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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). |
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Answer» Solution :`[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` |
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