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:
$pOH = -log[OH^-]$
$pOH = -log( 2.972 \times 10^{- 3})$
$pOH = 2.527$
$pH + pOH = 14$
$pH + 2.527 = 14$
$pH = 11.473$
