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A solution of 0.2g of a compoundcontaining Cu^(2+) and C_(2)O_(4)^(2-) ions on titration with 0.02" M " KMnO_(4) in presence of H_(2)SO_(4) consumes 22.6 mL of theoxidant .The resultant solution is neutralised with Na_(2)CO_(2) acidfied with dilute acetic acid and treated with excess KI . The liberated I_(2) required 11.3 mL of0.05Na_(2)S_(2)O_(3) solution for complete reduction . Findout the mole ratio ofCu^(2+) " to " C_(2)O_(4)^(2-)in the compound. Writedown the balanced redoxreactions involvedin the above titrations . |
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Answer» Solution :At the first stage , `C_(2)O_(4)^(2-) ` is oxidisedto `CO_(2)` by the OXIDANT `KMnO_(4)` `{:(5C_(2)O_(4)^(2-)+2MnO_(4)^(-)+16H^(+)to,10CO_(2)+2Mn^(2+)+3H_(2)O),("+7","+2"):}` ` :. ` normalityof `KMnO_(4)` solution= `0.02 xx (7-2) = 0.1 N ` At the second stage `Cu^(2+)`liberates `I_(2)` from KI and this LIBERATED `I_(2)` requires 11.3 mL of `0.05 " M " Na_(2)S_(2)O_(3)` solution for complete reaction . `{:(2Cu^(2+)+2I^(-)to,2Cu^(+)+I_(2)),(2S_(2)O_(3)^(2-)+I_(2) to ,S_(4)O_(6)^(2-)+2I^(-)),(" +4"," +5"):}` Normalityof `Na_(2)S_(2)O_(3)` solution ` = 0.05 (5-4) = 0.05 N ` Now , `(" m.e of "Cu^(2+))/( "m.e of " C_(2)O_(4)^(2-))= (" m.e of"Na_(2)S_(2)O_(3))/("m.e of"KMnO_(4)) = (0.05 xx 11.3 )/(0.1 xx 22.3) = 1/4 ` As `{:(Cu^(2+)to,Cu^(+)and,C_(2)O_(4)^(2-) to,2CO_(2)),(+2,+1,+6,+8):}` ` :. ("mmol of "Cu^(2+)xx1)/("mmol of"C_(2)O_(4)^(2-)xx2) =1/4` or `("mole of "Cu^(2+))/(" mole of " C_(2)O_(4)^(2-)) = 1/2 ` |
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