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The conductivity of 0.001028 mol L^(-1) ...

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)`

Text Solution

Verified by Experts

The correct Answer is:
A
`Lambda_m=k/C=(4.95xx10^(-5)Scm^(-1))/(0.001028molL^(-1))xx(1000cm^3)/L`
`=48.15 S cm^(2) mol^(-1)`
`alpha=Lambda_m/Lambda_m^@=48.15/390.5=0.1233`
`K_a=(Calpha^2)/(1-alpha)=(0.001028xx(0.1233)^2)/(1-0.1233)=1.78xx10^(-5)mol L^(-1)`
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