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:
- 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:
- 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.