Question

A 25.0 mL aliquot of 0.0440 M EDTA was added to a 49.0 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 0.0340 M Ga3+ solution until all of the EDTA reacted, requiring 11.0 mL of the Ga3+ solution. What was the original concentration of the V3+ solution?

Answer 1

concentration EDTA = 0.0440 M

volume EDTA = 25.0 mL

total moles EDTA = (concentration EDTA) * (volume EDTA)

total moles EDTA = (0.0440 M) * (25.0 mL)

total moles EDTA = 1.10 mmol

moles Ga³⁺ added = (concentration Ga³⁺) * (volume Ga³⁺)

moles Ga³⁺ added = (0.0340 M) * (11.0 mL)

moles Ga³⁺ added = 0.374 mmol

moles EDTA consumed by Ga³⁺ = moles Ga³⁺ added

moles EDTA consumed by Ga³⁺ = 0.374 mmol

moles EDTA consumed by V³⁺ = (total moles EDTA) - (moles EDTA consumed by Ga³⁺)

moles EDTA consumed by V³⁺ = (1.10 mmol) - (0.374 mmol)

moles EDTA consumed by V³⁺ = 0.726 mmol

moles V³⁺ present = moles EDTA consumed by V³⁺

moles V³⁺ present = 0.726 mmol

concentration V³⁺ = (moles V³⁺ present) / (volume V³⁺)

concentration V³⁺ = (0.726 mmol) / (49.0 mL)

concentration V³⁺ = 0.0148 M