Prairie View A & M Univer x If a buffer solution is 0.43( X C WWW sapling learni

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Prairie View A & M Univer x If a buffer solution is 0.43( X C WWW sapling learning.com /ibiscms/mod/ibis/view.php?id-1973899 O 9/21/2015 06:00 PM CA 91.1/100 9/12/2015 04:18 AM Gradebook Attempts Score Print Calculator Periodic Table Questi 21 of 27 Map You have 225 mL of an 0.41 M acetic acid solution. What volume (vy of 1.90 M NaOH solution must you add in order to prepare an acetate buffer of pH 5.22? The pKa of acetic acid is 4.76 Number mL, NaO 99 20 95 23 24 25 90 26 95 Previous Give Up & View Solution Check Answer Next Exit 27 Copyright 2011 2015 Sapling Learning, Inc. 113 about us careers partners privacy policy terms of use tact us help a Search the web and Windows Assignment Information Available From: 8/24/2015 06:00 PM 9/21/2015 06:00 PM Due Date Points Possible 100 Grade Category: Graded Description: Policies: Homework n check your answers. n view solutions when you complete or give up on any question You can keep trying to answer each question until you get it right or give up You lose 5% of the points available to each answer in your question for each incorrect a at that O eTextbook OHelp With This Topic Web Help & Videos O Technical Support and Bug Reports 35 PM 9/15/2015

Explanation / Answer

CH3COOH + NaOH ---> CH3COONa + H2O
All NaOH will react to form CH3COONa.

Let the volume of NaOH be v mL
Then,
number of moles of NaOH = 1.9 * v mmol=19v mmol
number of moles of CH3COOH =M*V = 0.41*225 = 92.25 mmol

19v mmol of each will react to form 19v mmol of CH3COONa.

After reaction:
mol of CH3COONa = 19v mmol
mol of CH3COOH = 92.25 - 19 v   mmol
Total volume = 225+v mL

[CH3COONa] = mol/ volume = 19v / (225+v)
[CH3COOH] = mol/ volume = (92.25-19v) / (225+v)

use:
pH = pKa + log([CH3COONa] / [CH3COOH])
5.22 = 4.76 + log (19v / (225+v) / (92.25-19v) / (225+v))
5.22 = 4.76 + log (19v / (92.25-19v))
log (19v / (92.25-19v)) = 0.46
19v / (92.25-19v) = 2.884
19v = 266.05 - 54.8 v
v = 3.6 mL

Answer: 3.6 mL

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