A 1.000-mL aliquot of a solution containing Cu and Ni2 is treated with 25.00 mL
ID: 557984 • Letter: A
Question
A 1.000-mL aliquot of a solution containing Cu and Ni2 is treated with 25.00 mL of a 0.03679 M EDTA solution. The solution is then back titrated with 0.02166 M Zn2 solution at a pH of 5. A volume of 18.62 mL of the Zn2 solution was needed to reach the xylenol orange end point. A 2.000-mL aliquot of the Cu2 and Ni solution is fed through an ion-exchange column that retains Ni2*. The Cu2 that passed through the column is treated with 25.00 mL 0.03679 M EDTA. This solution required 18.67 mL of 0.02166 M Zn2* for back titration. The Ni2+ extracted from the column was treated witn 25.00 mL of 0.03679 M EDTA 2+ Number mLExplanation / Answer
mmol of EDTA = 25 x 0.03679 = 0.9198
mmol of Zn+2 = 18.62 x 0.02166 = 0.4033
mmol Cu+2 + Ni+2 = 0.9198 -0.4033 = 0.5165
mmol of EDTA = 0.9198
mmol Zn+2 = 18.67 x 0.02166 = 0.4044
mmol of Cu+2 in 2 mL = 0.9198 - 0.4044 = 0.5154
mmol of Ni+2 = 2 x 0.5165 - 0.5154 = 0.5176
mmol of EDTA reamins = 0.9198 - 0.5176 = 0.4022
mmol of EDTA = mmol Zn+2 =0.4022
volume = 0.4022 / 0.02166 = 18.57 mL
volume of Zn+2 = 18.57 mL
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