A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL
ID: 500156 • Letter: A
Question
A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of a 0.05026 M EDTA solution. The solution is then back titrated with 0.02253 M Zn2 solution at a pH of 5. A volume of 20.26 mL of the Zn2 solution was needed to reach the xylenol orange end point. A 2.000-mL aliquot of the Cu2 and Ni2 solution is fed through an ion-exchange column that retains Ni2 . The Cu2 that passed through the column is treated with 25.00 mL 0.05026 M EDTA. This solution required 22.39 mL of 0.02253 M Zn2 for back titration. The Ni2 extracted from the column was treated witn 25.00 mL of 0.05026 M EDTA. How many milliliters of 0.02253 M Zn2 is required for the back titration of the Ni2 solution?
Explanation / Answer
mmol of EDTA = 25 x 0.05026 = 1.2565
mmol of Zn+2 = 20.26 x 0.02253 = 0.45646
mmol Cu+2 + Ni+2 = 1.2565 - 0.45646 = 0.8000
mmol of EDTA = 1.2565
mmol Zn+2 = 22.39 x 0.02253 = 0.5044
mmol of Cu+2 in 2 mL = 1.2565 - 0.5044 = 0.7521
mmol of Ni+2 = 2 x 0.8000 - 0.7521 = 0.8479
mmol of EDTA reamins = 1.2565 - 0.8479 = 0.4086
mmol of EDTA = mmol Zn+2 = 0.4086
volume = 0.4086 / 0.02253 = 18.14 mL
volume of Zn+2 = 18.14 mL
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