Answer
$$pH = 5.08$$
Work Step by Step
1. Identify the acids and bases.
$NH_4^+$ is the conjugate acid of $NH_3$.
$$K_a = \frac{10^{-14}}{K_b} = \frac{10^{-14}}{1.8 \times 10^{-5}} = 5.6 \times 10^{-10}$$
$Cl^-$ does not act as a base.
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ N{H_4}^{+} ]& [ NH_3 ]& [ H_3O^+ ]\\
Initial& 0.123 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 0.123 -x& 0 +x& 0 +x\\
\end{vmatrix}$$
2. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each concentration is equal to its balance coefficient.
$$K_C = \frac{[Products]}{[Reactants]} = \frac{[ NH_3 ][ H_3O^+ ]}{[ N{H_4}^{+} ]}$$
$$K_a = \frac{(x)(x)}{[ N{H_4}^{+} ]_{initial} - x}$$
3. Assuming $ 0.123 \gt\gt x:$
$$K_a = \frac{x^2}{[ N{H_4}^{+} ]_{initial}}$$
$$x = \sqrt{K_a \times [ N{H_4}^{+} ]_{initial}} = \sqrt{ 5.6 \times 10^{-10} \times 0.123 }$$
$x = 8.3 \times 10^{-6} $
4. Test if the assumption was correct:
$$\frac{ 8.3 \times 10^{-6} }{ 0.123 } \times 100\% = 6.7 \times 10^{-3} \%$$
5. The percent is less than 5%. Thus, it is correct to say that $x = 8.3 \times 10^{-6} $
6. $$[H_3O^+] = x = 8.3 \times 10^{-6} $$
7. Calculate the pH:
$$pH = -log[H_3O^+] = -log( 8.3 \times 10^{-6} ) = 5.08 $$
