mL aliquot of a solution containing Cu and N2* is treated with 25.00 mL of a 0.0
ID: 553877 • Letter: M
Question
mL aliquot of a solution containing Cu and N2* is treated with 25.00 mL of a 0.05640 M EDTA solution. The solution is then back titrated with 0.02389 M Zn2 solution at a pH of 5. A volume of 21.36 mL of the Zn2 solution was needed to reach the x Ni" solution is fed through an ion-exchange column that retains N*. The Cu?" that passed through the ylenol orange end point. A 2.000-mL aliquot of the Cu? and column is treated with 25.00 mL 0.05640 M EDTA. This solution required 21.90 mL of 0.02389 M Zn2 for back titration. The Ni2 extracted from the column was treated witn 25.00 mL of 0.05640 M EDTA How many mililiters of 0.02389 M Zn2 is required for the back titration of the Ni2 solution? Number mLExplanation / Answer
mmol of EDTA = 25 x 0.05640 = 1.41
mmol of Zn+2 = 21.36 x 0.02389 = 0.5103
mmol Cu+2 + Ni+2 = 1.41 - 0.5103 = 0.8997
mmol of EDTA = 1.41
mmol Zn+2 = 21.90 x 0.02030 = 0.5232
mmol of Cu+2 in 2 mL = 1.41 - 0.5232 = 0.8868
mmol of Ni+2 = 2 x 0.8997 - 0.8868 = 0.9126
mmol of EDTA reamins = 1.41 - 0.9126 = 0.4974
mmol of EDTA = mmol Zn+2 = 0.4974
volume = 0.4974 / 0.02389 = 20.82 mL
volume of Zn+2 = 20.82 mL
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.