Question

A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of a 0.04728 M EDTA solution. The solution is then back titrated with 0.02103 M Zn2 solution at a pH of 5. A volume of 19.11 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 0.04728 M EDTA. This solution required 15.49 mL of 0.02103 M Zn2 for back titration. The Ni2 extracted from the column was treated witn 25.00 mL of 0.04728 M EDTA. How many milliliters of 0.02103 M Zn2 is required for the back titration of the Ni2 solution?

Answer 1

Total moles of EDTA used for Cu2 + Ni2 = 0.04728 M x 25 ml = 1.182 mmol

moles of Cu2 + Ni2 in 1.0 ml sample = 1.182 - (0.02103 M x 19.11 ml) = 0.782 mmol

moles of Cu2 in 1 ml sample = (0.04728 M x 25 ml - 0.02103 M x 15.49 ml)/2 = 0.428 mmol

moles of Cu2 in 2 ml sample = 0.856 mmol

moles of Ni2 in 1 ml sample = 0.782 - 0.428 = 0.354 mmol

moles of Ni2 in 2 ml sample = 0.708 mmol

Volume of Zn2 required for back titration of Ni2 = 0.708 mmol/0.02103 M = 33.67 ml