`^^_(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. `(0.25)`
B. `(0.125)`
C. `(0.50)`
D. `(0.75)`

Explanation

`wedge_(HA)^(0)=wedge_(HCl)^(0)+wedge_(NaA)^(0)-wedge_(NaCl)^(0)`
`-425.9xx100.5-126.4=400 S cm^(2) m "ole"^(-1)`
`wedge=(Kxx1000)/(M)=(5xx10^(-5)xx1000)/(0.001)=50`
`alpha =(wedge)/(wedge^(@))=50/400=0.125`