Chile salt peter a source of `NaNO_(3)` also contains `NaIO_(3)` the `NaIO_(3)` is a source of `I_(2)` produced as shown in the following equation:
Step I: `IO_(3)^(ɵ)+3HSO_(3)^(ɵ)+3SO_(4)^(2-)`
Step II: `5I^(ɵ)+IO_(3)^(ɵ)+6H^(o+)to3I_(2)(s)+3H_(2)O`
One litre sample of chile salt peter solution containing 6.6 g `NaIO_(3)` is treated with `NaHSO_(3)` Now an additional amount of same solution is added to the reaction mixture to bring about the second titration. Calculate the weight of `NaHSO_(3)` requried in step I and what additional volume of chile salt peter mist be added in step II to bring out complete conversion of `I^(ɵ)` to `I_(2)`.

Explanation

`[Mw of NaIO_(3)=198,Mw of NaHSO_(3)=104]`
(a). `6e^(-)+IO_(3)^(ɵ)toI^(ɵ)(n=6)`
(b). `10e^(-)+2IO_(3)^(ɵ)toI_(2)(n=(10)/(2)=5)`
(c). `HSO_(3)^(ɵ)toSO_(4)^(2-)+2e^(-)(n=2)`
(d). `2I^(ɵ)toI_(2)+2e^(-)(n=(2)/(2)=1)`
`m" Eq of "NaHSO_(3)=m" Eq of "NaIO_(3)`
`=NxxV=(6.6xx10^(3))/((198)/(6))=200`
`m" Eq of "NaHSO_(3)=200`
`(Wxx10^(3))/((104)/(2))=200`
`W_(NaHSO_(3))=10.4g`
Also m" Eq of "`I^(ɵ)` formed in step I using n-factor of `6=200`
`6e^(-)+IO_(3)^(ɵ)toI^(ɵ)`
In step II, the valence factor or n-factor of `IO_(3)^(ɵ)` IS 5.
`IOe^(-)+2IO_(3)^(ɵ)toI_(2)(n=(10)/(2)=5)`
Thus m" Eq of "`I^(ɵ)` formed using n-factor of `1=(200)/(6)`
m" Eq of "`NaIO_(3)` used in step `II=(200)/(6)`
`NxxV=(200)/(6)`
`(6.6)/((198)/(5))xxV_(mL)=(200)/(6)`
`V_(NaIO_(3))=(200xx198)/(6xx5xx6.6)=200mL`