`0.1M NaOH` is titrated with `0.1M HA` till the end point. `K_(a)` of HA is `5.6xx10^(-6)` and degree of hydrolysis is less compared to 1. Calculate pH of the resulting solution at the end point ?

Explanation

`{:(NaOH+,HArarr,NaA+,H_(2)O),(0.1,0.1,0,0),(0,0,0.1,0.1):}`
`[NaA] = (0.1)/(2)` because volume is doubled on mixing.
`NaA` is salt of weak acid `HA` and stron base `NaOH`.
For such salt solution:
`pH = 7 +(1)/(2) pK_(a) +(1)/(2)logC`
`= 7+(1)/(2)(-log 5.6 xx 10^(-5))+(1)/(2)log((0.1)/(2))`
`= 7+(1)/(2) (5-log 5.6) +(-0.65)`
`= 7 +(5.25)/(2) - 0.65 = 8.975`