A solution of a weak acid was titrated was NaOH. The equivalence point was reached when `36.12` ml of `0.10` (N) NaOH have been added . Now `18.06` ml of `0.10` (N) HCl was addd to the titrated solution . The pH was found to be `4.92 ` .What is `K_(a)` of the acid ?

Explanation

At the end point,
m.e. (or millimoles) of the salt produced
=m.e. of NaOH
`=0.1 xx 36.12`
=3.612
Since salt formed will be uniunivalent and so for such salts m.m.=m.e.
m.e. (or millimole) of HCl added `=0.1 xx 18.06=1.806`
The addition of 1.806 m.e. fo HCl will produce the same number of m.e. of the unknown acid and reduce the amount of the salt by 1.806 m.e.
m.e. (or millimole) of the unknown acid =1.806
and m.e. (or millimole) of the salt =3.612 -1.806=1.806