Chemistry 10th Edition

Published by Brooks/Cole Publishing Co.
ISBN 10: 1133610668
ISBN 13: 978-1-13361-066-3

Chapter 18 - Ionic Equilibria I: Acids and Bases - Exercises - The pH and pOH Scales - Page 744: 42

Answer

$[H_3O^+] = [C_6H_5COO^-] = 5.724 \times 10^{- 3}M $ $[C_6H_5COOH] \approx 0.52M$

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. -$[H_3O^+] = [C_6H_5COO^-] = x$ -$[C_6H_5COOH] = [C_6H_5COOH]_{initial} - x = 0.52 - x$ For approximation, we consider: $[C_6H_5COOH] = 0.52M$ 2. Now, use the Ka value and equation to find the 'x' value. $Ka = \frac{[H_3O^+][C_6H_5COO^-]}{ [C_6H_5COOH]}$ $Ka = 6.3 \times 10^{- 5}= \frac{x * x}{ 0.52}$ $Ka = 6.3 \times 10^{- 5}= \frac{x^2}{ 0.52}$ $ 3.276 \times 10^{- 5} = x^2$ $x = 5.724 \times 10^{- 3}$ Percent ionization: $\frac{ 5.724 \times 10^{- 3}}{0.52} \times 100\% = 1.101\%$ %ionization < 5% : Right approximation. Therefore: $[H_3O^+]$ and $[C_6H_5COO^-] = x = 5.724 \times 10^{- 3} $ And $[C_6H_5COOH] \approx 0.52M$
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