Answer
$[H^+] = 7.58 \times 10^{- 13}M$
$pH = 12.12$
$pOH = 1.88$
$[OH^-] = 1.3 \times 10^{-2}M$
Work Step by Step
$Kb (CH_3NH_2): 4.4 \times 10^{-4}$
1. Drawing the ICE table, we get:
-$[OH^-] = [B^+] = x$
-$[BOH] = [BOH]_{initial} - x$
For approximation, we consider: $[BOH] = [BOH]_{initial}$
2. Now, use the Kb value and equation to find the 'x' value.
$Kb = \frac{[OH^-][B^+]}{ [BOH]}$
$Kb = 4.4 \times 10^{- 4}= \frac{x * x}{ 3.86 \times 10^{- 1}}$
$Kb = 4.4 \times 10^{- 4}= \frac{x^2}{ 3.86 \times 10^{- 1}}$
$ 1.69 \times 10^{- 4} = x^2$
$x = 1.3 \times 10^{- 2}$
5% test: $\frac{ 1.3 \times 10^{- 2}}{ 3.86 \times 10^{- 1}} \times 100\% = 3.37\%$
%ionization < 5% : Right approximation.
Therefore: $[OH^-] = 1.3 \times 10^{-2}M$
3. Find $[H_3O^+]$, pH and pOH:
$pOH = -log[OH^-]$
$pOH = -log( 1.3 \times 10^{- 2})$
$pOH = 1.88$
$pH + pOH = 14$
$pH + 1.88 = 14$
$pH = 12.12$
$[H^+] = 10^{-pH}$
$[H^+] = 10^{- 12.12}$
$[H^+] = 7.58 \times 10^{- 13}$
