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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? |
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Answer» Solution :Step 1 To write the balanced equationfor 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)` |
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