A particular acid rain water water contains sulphite `(SO_(3)^(2-))` ions if a 25.0 `cm^(3)` sample of this water requires `cm^(3)` of 0.02 M `KMnO_(4)` solution for titeation what is the amount of `SO_(3)^(2-)` ions per litre in rain water?

Explanation

Step 1 To write the balanced equation for the redox reaction
`MnO_(4)^(-)+8^(+)+5e^(-)rarrMn^(2+)+4H_(2)O]xx2`
`SO_(3)^(2-)+H_(2)OrarrSO_(4)^(2-)+2H^(+)+2 e^(-)]xx5`
`2MnO_(4)^(-)+5SP_(3)^(2-)+6H^(+)rarr2mn^(2+)+5SO_(4)^(2-)+3H_(2)O`
Step 2 To determine the molarity of `SO_(3)^(2-)` ion solution
Let `M_(1)` be the molaity of `SO_(3)^(2-)` ions in acid rain water applying molarity equaiton
`(M_(1)V_(1))/(n_(1))(SO_(3)^(2-))=(M_(2)V_(2))/(n_(2))(MnO_(4)^(-))` Thus the molariyt of `SO_(3)^(2-)` ions in acid rain water =0.07 M
Mol wt of `SO_(3)_^(2-)` ions =32 +48 =80
`therefore` Amount of `SO_(3)^(2-)` ions in rain water `=0.07xx80=0.56 gL^(-1)`