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Given the following limiting molar condu...

Given the following limiting molar conductivies at `25^(@)C,, Hcl, 426Omega^(-1)cm^(2)mol^(-1),NaCl, 126Omega^(-1)cm^(2)mol^(-1),NaC`(sodium crotonate) ,`83Omega^(-1)cm^(2)mol^(-1)`. What is the ionization constant of crotonic acid? If the conductivity of a `0.001M` crotonic acid `(HC)` solution is `3.83xx10^(-5)Omega^(-1)cm^(-1)`?

A

`1.11xx10^(-6)`

B

`1.11xx10^(-5)`

C

`1.11xx10^(-4)`

D

`1.11xx10^(-7)`

Text Solution

Verified by Experts

The correct Answer is:
B
The molar conductivity of the dissociated form of crotonic acid is
`^^_(m)^(oo)(HC)=^^_(m)(HCl)+^^_(m)(NaC)-^^_(m)(NaCl)`
`=(426+83-126)Omega^(-1)cm^(2)mol^(-1)`
`=383Omega^(-1)cm^(2)mol^(-1)`
The molar conductivity of `HC`,
`^^_(m)(HC)(1000xxK)/(C_(M))=(3.83xx10^(-5)Omega^(-1)cm^(2))/(0.001)xx1000`
`=383Omega^(-1)cm^(2)mol^(-1)`
The degree of dissociation.
`alpha=(^^_(m)(HC))/(^^_(m)^(oo)(HC))=((38.3Omega^(-1)cm^(2)mol^(-1)))/((383Omega^(-1)cm^(2)mol^(-1)))=0.1`
`k_(a)=(Calpha^(2))/(1-alpha)=((10^(-3))(0.1)^(2))/(1-0.1)=1.11xx10^(-5)`
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