`0.1M NaOH` is titrated with `0.1M, 20mL HA` till the point. `K_(a)(HA) = 6 xx10^(-6)` and degree of dissociation of `HA` is neglible (small) as compared to unity. Calculate the `pH` of the resulting solution at the end point [Use `log 6 ~~ 0.8]`

Explanation

`{:(,NaOH+,HArarr,NaA+,H_(2)O),("At end point",2,2,-,),((mmol),,(-=0.1xx20),,),(,-,-,2,-):}`
Note: `20mL` of `NaOH` is required for the complete metralisation of `HA`.
`NaA` is a salt of strong base and weak acid. Thus, will undergo hydrolysis and solution will become basic.
Here `C = [NaA] = (2)/(20+20) = 0.05M`
and `pK_(a) =- log (6 xx 10^(-6)) = 5.2`
`pH_("an end point") = 7 +(1)/(2) (pK_(a) + logC)`
`= 7 +(1)/(2) (5.2 + log 0.05 ) = 8.95`