contestada

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?

Respuesta :

Answer:

[tex]0.0151~M[/tex]

Explanation:

The reaction between [tex]V^+^3[/tex] and EDTA would be:

[tex]V^+^3~+~HY^-^3~=~VY^-~+~HY^-^3[/tex]

The EDTA in excess ([tex]HY^-^3[/tex]) would react with the [tex]Ga^+^3[/tex]:

[tex]Ga^+^3~+~HY^-^3~=~GaY^-~+~H^+[/tex]

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

Total moles of EDTA:

[tex]mol~EDTA=~0.025L*0.043M=~0.00107~mol~EDTA[/tex]

Moles that react with [tex]Ga^+^3[/tex]:

[tex]mol~EDTA=~0.011L*0.04M=~0.00044~mol~EDTA[/tex]

Moles that react with [tex]V^+^3[/tex]:

[tex]mol~EDTA=~0.00107~mol~-0.00044~mol=0.000635[/tex]

If we have a 1:1 ratio:

[tex]0.000635~mol~EDTA=0.000635~mol~V^+^3[/tex]

Now we can calculate the concentration:

[tex]M~=~\frac{0.000635~mol}{0.042~L}=~0.0151~M[/tex].

Otras preguntas