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Chile salt peter a source of NaNO(3) als...

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)`.

Text Solution

Verified by Experts

Meq. of `NaHSO_(3) = " Meq. of " NalO_(3) = N xx V = (5.8)/(198) xx 6 xx 1000 `
`l^(5+) + 6e^(-) to l^(-) " ( For " KlO_(3), " n-factor = 6 ")` .
` :. " Meq. Of " NaHSO_(3) = 175.76`
`W_(NaHSO_(3))/(M//2) xx 1000 = 175.76`
`rArr W_(NaHSO_(3)) = (175.75 xx 104)/(2 xx 1000) = 9.14` gm
Also , Meq. of `l^(-)` formed using n-factor `6 = 175.76`
In step 2, n-factor of `l^(-) ` is 1 and n- factor for `lO_(3)^(-)` is 5
Meq. of `l^(-)` formed using n-factor `1 = (175.76)/6`
Meq. of `NalO_(3) " used in " 2^(nd) " step " = (175.76)/6`
`rArr (5.8)/(198) xx 5 xx V = (175.76)/6`
`V_(NalO_(3)) = 20 ` ml
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