A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL
ID: 497911 • Letter: A
Question
A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of a 0.03480 M EDTA solution. The solution is then back titrated with 0.02022 M Zn2 solution at a pH of 5. A volume of 19.68 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 0.03480 M EDTA. This solution required 17.08 mL of 0.02022 M Zn2 for back titration. The Ni2 extracted from the column was treated witn 25.00 mL of 0.03480 M EDTA. How many milliliters of 0.02022 M Zn2 is required for the back titration of the Ni2 solution?
Explanation / Answer
mmol of EDTA = 25 x 0.03480 = 0.870
mmol of Zn+2 = 19.68 x 0.02022 = 0.3979
mmol Cu+2 + Ni+2 = 0.870 - 0.3979 = 0.4721
for 2 mL of unknown
mmol of EDTA = 0.870
mmol Zn+2 = 17.08 x 0.02022 = 0.3454
mmol of Cu+2 in 2 mL = 0.870 - 0.3454 = 0.5246
mmol of Ni+2 = 2 x 0.4721 - 0.5246 = 0.4196
mmol of EDTA reamins = 0.870 - 0.4196 = 0.4504
mmol of EDTA = mmol Zn+2 = 0.4504
volume = 0.4504 / 0.02022 = 22.27 mL
volume of Zn+2 = 22.27 mL
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