A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL
ID: 1032493 • Letter: A
Question
A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of a 0.03800 M EDTA solution. The solution is then back titrated with 0.02119 M Zn2 solution at a pH of 5. A volume of 15.44 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 0.03800 M EDTA. This solution required 15.86 mL of 0.02119 M Zn2 for back titration. The Ni2 extracted from the column was treated witn 25.00 mL of 0.03800 M EDTA. How many milliliters of 0.02119 M Zn2 is required for the back titration of the Ni2 solution?
Explanation / Answer
M1V1 =M2V2
M = Concentration
V= Volume
1.000 mL x M1 = 25.00 mL x 0.03800 M
M1 = 25.00 mL x 0.03800 M / 1.000 mL
M1 = 0.95 M
1.000 mL aliquot solution having concentration is 0.95 M.
Similarly for 2.000 mL aliquot solution containing Cu2 and Ni2 is
2.000 mL x M1 = 25.00 mL x 0.03800 M
M1 = 25.00 mL x 0.03800 M / 2.000 mL
M1 = 0.475 M
The concentration of 2.000 mL aliquot solution is 0.475 M
Then Cu2 is treated with 25.00 mL of a 0.03800 M EDTA solution. This solution required 15.86 mL of 0.02119 M Zn2 for back titration.
0.475 M x V1 = 15.86 mL x 0.02119 M
V1 = 15.86 mL x 0.02119 M / 0.475 M
V1 = 0.7075 mL
Total aliquot volume is 2.000 mL. Here 0.7075 mL Cu2 is used therefore remaining solutionis containing Ni2.
Volume of Ni2 is =2.000 mL - 0.7075 mL
= 1.2925 mL
The volume of Zn2 irequired is
0.475 M x 1.2925 mL = V2 x 0.02119 M
V2 = 0.475 M x 1.2925 mL / 0.02119 M
V2 = 28.97 mL
28.97 mL of 0.02119 M Zn2 is required for the back titration of the Ni2 solution.
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