Equivalent conductivity of acetic acid at infinite dilution is 39.7 and for 0.1 M acetic acid the equivalent conductance is `"5.2 mho.cm"^(2)."gm.equiv."^(-1)`. Calculate degree of dissociation, `H^(+)` ion concentration and dissociation constant of the acid.

Explanation

`alpha=(lambda_(C ))/(lambda_(oo))=0.01333=1.33%`
`{:(CH_(3)COOHhArrH^(+)+CHCOO^(-)),(C(1-alpha)" "Calpha" "Calpha):}`
`therefore[H^(+)]=Calpha=0.1xx0.0133=0.00133M`
`K=(alpha^(2)C)/(1-alpha)=(0.0133^(2)xx0.1)/((1-0.0133))=2.38xx10^(-5)M`