Answer
$pH = 12.013$
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.
-$[OH^-] = [C_2H_5N{H_3}^+] = x$
-$[C_2H_5NH_2] = [C_2H_5NH_2]_{initial} - x = 0.2 - x$
For approximation, we consider: $[C_2H_5NH_2] = 0.2M$
2. Now, use the Kb value and equation to find the 'x' value.
$Ka = \frac{[OH^-][C_2H_5N{H_3}^+]}{ [C_2H_5NH_2]}$
$Ka = 5.6 \times 10^{- 4}= \frac{x * x}{ 0.2}$
$Ka = 5.6 \times 10^{- 4}= \frac{x^2}{ 0.2}$
$ 1.12 \times 10^{- 4} = x^2$
$x = 1.058 \times 10^{- 2}$
Percent ionization: $\frac{ 1.058 \times 10^{- 2}}{ 0.2} \times 100\% = 5.292\%$
%ionization < 5% : Inappropriate approximation, so, we will have to consider the '-x' in the acid concentration:
$Ka = 5.6 \times 10^{- 4}= \frac{x^2}{ 0.2- x}$
$ 1.12 \times 10^{- 4} - 5.6 \times 10^{- 4}x = x^2$
$ 1.12 \times 10^{- 4} - 5.6 \times 10^{- 4}x - x^2 = 0$
$\Delta = (- 5.6 \times 10^{- 4})^2 - 4 * (-1) *( 1.12 \times 10^{- 4})$
$\Delta = 3.136 \times 10^{- 7} + 4.48 \times 10^{- 4} = 4.483 \times 10^{- 4}$
$x_1 = \frac{ - (- 5.6 \times 10^{- 4})+ \sqrt { 4.483 \times 10^{- 4}}}{2*(-1)}$
or
$x_2 = \frac{ - (- 5.6 \times 10^{- 4})- \sqrt { 4.483 \times 10^{- 4}}}{2*(-1)}$
$x_1 = - 1.087 \times 10^{- 2} (Negative)$
$x_2 = 1.031 \times 10^{- 2}$
- The concentration can't be negative, so $[OH^-]$ = $x_2$
3. Calculate the pH value:
$pOH = -log[OH^-]$
$pOH = -log( 0.01031)$
$pOH = 1.987$
$pH + pOH = 14$
$pH + 1.987 = 14$
$pH = 12.013$
