## Chemistry: Principles and Practice (3rd Edition)

Published by Cengage Learning

# Chapter 15 - Solutions of Acids and Bases - Questions and Exercises - Exercises - Page 675: 15.59

#### Answer

$Ka = 1.397\times 10^{- 5}$

#### 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. -$[H_3O^+] = [CH_3CH_2COO^-] = x$ -$[CH_3CH_2COOH] = [CH_3CH_2COOH]_{initial} - x$ 2. Calculate the $[H_3O^+]$ $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 2.93}$ $[H_3O^+] = 1.175 \times 10^{- 3}$ Therefore : $x = 1.175 \times 10^{-3}$ 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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