Answer
$[OH^-] = 3.223 \times 10^{- 3}M$
$pH = 11.508$
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.
$Kb = \frac{[OH^-][Conj. Acid]}{ [Methylamine]}$
$Kb = 5 \times 10^{- 4}= \frac{x * x}{ 0.024}$
$Kb = 5 \times 10^{- 4}= \frac{x^2}{ 0.024}$
$ 1.2 \times 10^{- 5} = x^2$
$x = 3.464 \times 10^{- 3}$
Percent ionization: $\frac{ 3.464 \times 10^{- 3}}{ 0.024} \times 100\% = 14.43\%$
%ionization > 5% : Inappropriate approximation, so, we will have to consider the '-x' in the base concentration:
$Kb = 5 \times 10^{- 4}= \frac{x^2}{ 0.024- x}$
$ 1.2 \times 10^{- 5} - 5 \times 10^{- 4}x = x^2$
$ 1.2 \times 10^{- 5} - 5 \times 10^{- 4}x - x^2 = 0$
$\Delta = (- 5 \times 10^{- 4})^2 - 4 * (-1) *( 1.2 \times 10^{- 5})$
$\Delta = 2.5 \times 10^{- 7} + 4.8 \times 10^{- 5} = 4.825 \times 10^{- 5}$
$x_1 = \frac{ - (- 5 \times 10^{- 4})+ \sqrt { 4.825 \times 10^{- 5}}}{2*(-1)}$
or
$x_2 = \frac{ - (- 5 \times 10^{- 4})- \sqrt { 4.825 \times 10^{- 5}}}{2*(-1)}$
$x_1 = - 3.723 \times 10^{- 3} (Negative)$
$x_2 = 3.223 \times 10^{- 3}$
- The concentration can't be negative, so $[OH^-]$ = $x_2 = 3.223 \times 10^{- 3}M$
3. Find the pH value.
$pOH = -log[OH^-]$
$pOH = -log( 3.223 \times 10^{- 3})$
$pOH = 2.492$
$pH + pOH = 14$
$pH + 2.492 = 14$
$pH = 11.508$
