.

The 2.00 mL of aliquot of the sample mixture is taken, hence the millimoles of Cu⁺ in 1.00 mL aliquot of the sample mixture = (0.60315 millimoles)*(1.00 mL/2.00 mL) = 0.30158 millimoles

Part 3:

The millimoles of Ni²⁺ in the mixture = millimoles of EDTA reacted with EDTA = 0.5969 - 0.30158

= 0.29532 milli moles

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

= (0.29532 mmole)*(2.00 mL/1.00 mL)

= 0.59064 milli moles

The millimoles of EDTA added = 1.082 mmole.

The millimoles of EDTA available for reaction with Zn²⁺ = millimoles of Zn²⁺ neutralized

= (1.082 – 0.59064)

= 0.49136 millimoles

The millimoles of 0.02246 M Zn²⁺ required for the titration = ( 0.49136 millimoles)/(0.02246 M)

= 20.088 mL