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

Published by Cengage Learning

# Chapter 15 - Solutions of Acids and Bases - Questions and Exercises - Exercises - Page 676: 15.75

#### Answer

$[OH^-] = 2.972 \times 10^{- 3}M$ $pH = 11.473$

#### 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. -$[OH^-] = [Conj. Acid] = x$ -$[Methylamine] = [Methylamine]_{initial} - x = 0.024 - x$ For approximation, we consider: $[Methylamine] = 0.024M$ 2. Now, use the Kb value and equation to find the 'x' value. $Kb = \frac{[OH^-][Conj. Acid]}{ [Methylamine]}$ $Kb = 4.2 \times 10^{- 4}= \frac{x * x}{ 0.024}$ $Kb = 4.2 \times 10^{- 4}= \frac{x^2}{ 0.024}$ $1.008 \times 10^{- 5} = x^2$ $x = 3.175 \times 10^{- 3}$ Percent ionization: $\frac{ 3.175 \times 10^{- 3}}{ 0.024} \times 100\% = 13.23\%$ Since the percent ionization is more than 5 percent, this is a bad approximation. Thus, we find: $Ka = 4.2 \times 10^{- 4}= \frac{x^2}{ 0.024- x}$ $1.008 \times 10^{- 5} - 4.2 \times 10^{- 4}x = x^2$ $1.008 \times 10^{- 5} - 4.2 \times 10^{- 4}x - x^2 = 0$ $\Delta = (- 4.2 \times 10^{- 4})^2 - 4 * (-1) *( 1.008 \times 10^{- 5})$ $\Delta = 1.764 \times 10^{- 7} + 4.032 \times 10^{- 5} = 4.05 \times 10^{- 5}$ $x_1 = \frac{ - (- 4.2 \times 10^{- 4})+ \sqrt { 4.05 \times 10^{- 5}}}{2*(-1)}$ or $x_2 = \frac{ - (- 4.2 \times 10^{- 4})- \sqrt { 4.05 \times 10^{- 5}}}{2*(-1)}$ $x_1 = - 3.392 \times 10^{- 3} (Negative)$ $x_2 = 2.972 \times 10^{- 3}$ - The concentration can't be negative, so $[OH^-]$ = $x_2$ 3. Now, calculate the pH: $pOH = -log[OH^-]$ $pOH = -log( 2.972 \times 10^{- 3})$ $pOH = 2.527$ $pH + pOH = 14$ $pH + 2.527 = 14$ $pH = 11.473$

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