Question

Download the graphing paper posted on Moodle under COURSE MATERIALS/Handouts. Plots utilizing free-hand sketched and/or not-to-scale axes will not be graded. Period. Consider a titration of a 25.00 mL propionic acid solution [KCH COOH) = 1.3x10-1 with 0.1093 M solution of sodium hydroxide. The volume of 23.56 mL of NaOH(aq) solution was needed to reach the equivalence point. Calculate (SHOW ALL WORK for full marks): a) The pH of the propionic acid solution before the titration. b) The pH after 5.89 mL of NaOH were added. c) The pH of the solution at the half-equivalence point. d) The pH after 12.00 mL of NaOH were added. e) The pH of the solution after 17.00 mL of the NaOH solution were added. 1) The pH of the solution after 20.00 mL of the NaOH solution were added. g) The pH of the solution after 23.25 mL of the NaOH solution were added. h) The pH of the solution at the equivalence point. i) The pH of the solution after 1.00 mL of NaOH solution was added after the equivalence point. j) Draw the titration curve (use graphing paper posted on Moodle under COURSE MATERIALS/Handouts).

Hello! please help me! need to know the answers to see If I’m right before i turn this in :)

Answer 1

Bokione acid 25mL of propionic acid 0.1093 mol of 23.56 mk Nooh 2.5 75 mmol NaOH ka (acid) = 1:3x10-5 pka = 4.89

COOH + MapH Naot oct H₂O d) 12 mx of 0.1093 mol Naos ca Mivi= M.UZ aid Banc X 25ML = Oe1893 M X 23. 56 my Mavid- 0.1030M 1.3116 mmol Nook NA-=1.3 116 mmol NHA = 2.575-1.3116 mor = 1.2634 mmol pH = 4.89 tlog / 1.3116) 1.2634 (₂ HgCool » SHgloo tht egno c-ca ca Cd TpH = 4.91 Ka=cd² d = 1 1 3x16 5 = 0.01123 e) 17ml of 0.1093 mol Moon OoBo (H+) = CH = -logolo. 2) = 1157x10 1.858 mmol SMA THA = 2.575-1.8561= 0.7169 pH = -log[et] [pH = 2.937) - 6) 5.89 ml of 0.1093 mol NaOH .O. 7169 I pH = 4.89 + log (1958) pH = 5:30 f) zome of 0.1093 mol Naoh - acid. I nu = 2.575-7 0 0.389 mmol 0.6438 mmol OH- 25ml of 0.1030 mol and 72.575mmal 2.186 mmol OH ANA I NHA = 2.575-21 186 7 8.389 mmol C₂ H glooH + NaOH CH3COONa+H₂O na tloot.2.166 pH = 4.89 +logi too 2.575 0.6438 epimi (2.575-0.6436) o 016430 [PH = 5.641 1.9312mmal 9 23.25 mL of 0.1093 mal nach PH= pka+log [Cathy Coonal 2.54 12 mmol = NA- NHA = 2.575-2.5412 = 0.0338mmal - 4.89+10g (0.6436) pH = 4.89+109 ( 2164323) 1.9312 (pH = 6.766 1pH = 4,4131 Then pH was found to be f. 766 c) At half eq.pt. [A7=1 HA) (salt) (and) pH=pka PH = 4.89 L TCGHS Cook