The ionization constant of propionic 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 in the solution of `0.01N HCI` ?

Explanation

`underset(c(1-alpha))(CH_(3)CH_(2)COOH) Leftrightarrow underset(c alpha)(CH_(3)CH_(2)COO^(-))+underset(c alpha)(H^(+))`
`=(c alpha .calpha)/(c(1-alpha))=c alpha^(2) (? 1-alpha approx )`
`therefore 0.05 xx alpha^(2)=1.32 xx 10^(-5) , alpha =1.63 xx 10^(-2)`
`pH =-log[H^(+)]=-log (c alpha)`
In 0.01 N HCl, `[H^(+)] =0.01` and thus
`1.32 xx 10^(-5) =(c alpha xx 0.01)/(c(1-alpha))=alpha xx 0.01 (1-alpha approx1)`
`therefore alpha=1.32 xx 10^(-3)`