Question

A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of a 0.03146 M EDTA solution. The solution is then back titrated with 0.02115 M Zn2 solution at a pH of 5. A volume of 16.97 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 0.03146 M EDTA. This solution required 21.80 mL of 0.02115 M Zn2 for back titration. The Ni2 extracted from the column was treated witn 25.00 mL of 0.03146 M EDTA. How many milliliters of 0.02115 M Zn2 is required for the back titration of the Ni2 solution?

Answer 1

millimoles of EDTA = 25.00 x 0.03146 = 0.7865

mmol of Zn+2 = 0.02115 x 16.97 = 0.3589

Cu+2 + Ni+2 mmol = 0.7865 - 0.3589

                               = 0.4276

EDTA mmol = 25.00 x 0.03146 = 0.7865

again Zn+2 mmol = 21.80 x 0.02115 = 0.4611

Cu+2 mmol = 0.7865 - 0.4611 = 0.3254

Ni+2 in 2.00 mL of unknown = 2 x 0.4276 - 0.3254

                                               = 0.5298 mmol of Ni+2

0.7865 - 0.5298 = 0.2567 mmol EDTA

0.2567 mmol Zn+2 / 0.02115 = 12 mL

volume of Zn+2 = 12 mL