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The conductivity of 0.001 M acetic acid ...

The conductivity of 0.001 M acetic acid is `5xx10^(-5)S cm^(-1)` and `^^^(@)` is 390.5 `S cm^(2) "mol"^(-1)` then the calculated value of dissociation constant of acetic acid would be

A

`81.78xx10^(-4)`

B

`81.78xx10^(-5)`

C

`18.78xx10^(-6)`

D

`18.78xx10^(-5)`

Text Solution

Verified by Experts

The correct Answer is:
C
`^^_(m)=(kxx1000)/(C)=(5xx10^(-5)xx1000)/(10^(-3))`
`50S cm^(2) "mol"^(-1)`
Degree of dissociation `alpha=(^^_(m))/(^^_(m)^(0))`
`=(50)/(390.5)=1.28xx10^(-1)`
`CH_(3)COOHhArrCH_(3)COO^(-)+H^(+)`
C(1-a)
`k=(Calpha.Calpha)/(C(1-alpha))=(Calpha^(2))/(1-alpha)`
=`(10^(-3)xx(1.2 8xx10^(-1))^(2))/(1-0.128)`
`=(10^(-3)xx1.64xx10^(-2))/(0.872)=(1.64xx10^(-3))/(0.872)`
`=1.88xx10^(-5)`
`=18.8xx10^(-6)`
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