Chemistry: The Molecular Science (5th Edition)

Published by Cengage Learning
ISBN 10: 1285199049
ISBN 13: 978-1-28519-904-7

Chapter 14 - Acids and Bases - Questions for Review and Thought - Topical Questions - Page 652c: 56

Answer

$K_a$(Butyric acid) $\approx 1.6\times 10^{- 5}$

Work Step by Step

1. Calculate the hydronium concentration: $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 3.21}$ $[H_3O^+] = 6.166 \times 10^{- 4}$ 2. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer. -$[H_3O^+] = [Conj. Base] = x = 6.166 \times 10^{-4}$ -$[Butyric Acid] = [Butyric Acid]_{initial} - x$ 3. Write the Ka equation, and find its value: $Ka = \frac{[H_3O^+][Conj. Base]}{ [Butyric Acid]}$ $Ka = \frac{x^2}{[InitialButyric Acid] - x}$ $Ka = \frac{( 6.166\times 10^{- 4})^2}{ 0.025- 6.166\times 10^{- 4}}$ $Ka = \frac{ 3.802\times 10^{- 7}}{ 0.02438}$ $Ka = 1.559\times 10^{- 5}$
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