Answer
- $NH_3$, $OH^-$, $N{H_4}^+$ and $H_2O$
- $[OH^-] = 1.643 \times 10^{- 3}M$ and $pH = 11.216$
Work Step by Step
- Since $NH_3$ is a weak base, the ionization reaction is an equilibrium reaction:
$NH_3(aq) + H_2O(l) \lt -- \gt N{H_4}^+(aq) + OH^-(aq)$
Therefore, both reactants and products are present in the solution.
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[OH^-] = [{NH_4}^+] = x$
-$[NH_3] = [NH_3]_{initial} - x = 0.15 - x$
For approximation, we consider: $[NH_3] = 0.15M$
2. Now, use the Kb value and equation to find the 'x' value.
$Ka = \frac{[OH^-][{NH_4}^+]}{ [NH_3]}$
$Ka = 1.8 \times 10^{- 5}= \frac{x * x}{ 0.15}$
$Ka = 1.8 \times 10^{- 5}= \frac{x^2}{ 0.15}$
$ 2.7 \times 10^{- 6} = x^2$
$x = 1.643 \times 10^{- 3}$
Percent ionization: $\frac{ 1.643 \times 10^{- 3}}{ 0.15} \times 100\% = 1.095\%$
%ionization < 5% : Right approximation.
Therefore: $[OH^-] = [{NH_4}^+] = x = 1.643 \times 10^{- 3}M $
$[NH_3] \approx 0.15M$
3. Calculate the pH:
$pOH = -log[OH^-]$
$pOH = -log( 1.643 \times 10^{- 3})$
$pOH = 2.784$
$pH + pOH = 14$
$pH + 2.784 = 14$
$pH = 11.216$
