InterviewSolution
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InterviewSolution
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A solution of `0.2 g` of a compound containing `Cu^(2+)` and `C_(2)O_(4)^(2-)` ions on titration with `0.02M KMnO_(4)` in presence of `H_(2)SO_(4)` consumes `22.6mL` oxidant. The resulting solution is neutralized by `Na_(2)CO_(3)`, acidified with dilute `CH_(3)COOH` and titrated with excess of `KI`. The liberated `I_(2)` required `11.3 mL "of" 0.05M Na_(2)S_(2)O_(3)` for complete reduction. Find out mole ratio of `Cu^(2+)` and `C_(2)O_(4)^(2+)` in compound. |
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Answer» 1st case : Only `C_(2)O_(4)^(2-)` ions are oxidised by `KMnO_(4)` solution. Normality of `KMnO_(4)` solution `=0.02xx5=0.1 N` 22.6 mL of 0.1 N `KMnO_(4)=22.6 mL " of" 0.1 N C_(2)O_(4)^(2-)` soln. `underset("in the solution")("Mass of" C_(2)O_(4)^(2-))` ions `=(N xx E xx V)/(1000)=(N xx M xx V)/(1000xx2)` No. of moles of `C_(2)O_(4)^(2-)` ions in the solution`=(N xx M xxV)/(1000xx2xxM)` `=(N xx V)/(2000)` `=(0.1xx22.6)/(2000)` `=11.3xx10^(-4)` 2nd case : Only `Cu^(2+)` ions are reduced by KI and iodine liberated is neutralised by `Na_(2)S_(2)O_(3)` solution. 11.3 mL of `0.05 M Na_(2)S_(2)O_(3)-=11.3 mL of 0.05 N Na_(2)S_(2)O_(3)` `=11.3 mL of 0.05 N I_(2)` `=11.3 mL " of " 0.05 N Cu^(2+)` Mass of `Cu^(2+)` ions in the solution`=(NxxExxV)/(1000)=(NxxMxxV)/(1000)` No. of moles of `Cu^(2+)` ions in the solution `=(N xx M xx V)/(1000xxM)` `=(N xx V)/(1000)` `=(0.05xx11.3)/(1000)` `=5.65xx10^(-4)` Molar ratio of `(Cu^(2+))/(C_(2)O_(4)^(2-))=(5.65xx10^(-4))/(11.3xx10^(-4))=(1)/(2)` |
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