Conductivity of 0.00241 M acetic acid is `7.896xx10^(-5)S" "cm^(-1)`. Calculate its molar conductivity. If `^^_(m)^(@)` for acetic acid is `390.5S" "cm^(2)" "mol^(-1)`, what is its dissociation constant?

Explanation

According to the given data :
Conductivity, `k= 7.896 xx 10^(-5) S cm^(-1)`
Conc. `c=0.00241 M , wedge^(0) = 390.5 S cm^(2)"mol"^(-1)`
`wedge =(k )/(c )=(k("S cm"^(-1))xx 1000 cm^(3) L^(-1))/( "molarity" ("mol L"^(-1)))`
`=(7.896 xx 10^(-5) xx1000)/(0.00241) S cm^(2) "mol"^(-1)`
`=32.76" S cm"^(2) "mol"^(-1)`
Degree of dissociation, `alpha=(wedge)/(wedge^(@)) = (32.76)/(390.5)=0.084 `
Dissociation constant,
`k=(c alpha^(2))/((1-alpha))=(0.00241xx(0.084)^(2))/((1-0.084))`
`=(0.00241 xx 7.56xx10^(-3))/((1.0-0.084))=1.85 xx 10^(-5)`