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?

A particular acid rain water water contains sulphite (SO_(3)^(2-)) ion

1.	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?
Answer» <html><body><p></p>Solution :Step 1 To write the balanced equationfor the redox reaction <br/> `MnO_(4)^(-)+<a href="https://interviewquestions.tuteehub.com/tag/8-336412" style="font-weight:bold;" target="_blank" title="Click to know more about 8">8</a>^(+)+5e^(-)rarrMn^(2+)+4H_(2)O]xx2` <br/> `SO_(3)^(2-)+H_(2)OrarrSO_(4)^(2-)+2H^(+)+2 e^(-)]xx5` <br/> `2MnO_(4)^(-)+5SP_(3)^(2-)+6H^(+)rarr2mn^(2+)+5SO_(4)^(2-)+3H_(2)O` <br/> Step 2 To determine the molarity of `SO_(3)^(2-)` ion solution<br/> Let `M_(1)` be the molaity of `SO_(3)^(2-)` ions in <a href="https://interviewquestions.tuteehub.com/tag/acid-847491" style="font-weight:bold;" target="_blank" title="Click to know more about ACID">ACID</a> rain water applying molarity equaiton <br/> `(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 <br/> Mol wt of `SO_(3)_^(2-)` ions =32 +48 =80 <br/> `therefore` Amount of `SO_(3)^(2-)` ions in rain water `=0.07xx80=0.56 gL^(-1)`</body></html>