`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

Let V mL of `0.1M NaOH` and `0.1M HA` reacts to give NaA.
`{:(,NaOH+,HAhArr,NaA+,H_(2)O),("Meq. before reaction",0.1xxV,0.1xxV,0,0),("Meq. at end point",0,0,0.1xxV,):}`
`[NaA]=(0.1xxV)/(2V)=0.05M`
(Total volume becomes 2 V mL)
Now `0.05M` NaA solution undergoes hydrolysis:
`{:(A^(-)+H_(2)OhArr,OH^(-)+,HA),(1,0,0),(1-h,h,h):}`
`:. [OH^(-)]= C.h=C sqrt((K_(H))/(C ))`
`= sqrt((K_(w).C)/(K_(a)))=sqrt((10^(-4)xx0.05)/(5.6xx10^(-6)))`
`=9.45xx10^(-6)`
`:. pOH= -log[OH^(-)]= -log 9.45xx10^(-6)=5.025`
`:. pH= 14-pOH=14-5.025`
`=8.975`