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The conductivity of 0.00241 M acetic aci...

The conductivity of `0.00241 M` acetic acid is `7.896xx10^(-5)Scm^(-1)`. Calculate its molar conductivity. If `wedge_(m)^(@)` for acetic acid is `390.5Scm^(2)mol^(-1)`, what is its dissociation constant ?

Text Solution

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`wedge_(m)^(@)=(kappaxx1000)/("Molarity")=((7.896xx10^(-5)" S "cm^(-1))xx1000cm^(3)L^(-1))/(0.00241" mol "L^(-1))=32.76" S "cm^(2)mol^(-1)`
`alpha=(wedge_(m)^(c))/(wedge_(m)^(@))=(32.76)/(390.5)=8.4xx10^(-2),K_(a)=(calpha^(2))/(1-alpha)=(0.00241xx(8.4xx10^(-2))^(2))/(1-0.084)=1.86xx10^(-5)`
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