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)`?

Options

A. `10^(-5)`
B. `1.11xx10^(-5)`
C. `1.11xx10^(-4)`
D. `0.01`

Answer

B

Explanation

The molar conductivity of the dissociated form of crotonic acid is
`Lambda_(m)^(@)(HC)=Lambda_(m)^(0)(HCl)+Lambda_(m)^(@)(NaC)-Lambda_(m)^(@)(NaCl)`
`=(426+83-126)Omega^(1)cm^(2)mol^(-1)`
`=383Omega^(-1)cm^(2)mol^(-1)`
The molar conductivity of HC, `Lambda_(m)(HC)=(K)/(C)`
`=(3.83xx10^(-5)Omega^(-1)cm^(-1))/(0.001)xx1000=38.3Omega^(-1)cm^(2)mol^(-1)`
The degree of dissociation,
`alpha=(Delta_(m)(HC))/(Delta_(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))/(1-01)=1.11xx10^(-5)`