25 mL of 0.017 " M " H_(2)SO_(3)^(-)in strongly acidic required the addition of 16.9mL of 0.01M MnO_(4)^(-) for itscomplete oxidation t SO_(4)^(2-)or HsO_(4)^(-) .in neutralsolution it required28.6 mL . Assign oxidation numbers to Mn in eachof the products .

25 mL of 0.017 " M " H_(2)SO_(3)^(-)in strongly acidic required the ad

1.	25 mL of 0.017 " M " H_(2)SO_(3)^(-)in strongly acidic required the addition of 16.9mL of 0.01M MnO_(4)^(-) for itscomplete oxidation t SO_(4)^(2-)or HsO_(4)^(-) .in neutralsolution it required28.6 mL . Assign oxidation numbers to Mn in eachof the products .
Answer» SOLUTION :` {:( HSO_(3)^(-),overset(" change in ON = 2") to,SO_(4)^(2-),or HSO_(4)^(-)),(+4,,+6,"+6"):}` ` :. 0.017 " M " HSO_(3)^(-) -= 2 xx 0.017 N "" ..(Eqn.6i)` ` = 0.034 N ` In the first case suppose the On of Mn in the PRODUCTIS X ` :. " " 0.01 " M " MnO_(4)^(-) = 0.01 (7-X) " N " MnO_(4)^(-) "" ...(Eqn.6i)` m.e of `HSO_(3)^(-) = " m.e of " MnO_(4)^(-)` `0.034 xx 25 = 0.01 ( 7 - X ) 16.9` ` :. 7 - X = (0.034 xx25)/(16.9 xx0.01)= 5.00` or X = 2 . Now in the second titration , suppose the oN of Mn in the product is Y . ` :. 0.01" M " MnO_(4)^(-) = 0.01 (7 - Y) N MnO_(4)^(-)` ` 0.034 xx 25 = 0.01 (7-Y) xx 28.6` ` 7 - Y = = (0.034 xx 25)/(0.01 xx 28.6) = 3 ` ` :. Y = 4 `