At a certain temperature, the ratio of ionisation constants of weak acids HA and HB is 100:1. if the molarity of the solution of HA is the samme as that of HB and the degrees of ionisation of HA and HB in their respective solutions are `alpha_(1) and alpha_(2)` respectively, then show that `alpha_(1)=10alpha_(2)`.

Explanation

Suppose, the ionisation constants of HA and HB are `K_(1) and K_(2)`, respectively.
If the concentration of each of the solutions of HA and HB be c `mol*L^(-1)`, then
`alpha_(1)=sqrt((K_(1))/(c)) and alpha_(2)=sqrt((K_(2))/(c))`, where `alpha_(1) and alpha_(2)` are the degrees of ionisation of HA and HB in their respectively solutions.
It is given that `K_(1):K_(2)=100:1`
So, `(alpha_(1))/(alpha_(2))=sqrt((K_(1))/(K_(2)))=sqrt(100)=10,` i.e., `alpha_(1)=10alpha_(2)`.