A 1 .000 - mL aliquot of a solution containing Cu^2+ and Ni^2+ is treated with 2
ID: 1058776 • 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 0.03998 M EDTA solution. The solution is then back titrated with 0.02014 M Zn^2+ solution at a pH of 5. A volume of 19.38 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 0.03998 M EDTA. This solution required 22.82 mL of 0.02014 M Zn^2+ for back titration. The Ni^2+ extracted from the column was treated with 25.00 mL of 0.03998 M EDTA. How many milliliters of 0.02014 M Zn^2+ is required for the back titration of the Ni^2+ solution?Explanation / Answer
mmol of EDTA = 25 x 0.03998 = 0.9995
mmol of Zn+2 = 19.38 x 0.02014 = 0.3903
mmol Cu+2 + Ni+2 = 0.9995 -0.3903 = 0.6092
mmol of EDTA = 0.9995
mmol Zn+2 = 22.82 x 0.02014 = 0.4596
mmol of Cu+2 in 2 mL = 0.9995 -0.4596 = 0.5399
mmol of Ni+2 = 2 x 0.6092 - 0.5399 = 0.6785
mmol of EDTA reamins = 0.995 - 0.6785 = 0.3210
mmol of EDTA = mmol Zn+2 = 0.3210
volume = 0.3210 / 0.02014 = 15.94 mL
volume of Zn+2 = 15.94 mL
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