A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL
ID: 1997724 • Letter: A
Question
A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of a 0.03993 M EDTA solution. The solution is then back titrated with 0.02094 M Zn2 solution at a pH of 5. A volume of 17.18 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 0.03993 M EDTA. This solution required 17.53 mL of 0.02094 M Zn2 for back titration. The Ni2 extracted from the column was treated witn 25.00 mL of 0.03993 M EDTA. How many milliliters of 0.02094 M Zn2 is required for the back titration of the Ni2 solution?
Explanation / Answer
Step 1 :
Total moles of EDTA used for Cu2 and Ni2 titration = 0.03993 x 0.025 = 9.9825 x 10^-4 mols
moles of Zn2 reacted with excess EDTA = 0.02094 x 0.01718 = 3.59 x 10^-4 mols
Actual moles of EDTA reacted with Cu2 and Ni2 = ( 9.9825 x 10^-3) - (3.59 x 10^-4) = 6.385 x 10^-4 mols
Concentration of Cu2 and Ni2 in original solution = 6.385 x 10^-4 / 0.001 = 0.6385 M
Step 2 :
Total moles of EDTA used ed for Cu2 = 0.03993 x 0.025 = 9.9825 x 10^-4 mols
moles of Zn2 required for excess EDTA = 0.02094 x 0.01753 = 3.67 x 10^-4 mols
Actual moles of EDTA used for Cu2 = 6.312 x 10^-4 mols
molarity of Cu2 in 2 ml solution = 6.312 x 10^-4 / 0.002 = 0.3156 M
Step 3 :
Molarity of Ni2 in solution = 0.6385 - 0.3156 = 0.323 M
Step 4 :
EDTA used for Ni2 titration = 0.03993 x 0.025 = 9.9825 x 10^-4 mols
moles of Ni2 in solution = 0.323 x 0.002 = 6.46 x 10^-4 mols
moles of Excess EDTA = 9.9825 x 10^-4 - 6.46 x 10^-4 = 3.52 x 10^-4 mols
Volume of Zn2 required to titrate excess EDTA = 3.52 x 10^-4/0.02094 = 0.01683 L = 16.83 ml
Thus we need 16.83 ml of 0.02094 M Zn2 for back titration.
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