Answer
$pH = 5.03$
Work Step by Step
1. $NH_4Cl$ is a salt formed by $N{H_4}^+$ and $Cl^-$
$Cl^-$ is a very weak base, so, we don't need to consider it.
$N{H_4}^+$ is a weak acid.
2. Now, use the $K_b$ of $NH_3$ to calculate the $K_a$ of $N{H_4}^+$:
$K_b ( NH_3) = 1.76 \times 10^{-5}$
$K_b * K_a = K_w = 10^{-14}$
$ 1.76\times 10^{- 5} * K_a = 10^{-14}$
$K_a = \frac{10^{-14}}{ 1.76\times 10^{- 5}}$
$K_a = 5.682\times 10^{- 10}$
3. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[H_3O^+] = [NH_3] = x$
-$[N{H_4}^+] = [N{H_4}^+]_{initial} - x = 0.155 - x$
For approximation, we consider: $[N{H_4}^+] = 0.155M$
4. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][NH_3]}{ [N{H_4}^+]}$
$Ka = 5.682 \times 10^{- 10}= \frac{x * x}{ 0.155}$
$Ka = 5.682 \times 10^{- 10}= \frac{x^2}{ 0.155}$
$ 8.807 \times 10^{- 11} = x^2$
$x = 9.384 \times 10^{- 6}$
Percent dissociation: $\frac{ 9.384 \times 10^{- 6}}{ 0.155} \times 100\% = 0.006054\%$
%dissociation < 5% : Right approximation.
Therefore: $[H_3O^+] = [NH_3] = x = 9.384 \times 10^{- 6}M $
And, since 'x' has a very small value (compared to the initial concentration): $[N{H_4}^+] \approx 0.155M$
5. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 9.384 \times 10^{- 6})$
$pH = 5.03$
