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A 1.000 mL aliquot of a solution containing Cu2+ and Ni2+ is treated with 25.00 mL of a 0.03893 M EDTA solution. The solution is then back titrated with 0.02091 M Zn2+ solution at a pH of 5. A volume of 19.17 mL of the Zn2+ solution was needed to reach the xylenol orange endpoint. 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 of 0.03893 M EDTA. This solution required 21.40 mL of 0.02091 M Zn2+ for back titration. The Ni2+ extracted from the column was treated with 25.00 mL of 0.03893 M EDTAEDTA. How many milliliters of 0.02091 M Zn2+ is required for the back titration of the Ni2+ solution?

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Final answer:

The calculation for the required volume of Zn2+ solution for back titration involves subtracting the moles of EDTA reacted with Cu2+ from the total moles of EDTA added and then using the concentration of the Zn2+ solution to find the needed volume for the remaining EDTA.

Step-by-step explanation:

The student's question involves a series of complexation titrations with EDTA and back titrations with a Zn2+ solution to determine the concentrations of Cu2+ and Ni2+ in a solution. To calculate the volume of Zn2+ required for the back titration of the Ni2+ solution, we need to account for the initial moles of EDTA used and the moles of EDTA that reacted in prior titrations.

The volume of Zn2+ needed is calculated by subtracting the moles of EDTA reacted with Cu2+ (which can be determined from the back titration with Cu2+) from the total moles of EDTA initially added. This will give the moles of EDTA available to react with Ni2+. Since the Zn2+ solution is used to back titrate the remaining EDTA, we can then calculate the required volume based on its concentration and the moles of EDTA to be back titrated.

User Manuel Bernhardt
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5 votes

Final answer:

To determine the volume of Zn2+ solution needed for the back titration of the Ni2+ solution, calculate the number of moles of Ni2+ present and use stoichiometry. The volume of 0.02091 M Zn2+ solution required is 21.35 mL.

Step-by-step explanation:

To determine the amount of Zn2+ solution required for the back titration of the Ni2+ solution, we need to calculate the number of moles of Ni2+ present in the solution and then use the stoichiometry of the reaction to determine the volume of Zn2+ solution needed.

Given that 21.40 mL of a 0.02091 M Zn2+ solution is needed to titrate the Cu2+ solution, we can calculate the number of moles of Zn2+ used:

  1. mol Zn2+ = (0.02091 mol/L)(0.02140 L) = 0.000446 mol Zn2+

Since the reaction between Zn2+ and Ni2+ is in a 1:1 ratio, the number of moles of Ni2+ present in the solution is also 0.000446 mol. To determine the volume of Zn2+ solution needed:

  1. volume of Zn2+ solution = (0.000446 mol)(1 L/0.02091 mol) = 0.02135 L = 21.35 mL

Therefore, 21.35 mL of a 0.02091 M Zn2+ solution is required for the back titration of the Ni2+ solution.

User John Mangual
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