A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL
ID: 576793 • Letter: A
Question
A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of a 0.04463 M EDTA solution. The solution is then back titrated with 0.02113 M Zn2 solution at a pH of 5. A volume of 20.72 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 0.04463 M EDTA. This solution required 23.15 mL of 0.02113 M Zn2 for back titration. The Ni2 extracted from the column was treated witn 25.00 mL of 0.04463 M EDTA.
A 1.000-mL aliquot of a solution containing Cu2 and Ni2+ is treated with 25.00 mL of a 0.04463 M EDTA solution. The solution is then back titrated with 0.02113 M Zn2 solution at a pH of 5. A volume of 20.72 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 Ni2t. The Cu2+ that passed through the column is treated with 25.00 mL 0.04463 M EDTA. This solution required 23.15 mL of 0.02113 M Zn2+ for back titration. The Ni2+ extracted from the column was treated witn 25.00 mL of 0.04463 M EDTA How many millters of 0.02113 M Zn2* is required for the back titration of the Ni+ solution? Number mLExplanation / Answer
mmol of EDTA = 25 x 0.04463 = 1.11575
mmol of Zn+2 = 20.72 x 0.02113 = 0.4378
mmol Cu+2 + Ni+2 = 1.11575 - 0.4378 = 0.6779
mmol of EDTA =1.11575
mmol Zn+2 = 23.15 x 0.02113 = 0.48916
mmol of Cu+2 in 2 mL = 1.11575 - 0.48916 = 0.62659
mmol of Ni+2 = 2 x 0.6779 - 0.62659 = 0.72921
mmol of EDTA reamins = 1.11575 - 0.72921 = 0.38654
mmol of EDTA = mmol Zn+2 = 0.38654
volume = 0.38654 / 0.02113 = 18.29 mL
volume of Zn+2 = 18.29 mL
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