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: 57

Answer

$K_a (CH_3CH_2COOH) \approx 1.4\times 10^{- 5}$

Work Step by Step

1. Calculate the hydronium concentration: $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 2.93}$ $[H_3O^+] = 1.175 \times 10^{- 3}M$ 2. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer. -$[H_3O^+] = [CH_3CH_2COO^-] = x = 1.175 \times 10^{-3}M$ -$[CH_3CH_2COOH] = [CH_3CH_2COOH]_{initial} - x$ 3. Write the Ka equation, and find its value: $Ka = \frac{[H_3O^+][CH_3CH_2COO^-]}{ [CH_3CH_2COOH]}$ $Ka = \frac{x^2}{[InitialCH_3CH_2COOH] - x}$ $Ka = \frac{( 1.175\times 10^{- 3})^2}{ 0.1- 1.175\times 10^{- 3}}$ $Ka = \frac{ 1.38\times 10^{- 6}}{ 0.09883}$ $Ka = 1.397\times 10^{- 5}$
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