Answer
$pH = 5.12$
Work Step by Step
- $Cl^-$ is a insignificant base, therefore, to calculate the pH of $0.10 M$ $N{H_4}Cl$, we can calculate the value for a $N{H_4}^+ 0.10M$ solution, because it has the same pH value.
1. Since $N{H_4}^+$ is the conjugate acid of $NH_3$ , we can calculate its ka by using this equation:
$K_b * K_a = K_w = 10^{-14}$
$ 1.76\times 10^{- 5} * K_a = 10^{-14}$
$K_a = \frac{10^{-14}}{ 1.76\times 10^{- 5}}$
$K_a = 5.682\times 10^{- 10}$
2. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[H_3O^+] = [NH_3] = x$
-$[N{H_4}^+] = [N{H_4}^+]_{initial} - x = 0.1 - x$
For approximation, we consider: $[N{H_4}^+] = 0.1M$
3. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][NH_3]}{ [N{H_4}^+]}$
$Ka = 5.682 \times 10^{- 10}= \frac{x * x}{ 0.1}$
$Ka = 5.682 \times 10^{- 10}= \frac{x^2}{ 0.1}$
$ 5.682 \times 10^{- 11} = x^2$
$x = 7.538 \times 10^{- 6}$
Percent dissociation: $\frac{ 7.538 \times 10^{- 6}}{ 0.1} \times 100\% = 0.007538\%$
%dissociation < 5% : Right approximation.
Therefore: $[H_3O^+] = [NH_3] = x = 7.538 \times 10^{- 6}M $
And, since 'x' has a very small value (compared to the initial concentration): $[N{H_4}^+] \approx 0.1M$
4. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 7.538 \times 10^{- 6})$
$pH = 5.12$
