Answer
$pH \approx 12.3$
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:
$pOH = -log[OH^-]$
$pOH = -log( 0.01993)$
$pOH = 1.701$
$pH + pOH = 14$
$pH + 1.701 = 14$
$pH = 12.299$
