A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL
ID: 590328 • Letter: A
Question
A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of a 0.05350 M EDTA solution. The solution is then back titrated with 0.02385 M Zn2 solution at a pH of 5. A volume of 22.00 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 0.05350 M EDTA. This solution required 15.92 mL of 0.02385 M Zn2 for back titration. The Ni2 extracted from the column was treated witn 25.00 mL of 0.05350 M EDTA.
How many milliliters of 0.02385 M Zn2 is required for the back titration of the Ni2 solution?
Explanation / Answer
mmol of EDTA = 25 x 0.05350 = 1.3375
mmol of Zn+2 = 22.00 x 0.02385 = 0.5247
mmol Cu+2 + Ni+2 = 1.3375 - 0.5247 = 0.8128
mmol of EDTA = 1.3375
mmol Zn+2 = 15.92 x 0.02385 = 0.37969
mmol of Cu+2 in 2 mL = 1.3375 - 0.37969 = 0.9578
mmol of Ni+2 = 2 x 0.8128 - 0.9578 = 0.6678
mmol of EDTA reamins = 1.3375 - 0.6678 = 0.6697
mmol of EDTA = mmol Zn+2 = 0.6697
volume = 0.6697 / 0.02385 = 28.08 mL
volume of Zn+2 = 28.08 mL
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.