A 1.000-mL aliquot of a solution containing Cu2+ and Ni2+ is treated with 25.00
ID: 1047931 • Letter: A
Question
A 1.000-mL aliquot of a solution containing Cu2+ and Ni2+ is treated with 25.00 mL of a 0.05665 M EDTA solution. The solution is then back titrated with 0.02386 M Zn2 solution at a pH of 5. A volume of 15.22 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 0.05665 M EDTA. This solution required 23.77 mL of 0.02386 M Zn2+ for back titration. The Ni2+ extracted from the column was treated witn 25.00 mL of 0.05665 M EDTA. How many milliliters of 0.02386 M Zn2+ is required for the back titration of the Ni2+ solution?
Explanation / Answer
Total EDTA added (Cu2 + Ni2) = 0.05665 M x 25 ml = 1.41625 mmol
Total (Cu2 + Ni2) present in 1 ml sample = 1.41625 - 0.02386 M x 15.22 ml = 1.053101 mmol
For Cu2 analysis
Total EDTA added (Cu2 + Ni2) = 0.05665 M x 25 ml = 1.41625 mmol
Total Cu2 present in 1 ml sample = (1.41625 - 0.02386 M x 23.77 ml)/2 = 0.42455 mmol
Total Ni2 present in 1 ml sample = 1.053101 - 0.42455 = 0.62855 mmol
For Ni2 analysis
Total EDTA added (Cu2 + Ni2) = 0.05665 M x 25 ml = 1.41625 mmol
Total Ni2 present in 2 ml sample = 2 x 0.62855 = 1.2571 mmol
Total Cu2 remained = 1.41625 - 1.2571 = 0.15915 mmol
Volume of Zn2 needed for back titration = 0.15915 mmol/0.02386 M = 6.67 ml
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