Chemistry: Principles and Practice (3rd Edition)

Published by Cengage Learning
ISBN 10: 0534420125
ISBN 13: 978-0-53442-012-3

Chapter 15 - Solutions of Acids and Bases - Questions and Exercises - Exercises - Page 675: 15.57

Answer

$Ka = 1.434\times 10^{- 5}$ $pKa = 4.843$

Work Step by Step

1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium: ** The image is at the end of this answer. -$[H_3O^+] = [Conj. Base] = x$ -$[Caproic acid] = [Caproic acid]_{initial} - x$ 2. Calculate the $[H_3O^+]$ $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 3.43}$ $[H_3O^+] = 3.715 \times 10^{- 4}$ Therefore: $x = 3.715 \times 10^{- 4}$ 3. Write the Ka equation, and find its value: $Ka = \frac{[H_3O^+][Conj. Base]}{ [Caproic acid]}$ $Ka = \frac{x^2}{[InitialCaproic acid] - x}$ $Ka = \frac{( 3.715\times 10^{- 4})^2}{ 0.01- 3.715\times 10^{- 4}}$ $Ka = \frac{ 1.38\times 10^{- 7}}{ 9.628\times 10^{- 3}}$ $Ka = 1.434\times 10^{- 5}$ 4. Calculate the pKa Value $pKa = -log(Ka)$ $pKa = -log( 1.434 \times 10^{- 5})$ $pKa = 4.843$
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