Conductivity of 0.00241 M acetic acid solution is `7.896 xx 10^(-5) S cm^(-1)`. Calculate its molar conductivity in this solution. If `Lambda_(m)^(@)` for acetic acid be 390.5 S `cm^(2) "mol"^(-1)`, what would be its dissociation constant?
(i) First , find molar conductivity using the formula,,
`Lambda_(m) = ( K xx 1000)/( C )`
(ii) Then find degree of dissociation `( alpha)` and dissociation constant `( K_(a))`
by using formula `alpha = ( Lambda_(m))/( Lambda_(m)^(@)) ` and `K_(a) = ( C alpha^(2))/( 1-alpha)` respectively.

Explanation

Given, `K = 7.896 xx 10^(-5) S cm^(-1)`,
`Lambda_((CH_(3)COOH))^(@) = 390.5 S cm^(2)"mol"^(-1)`
and molarity `= 0.00241 M`
Molar conductivity, `Lambda_(m)^(C ) = ( K xx 1000)/( "Molarity")`
`= (7.896 xx 10^(-5) S cm^(-1) xx 1000)/( 0.00241 )`
`Lambda_(m)^(C ) = 32.76 S cm^(2) "mol"^(-1)`
Degree of dissociation,
`alpha = ( Lambda_(m)^(C ))/( Lambda_(m)^(@)) = ( 32.76)/( 390.5) = 8.4 xx 10^(-2)`
Dissociation constant of acetic acid,
`K_(a) = ( C alpha^(2))/( 1-alpha)`
`K_(a) = ( 0.00241 ( 8.4 xx 10^(-2))^(2))/( ( 1- 0.084))`
`= 1.86 xx 10^(-5)`