0.01 M solution of an organic acid is found to have a pH of 4.15. Calculate the concentration of the anion, the ionization constant of the acid and its pK_(a).

0.01 M solution of an organic acid is found to have a pH of 4.15. Calc

1.	0.01 M solution of an organic acid is found to have a pH of 4.15. Calculate the concentration of the anion, the ionization constant of the acid and its pK_(a).
Answer» <html><body><p><br/></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/ph-1145128" style="font-weight:bold;" target="_blank" title="Click to know more about PH">PH</a> = 4.15 means `- log [<a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</a>^(+)] = 4.15 or log [H^(+)] = - 4.15 = bar(5) . 85 or [H^(+)] = 7.08 xx <a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>^(-5)M` <br/> `<a href="https://interviewquestions.tuteehub.com/tag/ha-479237" style="font-weight:bold;" target="_blank" title="Click to know more about HA">HA</a> hArr H^(+) + A^(-)` <br/> <a href="https://interviewquestions.tuteehub.com/tag/hence-484344" style="font-weight:bold;" target="_blank" title="Click to know more about HENCE">HENCE</a>, at equilibrium `[H^(+)]=[A^(-)] = 7.08 xx 10^(-5)M ~= 7.1 xx 10^(-5)M` <br/> `[HA] = (0.01-7.1xx10^(-5))=(0.01-0.000071)M = 0.009929 M` <br/> `K_(a) = ([H^(+)][A^(-)])/([HA])=((7.1xx10^(-5))^(2))/(9.929xx10^(-3))=5.08 xx 10^(-7)` <br/> `pK_(a) = - log K_(a) = - log (5.08 xx 10^(-7))= 6.29`.</body></html>