0.1 M HA is tritrated against 0.1 M NaOH . Find the pH the end point . Dissociation constant for the end acid HA is `5 xx 10^(-6)` and degree of hydrolysis , `h lt 1 `

Explanation

`underset("weak")HA+NaOH rarr NaA+H_(2)O`
At the end point, their equivalent amounts react together
`:.` In the final solution. `[NaA]=(0.1)/(2)=0.05M`
As NaA is a salt of weak acid and strong base, it hydrolyses as
`a^(-) + H_(2)O hArr HA+ OH^(-)`
For such a salt,
`pH = 7 + (1)/(2) [pK_(a) + log c]=7+(1)/(2) [-log (5xx10^(-6))+log 0.05]`
`=7+(1)/(2) [ 6-0.6990+0.6990-2]=9`