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)(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)=(383xx10^(-5)Omega^(-1)cm^(-1))/(0.001)xx1000`
`=38.3Omega^(-1)cm^(2)mol^(-1)`
The degree of dissocitation,
`alpha=(Lambda_(m)(HC))/(Lambda_(m)(HC))=((38.3Omega^(-1)cm^(2)mol^(2)))/((383Omega^(-1))cm^(2)mol^(-1))=0.1`
`K_(a)=(Calpha^(2))/(1-alpha)=((10^(-3))(0.1)^(2)=1.11xx10^(-5))/(1-0.1)`