0.1 M HA is titrated against 0.1 M NaOH. Find the pH at the end point. Dissociation constant for the acid HA is 5xx10^(-6) and the degree of hydrolysis, h lt 1.

0.1 M HA is titrated against 0.1 M NaOH. Find the pH at the end point.

1.	0.1 M HA is titrated against 0.1 M NaOH. Find the pH at the end point. Dissociation constant for the acid HA is 5xx10^(-6) and the degree of hydrolysis, h lt 1.
Answer» Solution :`UNDERSET("weak")HA+NaOH rarr NAA+H_(2)O` At the END POINT, theirequivalent AMOUNTS react together `:.` In thefinal solution. `[NaA]=(0.1)/(2)=0.05M` AsNaA 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`