Chemistry question on EDTA titrations

Question

The Ni2+ and Zn²+ in a 25.00mL sample were complexed by addition of 50.00mL of 0.0125M EDTA, and the excess chelating agent was titrated with 7.50 mL of 0.0104 M Mg²⁺ solution. Introduction of an excess of 2,3-dimercapto-1-propanol displaced the zinc from its EDTA complex. Titration of the liberated EDTA required an additional 21.50mL of the 0.0104M Mg? solution. Calculate the concentration, in mol/L, of Ni2+ and Zn* in the original sample.

Answer 1

The divalent ions react on a 1:1 molar basis with EDTA.

By adding the 50 mL of 0.0125 M EDTA, all of the Zn2+ and Ni2+ was complexed leaving excess EDTA.

The titration of the excess EDTA with 7.50 mL of 0.0104 M Mg2+ will give you the moles of the excess EDTA and by subtraction from 50 mL of 0.0125M EDTA, the total moles of Zn2+ and Ni2+ can be determined.

The addition of excess 2,3 dimercapto-1-propanol liberates the Zn2+ from its EDTA complex. Titration of the liberated Zn2+ gives the total moles of Zn2+ present.

As we have already determined the total moles of Zn2+ and Ni2+, the moles of Ni2+ can be calculated by difference. At this point you have enough information to determine the molarity of each ion in the sample.