The specific conductivity of `CH_(3)COOH`(0.001N) at 291 K is `4.09 xx 10^(-9)Omega^(-1)m^(-1)`. Calculate its degree of dissociation if equivalent conductivities at infinite dilution of KCl, HCl and `CH_(3)COOK` are 0.01301, 0.03794 and 0.00956 `Omega^(-1)m^(2)g. eq^(-1)` respectively at 291K.

Explanation

Specific conductance, k, of 0.001 N `CH_(3) COOH = 4.09 xx 10^(-3) Omega^(-1)m^(-1)`
`:.` Equivalent conductance, `wedge_(eq)`, of 0.001N `CH_(3)COOH`
`=(k)/(c(g. eq//m^(3)))=(4.09 xx10^(-3) Omega^(-1) m^(-1))/( 0.001 xx10^(3) g.eq m^(-3))`
`=4.09 " "10^(-3)Omega^(-1) m^(3) (g.eq.)^(-)`
By Kohlrausch.s Law, we have
`wedge^(0) (CH_(3)COOH) =wedge^(0) (CH_(3)COOK)+wedge^(0) (HCl) -wedge^(0)(KCl)`
`=(0.00956 +0.03794-0.01301)Omega^(-1)m^(2)("g eq"^(-1))`
`=0.03449Omega^(-1)m^(2) ("g.eq")^(-)`
`:.` The degree of dissociation , `alpha` of `CH_(3)COOH`
`=(wedge^(c ))/(wedge^(0)) =(4.09 xx 10^(-3) Omega^(-1) m^(2)("g eq")^(-))/(0.03449Omega^(-1)m^(2)("g eq")^(-))=0.1185`.