A definite volume of an aqueous `N//20` acetic acid `(pK_(a) = 4.74)` is titrated with a strongs base. It is found that `75` equal-sized drops of `NaOH` added from a burette effect the complete neutralisation. Find the `pH` when an acid solution is neutralised to the extent of `20%, 40%`, and `80%`, respectively.

Explanation

During the titration of weak acid with swtrong base, if weak acid is left bwhind, acidic buffer is formed.
i. `20%` neutralistion of `W_(A)` means `20%` salf of `W_(A)//S_(B)` is formed and `80% W_(A)` is left.
`:. pH = pK_(a) + "log"(["Salt"])/(["Acid")]`
`= 4.74 +log ((20)/(80))`
`=4.74 + log ((1)/(4))`
`= 4.74 - log2^(2)`
`= 4.74 - 2 xx 0.3 = 4.14`
ii. `40%` neutralisation of `W_(A)` means `40%` salt of `W_(A)//S_(B)` is formed and `60% W_(A)` is left.
`:. pH = 4.64 +log ((40)/(60))`
`= 4.74 +log 2 - log3`
`= 4.74 +0.3 - 0.48 = 4.56`
iii. `90%` neutralisation of `W_(A)` means `90%` salt of `W_(A)//S_(B)` is formed and `10% W_(A)` is left.
`:. pH = 4.74 + log ((90)/(10))`
`= 4.74 + log 3^(2)`
`= 4.74 +2 xx 0.48 = 5.70`