Map Learning Sapling macmillan learning A 1.000-mL aliquot of a solution contain
ID: 499031 • Letter: M
Question
Map Learning Sapling macmillan learning A 1.000-mL aliquot of a solution containing Cu and N 2+ s treated with 25.00 mL of a 0.03639 M EDTA solution. The solution is then back titrated with 0.02317 M Zn 2+ solution at a pH of 5. A volume of 19.28 mL of the Zn solution was needed to reach the xylenol orange end point. A 2.000-mL aliquot of the Cu and 2+ Ni2 solution is fed through an ion-exchange column that retains N 2+ The Cu that passed through the column is treated with 25.00 mL 0.03639 M EDTA. This solution required 24.42 mL of 0.02317 M Zn 2+ for back titration. The Ni2 extracted from the column was treated witn 25.00 mL of 0.03639 M EDTA. 2+ How many m ters of 0.02317 M Zn s required for the back titration of the Ni solution? Number 24.42 mLExplanation / Answer
mmol of EDTA = 25 x 0.03639 = 0.9098
mmol of Zn+2 = 19.28 x 0.02317 = 0.4467
mmol Cu+2 + Ni+2 = 0.9098 - 0.4467 = 0.4631
mmol of EDTA = 0.9098
mmol Zn+2 = 24.42 x 0.02317 = 0.5658
mmol of Cu+2 in 2 mL = 0.9098 - 0.5658 = 0.344
mmol of Ni+2 = 2 x 0.4631 - 0.344 = 0.5822
mmol of EDTA reamins = 0.9098 - 0.5822 = 0.3276
mmol of EDTA = mmol Zn+2 = 0.3276
volume = 0.3276 / 0.02317 = 14.14mL
volume of Zn+2 = 14.14 mL