Question

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?

Answer 1

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.

Answer 2

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.