Chemistry: A Molecular Approach (3rd Edition)

Published by Prentice Hall
ISBN 10: 0321809246
ISBN 13: 978-0-32180-924-7

Chapter 15 - Sections 15.1-15.12 - Exercises - Problems by Topic - Page 747: 73d

Answer

Percent ionization = $1.897 \%$

Work Step by Step

1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer. -$[H_3O^+] = [Conj. Base] = x$ -$[Acetic Acid] = [Acetic Acid]_{initial} - x = 0.05 - x$ For approximation, we consider: $[Acetic Acid] = 0.05M$ 2. Now, use the Ka value and equation to find the 'x' value. $Ka = \frac{[H_3O^+][Conj. Base]}{ [Acetic Acid]}$ $Ka = 1.8 \times 10^{- 5}= \frac{x * x}{ 0.05}$ $Ka = 1.8 \times 10^{- 5}= \frac{x^2}{ 0.05}$ $ 9 \times 10^{- 7} = x^2$ $x = 9.487 \times 10^{- 4}$ Percent ionization: $\frac{ 9.487 \times 10^{- 4}}{ 0.05} \times 100\% = 1.897\%$ %ionization < 5% : Right approximation.
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