Question

A 25.0 mL aliquot of 0.0430 M EDTA was added to a 42.0 mL solution containing an unknown concentration of V3+. All of the V+ present in the solution formed a complex with EDTA, leaving an excess of EDTA in solution. This solution was back- titrated with a 0.0400 M Ga3+ solution until all of the EDTA reacted, requiring ії.0 mL of the Ga3+ solution. What was the original concentration of the Vs+solution?

Answer 1

`0.0151~M`

Step-by-step explanation:

The reaction between
`V^+^3` and EDTA would be:

`V^+^3~+~HY^-^3~=~VY^-~+~HY^-^3`

The EDTA in excess (
`HY^-^3`) would react with the
`Ga^+^3`:

`Ga^+^3~+~HY^-^3~=~GaY^-~+~H^+`

All the reactions have a 1:1 ratio. So:

Total moles of EDTA:

`mol~EDTA=~0.025L*0.043M=~0.00107~mol~EDTA`

Moles that react with
`Ga^+^3`:

`mol~EDTA=~0.011L*0.04M=~0.00044~mol~EDTA`

Moles that react with
`V^+^3`:

`mol~EDTA=~0.00107~mol~-0.00044~mol=0.000635`

If we have a 1:1 ratio:

`0.000635~mol~EDTA=0.000635~mol~V^+^3`

Now we can calculate the concentration:

`M~=~(0.000635~mol)/(0.042~L)=~0.0151~M`.