Part CIn Part A, a 35.00 ml aliquot of a 0.195 mol L 2-methylbutanoic acid solution was titrated to its equivalence point with 83.2 ml of 0.108 mol L- NaOH solution. At the equivalence point, all of the weak acid, 2-methylbutanoicacid, is converted to its weak conjugate base, 2-methylbutanoate. What is the pH at this equivalence point?O 4.800O 9.200O 8.821O 7.087O 5.179 2-Methylbutanoic acid is a weak organic acid that exists as two different optical isomers. One hasa sweet, fruity odour, and the other has a cheesy, sweaty odour. Like other small carboxylic acids,it is used to make pleasant smelling esters that are used in perfumes, cosmetics, and foodadditives.(5)-2-methylbutanoic (RI-2-methylbutanoicacidacidThe K, of 2-methylbutanoic acid is 1.58x10-5.

First we will draw ICE table of the salt formed, then calculate pH using Kb value.

Part C In Part A, a 35.00 ml aliquot of a 0.195 mol L 2-methylbutanoic acid solution was titrated to its equivalence point with 83.2 ml of 0.108 mol L- NaOH solution. At the equivalence point, all of the weak acid, 2-methylbutanoic acid, is converted to its weak conjugate base, 2-methylbutanoate. What is the pH at this equivalence point? O 4.800 O 9.200 O 8.821 O 7.087 O 5.179

Part C In Part A, a 35.00 ml aliquot of a 0.195 mol L 2-methylbutanoic acid solution was titrated to its equivalence point with 83.2 ml of 0.108 mol L- NaOH solution. At the equivalence point, all of the weak acid, 2-methylbutanoic acid, is converted to its weak conjugate base, 2-methylbutanoate. What is the pH at this equivalence point? O 4.800 O 9.200 O 8.821 O 7.087 O 5.179

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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