Calculate the pH at equivalence point when a solution of `0.10 M ` acetic acid is titrated with a solution of `0.10 ` M hydroxide
`(K_(a) " for acetic acid is " 1.9 xx10^(-5))`

Explanation

At the equivalence point, `CH_(3)CO ON a` is formed asn its concentration `=(0.1)/(2) M = 0.05 M`.
It is a slat of weak acid and strong base. The formula for finding the pH of such a salt is
`pH = - (1)/(2) [log K_(w) + log K_(a) - log c]`
`:. pH = - (1)/(2) [log 10^(-14)+log (1.9xx10^(-5))-log(5xx10^(-2))]`
`=-(1)/(2) [-14 + (-5+0.2788)-(-2+0.6990)]=(1)/(2) (14+5-0.2788-2+0.6990)=(17.42)/(2) = 8.71`