Answer
$pH = 11.871$
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.
-$[OH^-] = [(C_2H_5)_2N{H_2}^+] = x$
-$[(C_2H_5)_2NH] = [(C_2H_5)_2NH]_{initial} - x = 0.05 - x$
For approximation, we consider: $[(C_2H_5)_2NH] = 0.05M$
2. Now, use the Kb value and equation to find the 'x' value.
$Ka = \frac{[OH^-][(C_2H_5)_2N{H_2}^+]}{ [(C_2H_5)_2NH]}$
$Ka = 1.3 \times 10^{- 3}= \frac{x * x}{ 0.05}$
$Ka = 1.3 \times 10^{- 3}= \frac{x^2}{ 0.05}$
$ 6.5 \times 10^{- 5} = x^2$
$x = 8.062 \times 10^{- 3}$
Percent ionization: $\frac{ 8.062 \times 10^{- 3}}{ 0.05} \times 100\% = 16.12\%$
%ionization < 5% : Inappropriate approximation, so, we will have to consider the '-x' in the acid concentration:
$Ka = 1.3 \times 10^{- 3}= \frac{x^2}{ 0.05- x}$
$ 6.5 \times 10^{- 5} - 1.3 \times 10^{- 3}x = x^2$
$ 6.5 \times 10^{- 5} - 1.3 \times 10^{- 3}x - x^2 = 0$
$\Delta = (- 1.3 \times 10^{- 3})^2 - 4 * (-1) *( 6.5 \times 10^{- 5})$
$\Delta = 1.69 \times 10^{- 6} + 2.6 \times 10^{- 4} = 2.617 \times 10^{- 4}$
$x_1 = \frac{ - (- 1.3 \times 10^{- 3})+ \sqrt { 2.617 \times 10^{- 4}}}{2*(-1)}$
or
$x_2 = \frac{ - (- 1.3 \times 10^{- 3})- \sqrt { 2.617 \times 10^{- 4}}}{2*(-1)}$
$x_1 = - 8.738 \times 10^{- 3} (Negative)$
$x_2 = 7.438 \times 10^{- 3}$
- The concentration can't be negative, so $[OH^-]$ = $x_2$
3. Calculate the pH value:
$pOH = -log[OH^-]$
$pOH = -log( 7.438 \times 10^{- 3})$
$pOH = 2.129$
$pH + pOH = 14$
$pH + 2.129 = 14$
$pH = 11.871$
