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The resistance of 0.01 M CH(2)COOH solut...

The resistance of 0.01 M `CH_(2)COOH` solution was found to be 2220 ohm in a conductivity cell having cell constant 0.366 `cm^(-1)`. Calculate:
(i) molar conductivity `(wedge_(m))" of "0.01 M CH_(3)COOH`
(ii) `wedge_(m)^(oo)`
(iii) degree of dissociation, `alpha` and
(iv) dissociation constant of the acid.
`[lambda^(0) (H^(+))=349.1"ohm"^(-1)cm^(2)"mol"^(-1), lambda^(0)(CH_(3)COO^(-)) =40.9 "ohm"^(-1)cm^(2) "mol"^(-1)]`

Text Solution

Verified by Experts

`lambda_(m)^(@)(CH_(3)COOH)=lambda_(m)^(@)(H^(+))+lambda_(m)^(@)+lambda_(m)^(@)(CH_(3)COO^(-))`
`=3491.+40.9=390.0 s cm^(2) mol^(-1)`
(a) "conductivity" `=("cell constant")/("Resistance")=(0.366)/(2220) =1.648xx10^(-4) S cm^(-1)`
(b)`wedge_(m)=(conductivity xx1000)/(molarity)=(1.648xx10^(-4)xx1000)/(0.01)=16.48 S cm^(2) mol^(-1)`
(c )`alpha =(wedge_(m))/(wedge_(m))=(16.48)/(390)=0.0422`
(d)`K_(a)=(Calpha^(2))/(12-alpha) alphaa^(2)=0.01(0.0422)^(2)=1.78xx10^(-5)`
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