At 298 K, the specific conductance of 0.1 M acetic acid solution was found to be `0.00163Omega^(-1)cm^(-1)`. Calculate the degree of dissociation and dissociation constant of the acid if its molar conductance at infinite dilution is `390.7 Omega^(-1) cm^(2)mol^(-1)`.

Explanation

`k=0.00163Omega^(-1)cm^(-1)`
`C=0.1 M =0.1 " mol "L^(-1)=((0.1" mol"))/((10^(3)cm^(3)))=10^(-4)" mol "cm^(-3)`
Molar conductance at given concentration,
`Lambda_(m)^(oo)=(k)/(C )=((0.00163Omega^(-1)cm^(-1)))/((10^(-4)" mol "cm^(-3)))=16.3Omega^(-1)cm^(2)mol^(-1)`
Molar conductance at infinite dilution, `Lambda_(m)^(0)=390.7 Omega^(-1)cm^(2)mol^(-1)`
Degree of dissociation at given concentration,
`a=(Lambda_(m)^(c))/(Lambda_(m)^(0))=((16.3Omega^(-1)cm^(2)mol^(-1)))/((390.7 Omega^(-1)cm^(2)mol^(-1)))=0.0417`
`"Dissociation Constant" (K_(a))=(Ca^(2))/(1-alpha)=((10^(-4)" mol "cm^(-3))xx(0.0417)^(2))/((1-0.0417)~~1)=1.74xx10^(-7)" mol " cm^(3)`.