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 ?

Explanation

Step I. Calculation of molar conductance (`Lambda_(m)^(c )`)
`k=7.896xx10^(-5)" S "cm^(-1),C=0.00241" mol "L^(-1)=(0.00241" mol")/(10^(3)cm^(3))=241xx10^(-8)" mol "cm^(-3)`
`Lambda_(m)^(c )=(k)/(C )=((7.896xx10^(-5)" S "cm^(-1)))/((241xx10^(-8)" mol "cm^(-3)))=32.76" S "cm^(2)mol^(-1)`
Step II. Calculation of degree of dissociation of acetic acid.
`alpha=(Lambda_(m)^(c ))/(Lambda_(m)^(0))=((32.76" S "cm^(2)mol^(-1)))/((390.5" S "cm^(2)mol^(-1)))=0.084=8.4xx10^(-2)`
Step III. Calculation of dissociation constant `K_(c )`
`K_(c )=([CH_(3)COO^(-)][H^(+)])/([CH_(3)COOH])=(CalphaxxCalpha)/(C(1-alpha))=(Calpha^(2))/(1-alpha)`
`=((0.00241" mol "L^(-1))xx(0.084)^(2))/(1-0.084)=0.0000185" mol "L^(-1)=1.85xx10^(-5)" mol "L^(-1)`