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Answers to all problems are at the end of this book. Detailed solutions are available in the Student Solutions Manual, Study Guide, and Problems Book . How to Prepare a Triethanolamine Buffer Shown here is the structure of triethanolamine in its fully protonated form: Its p K a is 7.8. You have available at your kb bench 0.1 M solutions of HO, NaOH, and the uncharged (free base) form of triethanolamine, as well as ample distilled water. Describe the preparation of a 1-L solution of 0.05 M triethanolamine buffer, pH 7.6.

BuyFind

Biochemistry

6th Edition
Reginald H. Garrett + 1 other
Publisher: Cengage Learning
ISBN: 9781305577206
BuyFind

Biochemistry

6th Edition
Reginald H. Garrett + 1 other
Publisher: Cengage Learning
ISBN: 9781305577206

Solutions

Chapter
Section
Chapter 2, Problem 15P
Textbook Problem
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Answers to all problems are at the end of this book. Detailed solutions are available in the Student Solutions Manual, Study Guide, and Problems Book.

Chapter 2, Problem 15P, Answers to all problems are at the end of this book. Detailed solutions are available in the Student

How to Prepare a Triethanolamine Buffer Shown here is the structure of triethanolamine in its fully protonated form:

Its pKa is 7.8. You have available at your kb bench 0.1 M solutions of HO, NaOH, and the uncharged (free base) form of triethanolamine, as well as ample distilled water. Describe the preparation of a 1-L solution of 0.05 M triethanolamine buffer, pH 7.6.

Expert Solution
Interpretation Introduction

Describe the preparation of triethanolamine buffer

Introduction:

Triethanolamine free base contains an unprotonated tertiary nitrogen. The pH of the free base solution is too basic. Therefore, to adjust the pH, HCl have to be added.

Explanation of Solution

Given information:

Concentration of triethanolamine = 0.1 M

Formula used:

  pH=pKa+log[Triethanolamine][Triethanolamine-H]

  [Triethanolamine]= concentration of unprotonated tertiary amine form

  TriethanolamineH= protonated tertiary amine form

Calculation:

Used triethanolamine (0.1 M) volume =0.05×03/0.1=500mL

  pH=pKa+log[Triethanolamine][Triethanolamine-H]7.6=7.8+log[Triethanolamine][Triethanolamine-H]log[Triethanolamine][Triethanolamine-H]=0

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