Chemistry (4th Edition)

Published by McGraw-Hill Publishing Company
ISBN 10: 0078021529
ISBN 13: 978-0-07802-152-7

Chapter 16 - Questions and Problems - Page 772: 16.56

Answer

pH = 5.18

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. -$[H_3O^+] = [C_6H_5O^-] = x$ -$[C_6H_5OH] = [C_6H_5OH]_{initial} - x = 0.34 - x$ For approximation, we consider: $[C_6H_5OH] = 0.34M$ 2. Now, use the Ka value and equation to find the 'x' value. $Ka = \frac{[H_3O^+][C_6H_5O^-]}{ [C_6H_5OH]}$ $Ka = 1.3 \times 10^{- 10}= \frac{x * x}{0.34}$ $Ka = 1.3 \times 10^{- 10}= \frac{x^2}{ 0.34}$ $ 4.42 \times 10^{- 11} = x^2$ $x = 6.64 \times 10^{- 6}$ 5% test: $\frac{ 6.64 \times 10^{- 6}}{ 0.34} \times 100\% = 0.00196\%$ %ionization < 5% : Right approximation. Therefore: $[H_3O^+] = x = 6.64 \times 10^{- 6} $ 3. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 6.64 \times 10^{- 6})$ $pH = 5.18$
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