A 1.000-mL aliquot of a solution containing Cu and Ni2* is treated with 25.00 mL
ID: 1083650 • Letter: A
Question
A 1.000-mL aliquot of a solution containing Cu and Ni2* is treated with 25.00 mL of a 0.05541 M EDTA solution. The solution is then back titrated with 0.02070 M Zn2 solution at a pH of 5. A volume of 22.24 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 0.05541 M EDTA. This solution required 16.34 mL of 0.02070 M Zn2t for back titration. The Ni2+ extracted from the column was treated witn 25.00 mL of 0.05541 M EDTA. How many milliliters of 0.02070 M Zn2* is required for the back titration of the Ni2 solution? Number mLExplanation / Answer
mmoles of EDTA = 25 x 0.05541 = 1.385
mmoles of Zn+2 = 22.24 x 0.02070 = 0.46037
mmoles of Cu+2 + Ni+2 = 0.9246
for 2.000 mL unknown :
mmoles of EDTA = 25 x 0.05541 = 1.385
mmoles off Zn+2 = 16.34 x 0.02070 = 0.3382
mmoles of Cu+2 in 2.000 mL = 1.0468
Ni+2 in 2.000 mL = 2 x 0.9246 - 1.0468 = 0.8024
mmoles of EDTA = 1.385 - 0.8024 = 0.5826
volume of Zn+2 = 0.5826 / 0.02070 = 28.14
volume of Zn+2 = 28.14 mL
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