Answer
$pOH = 2.216$
$pH = 11.784$
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.
-$[OH^-] = [N{H_4}^+] = x$
-$[NH_3] = [NH_3]_{initial} - x = 2.05 - x$
For approximation, we consider: $[NH_3] = 2.05M$
2. Now, use the Kb value and equation to find the 'x' value.
$Ka = \frac{[OH^-][N{H_4}^+]}{ [NH_3]}$
$Ka = 1.8 \times 10^{- 5}= \frac{x * x}{ 2.05}$
$Ka = 1.8 \times 10^{- 5}= \frac{x^2}{ 2.05}$
$ 3.69 \times 10^{- 5} = x^2$
$x = 6.075 \times 10^{- 3}$
Percent ionization: $\frac{ 6.075 \times 10^{- 3}}{ 2.05} \times 100\% = 0.2963\%$
%ionization < 5% : Right approximation.
Therefore: $[OH^-] = x = 6.075 \times 10^{- 3}M $
3. Calculate the pOH and pH:
$pOH = -log[OH^-]$
$pOH = -log( 6.075 \times 10^{- 3})$
$pOH = 2.216$
$pH + pOH = 14$
$pH + 2.216 = 14$
$pH = 11.784$
