General Chemistry: Principles and Modern Applications (10th Edition)

Published by Pearson Prentice Hal
ISBN 10: 0132064529
ISBN 13: 978-0-13206-452-1

Chapter 16 - Acids and Bases - Example 16-12 - Evaluating Ionization Constants for Hydrolysis Reactions - Page 726: Practice Example A

Answer

The codeine hydrochloride solution has higher pH.

Work Step by Step

Cocaine: $$K_b = 10^{-pKb} = 10^{-8.41} = 3.9 \times 10^{-9} $$ $$K_a (conj. acid)= \frac{1.0 \times 10^{-14}}{K_b} = \frac{1.0 \times 10^{-14}}{ 3.9 \times 10^{-9} } = 2.6 \times 10^{-6} $$ Codeine: $$K_b = 10^{-pKb} = 10^{-7.95} = 1.1 \times 10^{-8} $$ $$K_a(conj. acid) = \frac{1.0 \times 10^{-14}}{K_b} = \frac{1.0 \times 10^{-14}}{ 1.1 \times 10^{-8} } = 9.1 \times 10^{-7} \space$$ Codeine hydrochloride has the lower $K_a$, thus, the lower $H_3O^+$ concentration in solution, which means higher pH.
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