A 1.000-mL aliquot of a solution containing Cu^2+ and Ni^2+ is treated with 25.0
ID: 496958 • 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 0.05171 M EDTA solution. The solution is then back titrated with 0.02002 M Zn^2+ solution at a pH of 5. A volume of 18.83 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 0.05171 M EDTA. This solution required 21.75 mL of 0.02002 M Zn^2+ for back titration. The Ni^2+ extracted from the column was treated with 25.00 mL of 0.05171 M EDTA. How many milliliters of 0.02002 M Zn^2+ is required for the back titration of the Ni^2+ solution?Explanation / Answer
mmol of EDTA = 25 x 0.05171 = 1.293
mmol of Zn+2 = 18.83 x 0.02002= 0.3770
mmol Cu+2 + Ni+2 = 1.293 -0.3770 = 0.9.160
mmol of EDTA = 1.293
mmol Zn+2 = 21.75 x 0.02002 = 0.4354
mmol of Cu+2 in 2 mL = 1.293 - 0.4354= 0.8576
mmol of Ni+2 = 2 x 0.9.160 - 0.8576 = 0.9744
mmol of EDTA reamins = 1.293 - 0.9744= 0.3186
mmol of EDTA = mmol Zn+2= 0.3186
volume = 0.3186 / 0.02002 = 15.91 mL
volume of Zn+2 = 15.91 mL
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