`^^_(m)^(@)` for `NaCI,HCI` and `NaA` are `126.4,425.9` and `100.5Scm^2 m ol^(-1)`, respectively. If the conductivity of `0.001 MHA` is `5xx10^(-5)Scm^(-1)`, degree of dissociation of HA is :

Options

A. `0.75`
B. `0.125`
C. `0.25`
D. `0.50`

Answer

B

Explanation

`Lambda_(m)^(@) (HA) = Lambda_(m)^(@) (HCl) + Lambda_(m)^(@) (NaA) - Lambda_(m)^(@) (NaCl)`
`= 425.9 + 100.5 - 126.4`
`= 400 S cm^(2) mol^(-1)`
`Lambda_(m) = (1000 xx kappa)/(M) = (1000 xx 5 xx 10^(-5))/(10^(-3)) = 50 S cm^(2) mol^(-1)`
`alpha = (Lambda_(m))/(Lambda_(m)^(@)) = (50)/(400) = 0.125`