A 1.00-mL aliquot of a solution containing Cu^2+ and Ni^2* is treated with 25.00
ID: 1056426 • Letter: A
Question
A 1.00-mL aliquot of a solution containing Cu^2+ and Ni^2* is treated with 25.00 mL of a 0.04415 M EDTA solution. The solution is then back titrated with 0.02366 M Zn^2+ solution at a pH of 5. A volume of 15.26 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 0.04415 M EDTA. This solution required 16.97 mL of 0.02366 M Zn^2+ for back titration. The Ni^2+ extracted from the column was treated witn 25.00 mL of 0.04415 M EDTA. How many milliliters of 0.02366 M Zn^2+ is required for the back titration of the Ni^2+ solution?Explanation / Answer
with 1 ml aliquot
1 ml sample (Cu2+ + Ni2+) EDTA added = 0.04415 M x 25 ml = 1.10375 mmol
Total (Cu2+ + Ni2+) in aliquot = 1.10375 - (0.02366 M x 15.26 ml) = 0.7427 mmol
Total (Cu2+ + Ni2+) in 2 ml aliquot = 1.4854 mmol
2 ml aliquot
Amount of Cu2+ = 1.10375 - (0.02366 M x 16.97 ml) = 0.70225 mmol
Amount of Ni2+ present = 1.4854 - 0.70225 = 0.78315 mmol
Excess EDTA = 1.10375 - 0.78315 = 0.3206 mmol
Volume of 0.02366 M Zn2+ required for Ni2+ back titration = 0.3206 mmol/0.02366 M = 13.55 ml
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