Question

A 25.0 mL aliquot of 0.0630 M EDTA was added to a 30.0 mL solution containing an unknown concentration of V3+. All of the V3+ present in the solution formed a complex with EDTA, leaving an excess of EDTA in solution. This solution was back-titrated with a 0.0360 M Ga3+ solution until all of the EDTA reacted, requiring 13.0 mL of the Ga3+ solution. What was the original concentration of the V3+ solution?

Answer 1

0.0369M of V³⁺

Step-by-step explanation:

The moles of EDTA added to the solution containing V³⁺ are:

0.025L ₓ (0.0630mol / L) = 0.001575 moles of EDTA.

The excess of EDTA was titrated with Ga³⁺. Moles required of Ga³⁺ = Moles in excess of EDTA are:

0.013L ₓ (0.0360mol / L) = 0.000468 moles.

That means initial moles of EDTA that react with V³⁺ solution are:

0.001575 moles - 0.000468 moles = 0.001107 moles of EDTA = moles of V³⁺

As volume of the solution was 30.0mL = 0.0300L, the concentration of the solution of V³⁺ is:

0.001107 moles V³⁺ / 0.0300L = 0.0369M of V³⁺