Question

Weak-Acid Strong-Base Titrations. These next questions relate to a 25 mL aliquot of 0.35 M acetic acid (Ka = 1.77 x 10) that is titrated with 0.20 M potassium hydroxide (KOH). (f) What is the pH of the acetic acid solution before the titration begins? (g) What is the pH after 14 mL of 0.20 M KOH has been added to the solution? Use the Henderson-Hasselbalch equation. (h) What is the pH at the equivalence point?

Answer 1

413.75+ lug (8.35)] = 1,65 +) pH = b[ pka +logc) pka = -log(ka) = 409 (1.77X164) = 3.75 2om kot addition | Aaa7 - 4)^3+1) 0.35 x 25 0.224 M (25 +14 ) [koh} = my hoy → og box things 0.0718 m Acid left = 0.224 -0.0118 = 0.1522 m pH= pka +log ( Salty Ci Here salt = koH concentration ) = 3.75 +log ( 20.07518) aud = 3:42 h) At equivalence point MiV EML V2 0:35x25 0:35 x 25 = 0.20 xV2 Vy 0.2 43.75 ml 0.20x43-75 25+43-75) →0.127 M pH = 7+ *[pka+ logc] 57+ [3.75+ log(0:127) S 8.43