The ionization constant of propanoic acid is `1.32xx10^(-5)`. Calculate the degree of ionization of the acid in its `0.05 M` solution and also its `pH`. What will be its degree of ionization if the solution is `0.01 M` on `HCl` also?

Explanation

If `alpha` is the degree of dissociation ,
`alpha=sqrt(K_a/c)`
`=sqrt((1.32xx10^(-5))/0.05)=1.62xx10^(-2)`
`[H^+]= c alpha =0.05 xx 1.62 xx10^(-2) =8.12 xx10^(-4)`
pH=-log `[H^+]`
=-log `(8.12xx10^(-4))` =3.09
In the presence of 0.01 M HCl,
`alpha =1.32 xx10^(-3)`