A 1.000 mL aliquot of a solution containing Cu²+ and Ni²+ is treated with 25.00 mL of a 0.04333 M EDTA solution. The solution is then back titrated with 0.02071 M Zn²+ solution at a pH of 5. A volume of 20.01 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 of 0.04333 M EDTA. This solution required 18.20 mL of 0.02071 M Zn²+ for back titration. The Ni²+ extracted from the column was treated witn 25.00 mL of 0.04333 M EDTA. How many milliliters of 0.02071 M Zn²+ is required for the back titration of the Ni²+ solution? volume: mL

Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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In this experiment, a 1.000 mL aliquot of a solution containing Cu²⁺ and Ni²⁺ ions is treated with 25.00 mL of 0.04333 M EDTA solution. The remaining solution is back titrated with a 0.02071 M Zn²⁺ solution at a pH of 5, requiring 20.01 mL of the Zn²⁺ solution to reach the xylend orange end point.

Further, a 2.000 mL aliquot of the Cu²⁺ and Ni²⁺ solution is processed through an ion-exchange column that captures Ni²⁺ ions. The Cu²⁺ ions that pass through are treated with 25.00 mL of 0.04333 M EDTA, and the solution requires 18.20 mL of 0.02071 M Zn²⁺ for back titration. The Ni²⁺ ions extracted from the column are treated with 25.00 mL of 0.04333 M EDTA.

The problem requires calculating the volume of 0.02071 M Zn²⁺ needed for the back titration of the Ni²⁺ solution.

**Question:**
How many milliliters of 0.02071 M Zn²⁺ are required for the back titration of the Ni²⁺ solution?

You are to find the solution using the text's information for educational analysis. The answer should be input in the space provided in mL.
Transcribed Image Text:In this experiment, a 1.000 mL aliquot of a solution containing Cu²⁺ and Ni²⁺ ions is treated with 25.00 mL of 0.04333 M EDTA solution. The remaining solution is back titrated with a 0.02071 M Zn²⁺ solution at a pH of 5, requiring 20.01 mL of the Zn²⁺ solution to reach the xylend orange end point. Further, a 2.000 mL aliquot of the Cu²⁺ and Ni²⁺ solution is processed through an ion-exchange column that captures Ni²⁺ ions. The Cu²⁺ ions that pass through are treated with 25.00 mL of 0.04333 M EDTA, and the solution requires 18.20 mL of 0.02071 M Zn²⁺ for back titration. The Ni²⁺ ions extracted from the column are treated with 25.00 mL of 0.04333 M EDTA. The problem requires calculating the volume of 0.02071 M Zn²⁺ needed for the back titration of the Ni²⁺ solution. **Question:** How many milliliters of 0.02071 M Zn²⁺ are required for the back titration of the Ni²⁺ solution? You are to find the solution using the text's information for educational analysis. The answer should be input in the space provided in mL.
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