In the titration of a solution of a weak acid `HX` with `NaOH`, the `pH` is `5.8` after `10mL` of `NaOH` solution has been added and `6.40` after `20mL` of the `NaOH` has been affed. What is the ionisation constant of the acid `HX`?

Explanation

Let the mormality of `HX` and `NaOH = 1`.
Let `x mEq` or mmol of `HX` is present initially.
When `pH = 5.8`.
`["Salt"] = 10 xx 1 = 10 mmol`
`["Acid"] = (x - 10) mmol`
When `pH = 6.4`
`["Salt"] = 20 xx 1 = 20 mmol`
`["Acid"] = (x - 20) mmol`
`5.8 = pK_(a) +log (10//x - 10) ..(i)`
`6.4 = pK_(a) +log (20//x - 20)...(ii)`
SOlving equaitons (i) and (ii), calculate the value of `x` and `pK_(a)`.
`:. pK_(a) = 6.1`