Conductivity of 0.00241 M acetic acid is `7.896xx10^(-5)" S "cm^(-1)`. Calculate its molar conductivity and if `Lamda_(m)^(@)` for acetic acid is 390.5 S `cm^(-2)mol^(-1)`, what is its dissociation constant ?

Explanation

* The calculation of molar conductivity `Lamda_(m)^(c)`:
Where, `M=0.00241" "CH_(3)COOH`,
`Lamda_(m)^(c)=(kxx1000)/("molarity")" "k=7.896xx10^(-5)" S "cm^(-1)`
`=(7.896xx10^(-5)(" S "cm^(-1))xx1000(cm^(3)L^(-1)))/(0.00241(mol" "L^(-1)))`
* The calculation of dissociation degree `(alpha)` of `CH_(3)COOH`:
`therefore alpha=(Lamda_(m)^(c))/(Lamda_(m)^(@))," "alpha=(32.763)/(390.5)=0.0839|" Where,"alpha="dissociation degree," Lamda_(m)^(c)=32.763" S "cm^(2)mol^(-1)," "Lamda_(m)^(@)=390.5" S "cm^(2)mol^(-1)`
* The calculation of dissociation constant `(K_(a))`:
The equilibrium in acetic acid `(CH_(3)COOH)` solution is as follow:
`CH_(3)COOH_((aq))hArr CH_(3)COO_((aq))^(-)+H_((aq))^(+)`
`therefore K_(a)=(CalphaxxCalpha)/(C-Calpha)=(Calpha^(2))/((1-alpha))`, `=(0.00241xx(0.0839)^(2))/(1-0.0839)," "=(0.00241xx0.0839xx0.0839)/(0.9161)|" Where, dissociation degree "alpha=0.0839" Molarity "C=0.00241" mol "L^(-1)`
`=1.8518xx10^(-5)`