Question

A 15.44 mL sample of acetic acid requires 32.10 mL of 0.0500 M Ba(OH)2 to reach the endpoint.

a) Calculate the molarity of the unknown acid

b) what is the theoretical pH when 15.8 mL of Ba(OH)2 is added?

c) What is the theoretical pH at the point when a total of 32.10 mL Ba(OH)2 is added?

Answer 1

a) at end point, 2chcoon & Balon ) Balthcool + 2H20 2 zx moles Bacon? moles chroom » (охvси, свои Lх Ско осон), 2) ма соон * 15-49 mt - Lх о-о5оом х 32-toml 2) "th cooит | o- 208 м)

mmoles of Bacon, added 2 (mxu) 2 0-05oom x 15.8ml 2 0-790 mmoles. mmoles of a 2x0-790 2 1-58 mmoles. mnoles acene acda 0-208 m x 15-44 mi - Ou) 2 2X0-05com 3210 ml 93-21 mmoles zenycoon & zout 2 Ch 00 + 2H2O 3.21 mmoles 1-58 mnoles - 1.58 -1.58 11:58 1-63 mmoles o r 1.58 mmoley Solution contains both C₂cool and Chhooo, so 2 acts as buffer pica of chroon 24-ty ph a plea & log (chor] -cthcool a 4-tyt logf1-58 mnotes 1-63 mmoles lur 4-73. 2 c) 32-10mL Cifcoon is end convert point so to chroo all Ccrtcoo) 2 mmoles 24 (ooh Total volume 3.21 mmoles 15-44mL + 32-10 ml 0.0675 m. 2

e Cocoon ton CHC05 +H2O 0-065 м 3 <ь с соо 25-6x1010 Kyzko 10x1014 -8 х 10- -Х e o-o6 55 -х , neglect x as ky clo 6-б35 ке (с, ссон? Сон) Carla coo) 5-681610 a x² ут 2 Сог. 5-6 xid10хо-свэл 2 с. x 10 м Ә) рок 2 - tes (61; 16%) -21 2) Ри 14-оо -5-21 | 8-99 )