Answer
pH = 8.841
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.
$Ka = \frac{[OH^-][C_6H_5N{H_3}^+]}{ [C_6H_5NH_2]}$
$Ka = 4 \times 10^{- 10}= \frac{x * x}{ 0.12}$
$Ka = 4 \times 10^{- 10}= \frac{x^2}{ 0.12}$
$ 4.8 \times 10^{- 11} = x^2$
$x = 6.928 \times 10^{- 6}$
Percent ionization: $\frac{ 6.928 \times 10^{- 6}}{ 0.12} \times 100\% = 0.005774\%$
%ionization < 5% : Right approximation.
Therefore: $[OH^-] = [C_6H_5N{H_3}^+] = x = 6.928 \times 10^{- 6}M $
3. Calculate the pH:
$pOH = -log[OH^-]$
$pOH = -log( 6.928 \times 10^{- 6})$
$pOH = 5.159$
$pH + pOH = 14$
$pH + 5.159 = 14$
$pH = 8.841$
