A 1.000-mL aliquot of a solution containing Cu^2+ and Ni^2+ is treated with 25.0
ID: 1051118 • 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 0.03235 M EDTA solution. The solution is then back titrated with 0.02021 M Zn^2+ solution at a pH of 5. A volume of 16.22 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 0.03235 M EDTA. This solution required 17.67 mL of 0.02021 M Zn^2+ for back titration. The Ni^2+ extracted from the column was treated with 25.00 mL of 0.03235 M EDTA. How many milliliters of 0.02021 M Zn^2+ is required for the back titration of the Ni^2+ solution?Explanation / Answer
mmol of EDTA = 25 x 0.03235 = 0.80875
mmol of Zn+2 = 16.22 x 0.02021 = 0.3278
mmol Cu+2 + Ni+2 = 0.80875 - 0.3278 = 0.4809
mmol of EDTA = 0.80875
mmol Zn+2 = 17.67 x 0.02021 = 0.3571
mmol of Cu+2 in 2 mL = 0.80875 - 0.35711 = 0.4516
mmol of Ni+2 = 2 x 0.4809 - 0.4516 = 0.5102
mmol of EDTA reamins = 0.80875 - 0.5102 = 0.29855
mmol of EDTA = mmol Zn+2 = 0.29855
volume = 0.29855 / 0.02021 = 14.77 mL
volume of Zn+2 = 14.77 mL
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