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» <html><body><p></p>Solution :`<a href="https://interviewquestions.tuteehub.com/tag/underset-3243992" style="font-weight:bold;" target="_blank" title="Click to know more about UNDERSET">UNDERSET</a>("weak")HA+NaOH rarr <a href="https://interviewquestions.tuteehub.com/tag/naa-572547" style="font-weight:bold;" target="_blank" title="Click to know more about NAA">NAA</a>+H_(2)O` <br/> At the <a href="https://interviewquestions.tuteehub.com/tag/end-971042" style="font-weight:bold;" target="_blank" title="Click to know more about END">END</a> <a href="https://interviewquestions.tuteehub.com/tag/point-1157106" style="font-weight:bold;" target="_blank" title="Click to know more about POINT">POINT</a>, theirequivalent <a href="https://interviewquestions.tuteehub.com/tag/amounts-374805" style="font-weight:bold;" target="_blank" title="Click to know more about AMOUNTS">AMOUNTS</a> react together <br/> `:.` In thefinal solution. `[NaA]=(0.1)/(2)=0.05M` <br/> AsNaA is a salt of weak acid and strong base, it hydrolyses as <br/> `a^(-) + H_(2)O hArr HA+ OH^(-)` <br/> For such a salt, <br/> `pH = 7 + (1)/(2) [pK_(a) + log c]=7+(1)/(2) [-log(5xx10^(-6))+log 0.05]` <br/>`=7+(1)/(2) [ 6-0.6990+0.6990-2]=9`</body></html>