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The conductivity of 0.001 M actic acid i...

The conductivity of 0.001 M actic acid is `4 xx 10^(-5) S //cm` . Calculate the dissociation constant acetic acid, if molar conductivity at infinite dilution for acetic acid is 390 S `cm^(2) //` mol.

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Molar conductivity at 0.001 M, `Lambda_(m) = ( k xx 1000)/( C ) = ( 4 xx 10^(-5) xx 1000)/( 0.001) S cm^(2) //"mol" = 40 S cm^(2) //` mol
Degree od dissociation, `alpha = ( Lambda_(m))/( Lambda_(m)^(oo)) = ( 40)/( 390) = 0.10256`
For the dissociation of acetic acid, `CH_(3) COOH`
Dissociation constant,
`K_(a) = ( [ CH_(3)COO^(-)][H^(+)])/([CH_(3)COOH])`
`= ( C alpha . C alpha)/( C - C alpha) = ( C^(2) alpha^(2))/( C ( 1- alpha))= ( C alpha^(2))/( ( 1- alpha))`
`K_(a) = ( 0.001 xx ( 0.103)^(2))/( ( 1- 0.103))`
`= ( 1.061 xx 10^(-5))/( 0.897) = 1.18 xx 10^(-5)`
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