A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL
ID: 557905 • Letter: A
Question
A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of a 0.03294 M EDTA solution. The solution is then back titrated with 0.02399 M Zn2 solution at a pH of 5. A volume of 22.49 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 0.03294 M EDTA. This solution required 20.46 mL of 0.02399 M Zn2 for back titration. The Ni2 extracted from the column was treated witn 25.00 mL of 0.03294 M EDTA. How many milliliters of 0.02399 M Zn2 is required for the back titration of the Ni2 solution?
Explanation / Answer
mmol of EDTA = 25 x 0.03294 = 0.8235
mmol of Zn+2 = 22.49 x 0.02399 = 0.5395
mmol Cu+2 + Ni+2 = 0.8235 - 0.5395 = 0.28396
mmol of EDTA = 0.8235
mmol Zn+2 = 20.46 x 0.02399 = 0.4908
mmol of Cu+2 in 2 mL =0.8235 - 0.4908 = 0.3327
mmol of Ni+2 = 2 x 0.28396 - 0.3327 = 0.2352
mmol of EDTA reamins = 0.8235 - 0.2352 = 0.5883
mmol of EDTA = mmol Zn+2 = 0.5883
volume = 0.5883 / 0.02399 = 24.52 mL
volume of Zn+2 = 24.52 mL
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