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Conductivity of 0.00241M acetic acid is ...

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 ?

Text Solution

Verified by Experts

`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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Knowledge Check

  • The conductivity of 0.001028 mol L^(-1) acetic acid is 4.95xx10^(-5) S cm^(-1) . Calculate its dissociation constant, if Lambda_m^@ , for acetic acid is 390.5 S cm^(2) mol^(-1)

    A
    `1.78xx10^(-5)` mol `L^(-1)`
    B
    `1.87xx10^(-5)` mol `L^(-1)`
    C
    `0.178xx10^(-5)` mol `L^(-1)`
    D
    `0.0178xx10^(-5)` mol `L^(-1)`