20 " mL of " `(M)/(60)` `KBrO_3` was added to a sample of `SeO_3^(2-)`, The bromine evolved was removed and the excess of `KBrO_3` was titrated with 5.1 " mL of " `(M)/(25)` solution of `NaAsO_2`. Calculate the amound of `SeO_3^(2-)` and balance the equation.
`SeO_3^(2-)+BrO_3^(ɵ)+H^(o+)toSeO_4^(2-)Br_2+H_2O`
`BrO_3^(ɵ)+AsO_2^(ɵ)+H_2OtoBr^(ɵ)+AsO_4^(3-)+H^(o+)`
`(Br=80, K=39, As=75, Se=79)`

Explanation

This question cannot be solved by the equivalent method, since the numbers of electrons in the two reactions are different. The n-factors for `BrO_3^(ɵ)` to `Br_2` and `BrO_3^(ɵ)` to `Br^(ɵ)` are different. So it has to be solved by mol method by balancing the two equations.
First Reation:
`{:[(undersetunderset(x=4)(x-6=2)(SeO_3^(2-))toundersetunderset(x=6)(x-8)(SeO_4^(2-))+2e^(-)),(undersetunderset(2x=10)(2x-12=-2)(10e^(-)+2BrO_3^(-))tounderset(2x=10)(Br_2)):}]_((n=(10)/(2)=5))^((n-2))`
Second reaction:
`{:[(undersetunderset(x=5)(x-6=-1)(6e^(-)+BrO_3^(ɵ)tounderset(x=-1)(Br^(ɵ)))),(undersetunderset(x=3)(x-4=-1)(AsO_2^(ɵ))toundersetunderset(x=5)(x-8=-3)(AsO_4^(3-)+2e^(-))):}]_((n=2))^((n=6))`
First reaction:
`5SeO_3^(2-)+5H_2Oto5SeO_4^(2-)+10H^(o+)+10e^(-)`
`underline(2BrO_3^(ɵ)+12H^(o+)+10e^(-)toBr_2+6H_2O)`
`underline(5SeO_3^(2-)+2BrO_3^(ɵ)+2H^(o+)to5SeO_4^(2-)+Br_2+H_2O)`
Second `BrO_3^(ɵ)+6H^(o+)+6e^(-)toBr^(ɵ)+2H_2O`
`underline(3AsO_2^(ɵ)+6^(ɵ)+12H^(o+))`
`underline(BrO_3^(ɵ)+3AsO_2^(ɵ)+3H_2OtoBr^(ɵ)+3AsO_4^(3-)+6H^(o+))`
Excess of `KBrO_3=5.1xx(1)/(25)=0.204 ` mol
`=0.204` m" mol of "`NaAsO_2`
`3 mmol` of `AsO_2^(ɵ)=1 mmol` of `KBrO_3`
`0.204 mmol` of `AsO_2^(ɵ)=0.068 m` " mol of "`KBrO_3`
Total `KBrO_3=20xx(1)/(60)=0.333` mmol
`KbrO_3` used `=0.333-0.068=0.265 mmol`
`2mmol` of `BrO_3^(ɵ)=(5)/(2)xx0.265`
`=0.6625 mmol` of `SeO_3^(2-)`
(molecular weight of `SeO_3^(2-)` is 127)
`implies0.6625xx10^(-3)xx127g of SeO_3^(2-)`
`=0.8143 g of SeO_3^(2-)`