Ammonium vanadate on theating with oxalic acid forms a compound `(Z)`. A sample of `(Z)` was titrated with `1 M KMnO_(4)` solution in hot acidic medium. The resulting solution was reduced with `SO_(2)`, the excess of `SO_(2)` is boiled out and the solution was again titrated with `1 M KMnO_(4)`. The volume ratio of `KMnO_(4)` used in two titrations was `5:1`. Given that `KMnO_(4)` oxidised all oxidation states of vanadium to `+5` and `SO_(2)` reduced to `+4`. Find the oxidation state of `V` in Z`.

Answer

0

Explanation

`VO_(4)^(3-) underset(H_(2)C_(2)O_(4))overset(By)rarrV^(x+)`
or `V^(5+) + (5-x)e rarr V^(x)`
Now `V^(x+)` is oxidised by `KMnO_(4)` to `V^(5+)`.
`V^(x+) rarr V^(5+)+(5-x)`
The `V^(5+)` formed are reduced to `V^(4+)` by `SO_(2)`
`{:(V^(5+),+e,rarr,V^(4+),),(,S^(4+),rarr,S^(6+),+2e):}`
The `V^(4+)` formed are then titrated against `KMnO_(4)` to give
`{:(,V^(4+),rarr,+e,),(Mn^(7+) ,+5e,rarr,Mn^(2+),):}`
Let the sample constains a millie-mole of `VO_(4)^(3-)` and `V_(1) mL` of `KMnO_(4)` are used.
`"Meq. of" V_(3-)^(4) = "Meq. of" V^(x+)="Meq. of"KMnO_(4)`
`a xx (5-x) = 1 xx 5 xx V_(1)` ...(1)
Now In IInd titration,
`Meq. of" V^(4+) = "Meq. of" KMnO_(4)`
`a xx 1 = 1 xx 5 xx V_(2)` ...(2)
By eq. (1) and (2),
`:. (5-x)/(1) = (V_(1))/(V_(2)) = (5)/(1)`
`x = 0`