Answer
$[OH^-] = 1.6 \times 10^{-3} \space M$
pH = 11.21
pOH = 2.79
Work Step by Step
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ NH_3 ]& [ NH{_4}^{+} ]& [ OH^- ]\\
Initial& 0.15 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 0.15 -x& 0 +x& 0 +x\\
\end{vmatrix}$$
2. Write the expression for $K_b$, and substitute the concentrations:
- The exponent of each concentration is equal to its balance coefficient.
$$K_b = \frac{[Products]}{[Reactants]} = \frac{[ NH{_4}^{+} ][ H^+ ]}{[ NH_3 ]}$$
$$K_b = \frac{(x)(x)}{[ NH_3 ]_{initial} - x}$$
3. Assuming $ 0.15 \gt\gt x$:
$$K_b = \frac{x^2}{[ NH_3 ]_{initial}}$$
$$x = \sqrt{K_b \times [ NH_3 ]_{initial}} = \sqrt{ 1.8 \times 10^{-5} \times 0.15 }$$
$x = 1.6 \times 10^{-3} $
4. Test if the assumption was correct:
$$\frac{ 1.6 \times 10^{-3} }{ 0.15 } \times 100\% = 1.1 \%$$
5. Thus, it is correct to say that $x = 1.6 \times 10^{-3} $
6. $[OH^-] = x = 1.6 \times 10^{-3} $
7. Calculate the pH:
$$[H_3O^+] = \frac{1.0 \times 10^{-14}}{[OH^-]} = \frac{1.0 \times 10^{-14}}{ 1.6 \times 10^{-3} } = 6.2 \times 10^{-12} \space M$$
$$pH = -log[H_3O^+] = -log( 6.2 \times 10^{-12} ) = 11.21 $$
$$pOH = 14 - pH = 14 - 11.21 = 2.79$$
