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Conductivity of 0.00241M acetic acid is 7.896 xx 10^(-5)" S cm"^(-1). Calculate its molar conductivity and if Lambda_(m)^(@)m for acetic acid is "390.5 S cm"^(2)" mol"^(-1), what is its dissociation constant ?

Answer»

SOLUTION :`Lambda_(m)^(@)=(kxx1000)/(M)`
`=(7.896xx10^(-5)"S CM"^(-1)xx1000cm^(3)L^(-1))/("0.0024 mol L"^(-1))`
`=32.76" S cm"^(2)" mol"^(-1)`
`alpha=(Lambda_(m))/(Lambda_(m)^(@))=(32.76)/(390.5)=8.39xx10^(-2)`
`K_(a)=(CALPHA^(2))/(1-alpha)=(0.00241xx(8.39xx10^(-2))^(3))/(1-8.39xx10^(-2))`
`=1.86xx10^(-5)`


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