At `25^(@)C` pH of 0.01 (M) propionic acid solution is 3.4. determine the degree of ionisation and ionisation constnt of propionic acid at `25^(@)C`.

Explanation

Propionic acid is a weak monobasic acid. The equation required for the determination of pH of such an aqueous acidic solution is-
(a) `pH=-log_(10)(calpha) and ` (b) `pH=(1)/(2)pK_(a)-(1)/(2)logc`
Putting the values of pH and c in equation (a) we get,
`3.4=-log_(10)(0.01alpha)=-log_(10)(0.01)-logalpha`
or, `3.4=2-logalpha or, logalpha=-1.4 or, alpha=0.039`
`therefore`The degree of ionisation of propionic acid in its 0.01(M) solution=0.039.
Putting the values of pH and c in equation (b), we get,
`3.4=(1)/(2)pK_(a)-(1)/(2)log(0.01)=(1)/(2)pK_(a)+1`
or, `pK_(a)=4.8 or, -log_(10)K_(a)=4.8 or, K_(a)=1.58xx10^(-5)`
`therefore`Ionisation constant of the acid at `25^(@)C=1.58xx10^(-5)`.