Question

A 25.0 mL aliquot of 0.05600.0560 M EDTAEDTA was added to a 40.040.0 mL solution containing an unknown concentration of V3+V3+. All of the V3+V3+ present in the solution formed a complex with EDTAEDTA, leaving an excess of EDTAEDTA in solution. This solution was back-titrated with a 0.03300.0330 M Ga3+Ga3+ solution until all of the EDTAEDTA reacted, requiring 10.010.0 mL of the Ga3+Ga3+ solution. What was the original concentration of the V3+V3+ solution?

Answer 1

Concentration of EDTA = 0.0560 M

volume EDTA solution = 25.0 mL

Total moles EDTA = (concentration of EDTA) * (volume EDTA solution)

Total moles EDTA = (0.0560 M) * (25.0 mL)

Total moles EDTA = 1.40 mmol

Concentration Ga³⁺ = 0.0330 M

volume Ga³⁺ solution = 10.0 mL

moles Ga³⁺ = (Concentration Ga³⁺) * (volume Ga³⁺ solution)

moles Ga³⁺ = (0.0330 M) * (10.0 mL)

moles Ga³⁺ = 0.330 mmol

moles consumed by Ga³⁺ = moles Ga³⁺

moles consumed by Ga³⁺ = 0.330 mmol

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

moles consumed by V³⁺ = (1.40 mmol) - (0.330 mmol)

moles consumed by V³⁺ = 1.07 mmol

moles V³⁺ present = moles consumed by V³⁺

moles V³⁺ present = 1.07 mmol

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

original concentration V³⁺ = (1.07 mmol) / (40.0 mL)

original concentration V³⁺ = 0.0268 M