Question

A 25.0 mL aliquot of 0.0430 0.0430 M EDTA EDTA was added to a 56.0 56.0 mL solution containing an unknown concentration of V 3 + V3+ . All of the V 3 + V3+ present in the solution formed a complex with EDTA EDTA , leaving an excess of EDTA EDTA in solution. This solution was back-titrated with a 0.0490 0.0490 M Ga 3 + Ga3+ solution until all of the EDTA EDTA reacted, requiring 13.0 13.0 mL of the Ga 3 + Ga3+ solution. What was the original concentration of the V 3 + V3+ solution?

Answer 1

Check the explanation

Step-by-step explanation:

As we know the reaction of EDTA and
`Ga^3`+ and EDTA and
`V^3`+

Let us say that the ratio is 1:1

Therefore, the number of moles of
`Ga^3`+ = molarity * volume

= 0.0400M * 0.011L

= 0.00044 moles

Therefore excess EDTA moles = 0.00044 moles

Given , initial moles of EDTA = 0.0430 M * 0.025 L

= 0.001075

Therefore reacting moles of EDTA with
`V^3+` = 0.001075 - 0.00044 = 0.000675 moles

Let us say that the ratio between
`V^3+` and EDTA is 1:1

Therefore moles of
`V^3+` = 0.000675 moles

Molarity = moles / volume

= 0.000675 moles / 0.057 L

= 0.011 M (answer).