A 1.000-mL aliquot of a solution containing Cu^2+ and Nr^2+ is treated with 25 0
ID: 957908 • Letter: A
Question
A 1.000-mL aliquot of a solution containing Cu^2+ and Nr^2+ is treated with 25 00 mL of a 0.03528 M EDTA solution. The solution is then back titrated with 0.02030 M Zn^2+ solution at a pH of 5. A volume of 17 37 mL of the Zn^2+ solution was needed to reach the xylenol orange and point. A 2.000-mL aliquot of the Cu^2+ and Ni^2+ solution is fed through an ion-exchange column that retains Ni^2+. The Cu^2+ that passed through the column a treated with 25.00 mL 0.03528 M EDTA This solution required 1696 mL of 0.02030 M.Zn^2+ for hack titration. The Ni^2+ extracted from the column was treated with? mL of 0 03528 M EDTA. How many milliliters of 0.02030 M Zn^2+ is required for the back titration of the Ni^2+ solution?Explanation / Answer
mmol of EDTA = 25 x 0.03528 = 0.882
mmol of Zn+2 = 17.37 x 0.02030 = 0.3526
mmol Cu+2 + Ni+2 = 0.882 -0.3526 = 0.5294
mmol of EDTA = 0.882
mmol Zn+2 = 16.98 x 0.02030 = 0.34469
mmol of Cu+2 in 2 mL = 0.882 - 0.34469 = 0.5373
mmol of Ni+2 = 2 x 0.5294 - 0.5373 = 0.5215
mmol of EDTA reamins = 0.882 - 0.5215 = 0.3605
mmol of EDTA = mmol Zn+2 = 0.3605
volume = 0.3605 / 0.02030 = 17.78 mL
volume of Zn+2 = 17.78 mL
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