Map A Sapling Learning macmillan learning A 1.000-mL aliquot of a solution conta
ID: 498938 • Letter: M
Question
Map A Sapling Learning macmillan learning A 1.000-mL aliquot of a solution containing Cu and N 2+ s treated with 25.00 mL of a 0.03639 M EDTA solution. The solution is then back titrated with 0.02317 M Zn solution at a pH of 5. A volume of 19.28 mL of the Zn solution was needed to reach the xylenol orange end point. A 2.000 mL aliquot of the Cu and 2+ 2+ Ni solution is fed through an ion-exchange column that retains N 2+ The Cu that passed through the column is treated with 25.00 mL 0.03639 M EDTA. This solution required 24.42 mL of 0.02317 M Zn for 2+ back titration. The Ni extracted from the column was treated witn 25.00 mL of 0.03639 M EDTA. How many m ters of 0.02317 M Zn s required for the back titration of the Ni solution? 2+ 2+ Number m LExplanation / Answer
1 ml aliquot (Cu2+ + Ni2+) reacted with = 0.03639 M x 25 ml = 0.90975 mmol EDTA
Total (Cu2+ + Ni2+) present = 0.90975 - 0.02317 M x 19.28 ml = 0.46303 mmol
2 ml aliquot passed through ion-exchange column
total Cu2+ present in 1 ml = 0.46303 - (0.90975 - 0.02317 M x 24.42 ml)/2 = 0.29106 mmol
Ni2+ content in 1 ml = 0.46303 - 0.29106 = 0.17197 mmol
This would react with 2 x 0.17197 = 0.34394 mmol of EDTA
So ml of Zn2+ needed for back titration = (0.90975 - 0.34394)mmol/0.02317 M
= 24.42 ml