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

Answer:

$0.0151~M$

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:

$M~=~\frac{0.000635~mol}{0.042~L}=~0.0151~M$.