## Chemistry: The Molecular Science (5th Edition)

Published by Cengage Learning

# Chapter 14 - Acids and Bases - Questions for Review and Thought - Topical Questions - Page 652c: 58

#### Answer

$[H_3O^+] = [CH_3COO^-] \approx 1.9 \times 10^{- 3}M$ $[CH_3COOH] \approx 0.2M$

#### Work Step by Step

$K_a(CH_3COOH) = 1.8 \times 10^{-5}$ 1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer. -$[H_3O^+] = [CH_3COO^-] = x$ -$[CH_3COOH] = [CH_3COOH]_{initial} - x = 0.2 - x$ For approximation, we consider: $[CH_3COOH] = 0.2M$ 2. Now, use the Ka value and equation to find the 'x' value. $Ka = \frac{[H_3O^+][CH_3COO^-]}{ [CH_3COOH]}$ $Ka = 1.8 \times 10^{- 5}= \frac{x * x}{ 0.2}$ $Ka = 1.8 \times 10^{- 5}= \frac{x^2}{ 0.2}$ $3.6 \times 10^{- 6} = x^2$ $x = 1.897 \times 10^{- 3}$ Percent ionization: $\frac{ 1.897 \times 10^{- 3}}{ 0.2} \times 100\% = 0.9487\%$ %ionization < 5% : Right approximation. Therefore: $[H_3O^+] = [CH_3COO^-] = x = 1.897 \times 10^{- 3}M$ And, since 'x' has a very small value (compared to the initial concentration): $[CH_3COOH] \approx 0.2M$

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