At `25^(@)C` the equivalent conductance of butanoic acid at infinite dilution is `386.6 ohm^(-1)cm^(2)eq^(-1)`. If the ionization consatant is `1.4xx10^(-5)`, calculate equivalent conductance of `0.05` N butanoic acid solution at `25^(@)C(ohm^(-1)cm^(2)eq^(-1))`?

Explanation

`K_(a)=(C alpha^(2))/((1- alph))=C alpha^(2) ("if "alpha lt lt 1)`
`alpha =sqrt((K_(a))/(C))=sqrt((1.4 xx 10^(-5))/(0.05))=0.0167`
`alpha =(wedge_(C))/(wedge_(0))`
`wedge_(C)=(alpha xx wedge_(0))=0.0167 xx 386.6`
`=6.57 Ohm^(-1) cm^(2) "equiv"^(-1)`