Calculate the degree of hydrolysis of a mixture of aniline and acetic acid each of them being `0.01M. K_(a)` of acetic acid `= 1.8 xx 10^(-5)` and `K_(a) ("aniline") = 4.5 xx 10^(-10)`. Also calculate `pH` of the mixture.

Explanation

`C_(6)C_(5)NH_(2)+CH_(3)COOH hArr underset(("Acetanilide"))(C_(6)H_(5)NHCOCH_(3))+H_(2)O`
Now when we mix equal concentrations of aniline and acetic acid, they will neutralise each other to form acetanilide (salf). The salt is of weak base and weak acid, so hydrolysis taken place. The dergee of hydrolysis of the salf of weak acid and weak base is given by:
`h = (sqrt(K_(h)))/(1+sqrt(K_(h))`
Here, `K_(h) = (K_(w))/(K_(a)K_(b)) = (10^(-14))/(1.8 xx 10^(-5) xx 4.5 xx 10^(-10)) = 1.23`
`rArr h = sqrt(1.23)/(1+sqrt(1.23)) = (1.109)/(1+1.109) = (1.109)/(2.109) = 0.52`
However the `pH` of solution can be calculated by using the standared result:
`pH = 7+(1)/(2) (pK_(a) - pK_(b))`
`rArr pH = 7+(1)/(2)[ -log 1.8 xx 10^(-5) - (-log 4.5 xx 10^(-10))]`
`rArr pH = 7 +(1)/(2) log (4.5 xx10^(-10))/(1.8 xx 10^(-5)) = 4.7`
Note: Solution in this case is acidci. in such salts, `pH` depends upon which of teh two, i.e. weak acid or weak base, is stringer than the other, i.e by comparing `K_(a)` and `K_(b)`, one can theoretically get an idea about the `pH` of solution.
The formula `pH = 7 +(1)/(2) (pK_(a) - pK_(b))` is applicable even when 'h' is large as ho approximation of 'h' comparison with unity, has been taken in the derivation.