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

Published by Cengage Learning

# Chapter 14 - Acids and Bases - Questions for Review and Thought - Topical Questions - Page 652c: 60

#### Answer

$pH = 8.835$

#### 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. -$[OH^-] = [C_6H_5N{H_3}^+] = x$ -$[C_6H_5NH_2] = [C_6H_5NH_2]_{initial} - x = 0.12 - x$ For approximation, we consider: $[C_6H_5NH_2] = 0.12M$ 2. Now, use the Kb value and equation to find the 'x' value. $Kb = \frac{[OH^-][C_6H_5N{H_3}^+]}{ [C_6H_5NH_2]}$ $Kb = 3.9 \times 10^{- 10}= \frac{x * x}{ 0.12}$ $Kb = 3.9 \times 10^{- 10}= \frac{x^2}{ 0.12}$ $4.68 \times 10^{- 11} = x^2$ $x = 6.84 \times 10^{- 6}$ Percent ionization: $\frac{ 6.84 \times 10^{- 6}}{ 0.12} \times 100\% = 0.005701\%$ %ionization < 5% : Right approximation. Therefore: $[OH^-] = [C_6H_5N{H_3}^+] = x = 6.84 \times 10^{- 6}M$ $[C_6H_5NH_2] \approx 0.12M$ 3. Calculate the pH $pOH = -log[OH^-]$ $pOH = -log( 6.84 \times 10^{- 6})$ $pOH = 5.165$ $pH + pOH = 14$ $pH + 5.165 = 14$ $pH = 8.835$

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