A 1.000-mL aliquot of a solution containing Cu^2+ and Ni^2+ is treated with 25.0
ID: 1068459 • Letter: A
Question
A 1.000-mL aliquot of a solution containing Cu^2+ and Ni^2+ is treated with 25.00 mL of a 0.04893 M EDTA solution. The solution is then back titrated with 0.02281 M Zn^2+ solution at a pH of 5. A volume of 17.97 mL of the Zn^2+ solution was needed to reach the xylenol orange end point. A 2.000-mL aliquot of the Cu^2+ and Ni^2+ solution is fed through an ion-exchange column that retains Ni^2+. The Cu^2+ that passed through the column is treated with 25.00 mL 0.04893 M EDTA. This solution required 15.91 mL of 0.02281 M Zn^2+ for back titration. The Ni^2+ extracted from the column was treated within 25.00 mL of 0.04893 M EDTA. How many milliliters of 0.02281 M Zn^2+ is required for the back titration of the Ni^2+ solution?Explanation / Answer
1 ml of solution,
Total EDTA added = 0.04893 M x 25 ml = 1.22325 mmol
exces EDTA = 0.02281 M x 17.97 ml = 0.4099 mmol
Total Cu2+ + Ni2+ present in solution = 0.81335 mmol
2 ml aliquot
Total Cu2+ present = 1.22325 - 0.02281 M x 15.91 ml = 0.860343 mmol
Ni2+ in 2 ml sample would be = 2 x 0.81335 - 0.860343 = 0.76636 mmol
moles of EDTA would be left after reaction = 1.22325 - 0.76636 = 0.45689 mmol
Volume of Zn2+ required for back titration of this EDTA = 0.45689/0.02281
= 20.03 ml
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