A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL
ID: 556997 • Letter: A
Question
A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of a 0.03575 M EDTA solution. The solution is then back titrated with 0.02155 M Zn2 solution at a pH of 5. A volume of 22.15 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 0.03575 M EDTA. This solution required 23.72 mL of 0.02155 M Zn2 for back titration. The Ni2 extracted from the column was treated witn 25.00 mL of 0.03575 M EDTA. How many milliliters of 0.02155 M Zn2 is required for the back titration of the Ni2 solution?
A 1.000-mL aliquot of a solution containing Cu and Ni2" is treated with 25.00 mL of a o.03575 M EDTA solution. The solution is then back titrated with 0.02155 M Zn2 solution at a pH of 5. A volume of 22.15 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 0.03575 M EDTA. This solution required 23.72 mL of 0.02155 M Zn2+ for back titration. The Ni2+ extracted from the column was treated witn 25.00 mL of 0.03575 M EDTA. 2+ Number mLExplanation / Answer
Let EDTA be Y4-; it complexes 1:1 with M2+ ions:
M2+ + Y4- -> [MY]2- (M2+ = Co2+, Ni2+)
1st Step What was the number of mmol of Y4- added at the beginning:
MV = 25.00 x 0.03575 = 0.89375 mmol
How many mmol not reacted (back-titrated) = 22.15 x 0.02155 = 0.47733
Therefore mmol consumed in first titration = 0.89375 - 0.47733 = 0.41642 mmol = [Co2+] + [Ni2+] note this is in 1.000 mL
2nd Step Titration of Ni2+; mmol of Y4- added = 25.00 x 0.03575 = 0.89375 mmol
mmol of Y4- not reacted = 23.72 x 0.02155 = 0.51116 mmol
Therefore mmol of Ni2+ = 0.89375 - 0.51116 = 0.38259 mmol
From Step 1 in 2.000 mL of the Cu2+/Ni2+ solution there will be 2 x 0.41642 = 0.83284 mmol
Combining results from Step 1 and 2 we have the [Cu2+] = 0.83284 - 0.38259 = 0.45025 mmol
3rd Step mmol of Y4- added 25.00 x 0.03575 = 0.89375 mmol
Therefore Y4- not consumed = 0.89375 - 0.45025 = 0.4435 mmol
This will require 0.4435/0.02155 = 20.58 mL for the back titration
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.