# Chapter 15 - Solutions of Acids and Bases - Questions and Exercises - Exercises - Page 676: 15.73

$pH \approx 12.3$

#### Work Step by Step

1. Calculate the Kb value: $K_b = 10^{-pKb}$ $K_b = 10^{- 3.1}$ $K_b = 7.943 \times 10^{- 4}$ 2. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium: ** The image is in the end of this answer. -$[OH^-] = [C_8H_{17}NH^+] = x$ -$[C_8H_{17}N] = [C_8H_{17}N]_{initial} - x = 0.5 - x$ For approximation, we consider: $[C_8H_{17}N] = 0.5M$ 3. Now, use the Kb value and equation to find the 'x' value. $Kb = \frac{[OH^-][C_8H_{17}NH^+]}{ [C_8H_{17}N]}$ $Kb = 7.943 \times 10^{- 4}= \frac{x * x}{ 0.5}$ $Kb = 7.943 \times 10^{- 4}= \frac{x^2}{ 0.5}$ $3.972 \times 10^{- 4} = x^2$ $x = 1.993 \times 10^{- 2}$ Percent ionization: $\frac{ 1.993 \times 10^{- 2}}{ 0.5} \times 100\% = 3.986\%$ Since the percent ionization is less than 5%, this is the right approximation. Therefore: $[OH^-] = [C_8H_{17}NH^+] = x = 1.993 \times 10^{- 2}M$ $[C_8H_{17}N] \approx 0.5M$ 4. Calculate the pH: $pOH = -log[OH^-]$ $pOH = -log( 0.01993)$ $pOH = 1.701$ $pH + pOH = 14$ $pH + 1.701 = 14$ $pH = 12.299$

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