Question

A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of a 0.05832 M EDTA solution. The solution is then back titrated with 0.02380 M Zn2 solution at a pH of 5. A volume of 18.49 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 0.05832 M EDTA. This solution required 23.22 mL of 0.02380 M Zn2 for back titration. The Ni2 extracted from the column was treated witn 25.00 mL of 0.05832 M EDTA.

How many milliliters of 0.02380 M Zn2 is required for the back titration of the Ni2 solution?

Answer 1

Step 1 :

Total moles of EDTA used for Cu2 and Ni2 titration = 005832 x 0.025 = 1.458 x 10^-3 mols

moles of Zn2 reacted with excess EDTA = 0.02380 x 0.01849 = 4.40 x 10^-4 mols

Actual moles of EDTA reacted with Cu2 and Ni2 = 1.458 x 10^-3 - 4.40 x 10^-4 = 1.018 x 10^-3 mols

Concentration of Cu2 and Ni2 in original solution = 1.018 x 10^-3/0.001 = 1.018 M

Step 2 :

Total moles of EDTA used ed for Cu2 = 0.05832 x 0.025 = 1.458 x 10^-3 mols

moles of Zn2 required for excess EDTA = 0.02380 x 0.02322 = 5.53 x 10^-4 mols

Actual moles of EDTA used for Cu2 = 9.05 x 10^-4 mols

molarity of Cu2 in 2 ml solution = 9.05 x 10^-4/0.002 = 0.4525 M

Step 3 :

Molarity of Ni2 in solution = 1.018 - 0.4525 = 0.5655 M

Step 4 :

EDTA used for Ni2 titration = 0.05832 x 0.025 = 1.458 x 10^-3 mols

moles of Ni2 in solution = 0.5655 x 0.002 = 1.131 x 10^-3 mols

moles of Excess EDTA = 1.458 x 10^-3 - 1.131 x 10^-3 = 3.27 x 10^-4 mols

Volume of Zn2 required to titrate excess EDTA = 3.27 x 10^-4/0.02380 = 0.01374 L = 13.74 ml

Thus we need 13.74 ml of 0.02380 M Zn2 for back titration.