A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL
ID: 1047980 • Letter: A
Question
A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of a 0.05768 M EDTA solution. The solution is then back titrated with 0.02176 M Zn2 solution at a pH of 5. A volume of 21.12 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 0.05768 M EDTA. This solution required 19.29 mL of 0.02176 M Zn2 for back titration. The Ni2 extracted from the column was treated witn 25.00 mL of 0.05768 M EDTA.
How many milliliters of 0.02176 M Zn2 is required for the back titration of the Ni2 solution?
Explanation / Answer
For Cu2 + Ni2 analysis
Total EDTA added = 0.05768 M x 25 ml = 1.442 mmol
Total Zn2 added = 0.02176 M x 21.12 ml = 0.4596 mmol
Total Cu2 + Zn2 in 1 ml aliquot = 1.442 - 0.4596 = 0.9824 mmol
Total Cu2 + Ni2 in 2 ml aliquot = 1.9648 mmol
For Cu2 analysis
Total EDTA added = 0.05768 M x 25 ml = 1.442 mmol
Total Zn2 added = 0.02176 M x 19.29 ml = 0.41975 mmol
Total Cu2 in 2 ml aliquot = 1.442 - 0.41975 = 1.02225 mmol
Total Ni2 present in 2 ml aliquot = 1.9648 - 1.02225 = 0.94255 mmol
For Ni2 analysis
Total EDTA added = 0.05768 M x 25 ml = 1.442 mmol
Total Ni2 present in 2 ml aliquot = 0.94255 mmol = moles of Zn2
Excess EDTA = 1.442 - 0.94255 = 0.49945 mmol
Volume of Zn2 required = 0.49945 mmol/0.02176 M = 22.95 ml
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