To determine the pH of a 0.050 M benzoic acid (HA) solution with a given acid dissociation constant [tex]\(K_a = 6.3 \times 10^{-5}\)[/tex], follow these steps:
1. Expression of Equilibrium:
Benzoic acid dissociates in water according to the following reaction:
[tex]\[
\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-
\][/tex]
Here, initially, the concentration of HA is 0.050 M, while the initial concentrations of [tex]\(\text{H}^+\)[/tex] and [tex]\(\text{A}^-\)[/tex] are 0 M.
2. Setting up the Equilibrium Concentrations:
At equilibrium, let [tex]\(x\)[/tex] be the concentration of [tex]\(\text{H}^+\)[/tex] and [tex]\(\text{A}^-\)[/tex] ions due to the dissociation of HA. Then the equilibrium concentrations will be:
[tex]\[
[\text{HA}] = 0.050 - x
\][/tex]
[tex]\[
[\text{H}^+] = x
\][/tex]
[tex]\[
[\text{A}^-] = x
\][/tex]
3. Equilibrium Expression using [tex]\(K_a\)[/tex]:
The equilibrium expression for the dissociation can be written as:
[tex]\[
K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}
\][/tex]
Substituting the equilibrium concentrations into this expression gives:
[tex]\[
6.3 \times 10^{-5} = \frac{x \cdot x}{0.050 - x}
\][/tex]
4. Simplifying the Expression:
Since [tex]\(K_a\)[/tex] is small, we can assume [tex]\(x \ll 0.050\)[/tex], thus [tex]\(0.050 - x \approx 0.050\)[/tex]. This simplifies our equation to:
[tex]\[
6.3 \times 10^{-5} \approx \frac{x^2}{0.050}
\][/tex]
5. Solving for [tex]\(x\)[/tex]:
Now, solve for [tex]\(x\)[/tex] (which represents the [tex]\(\text{H}^+\)[/tex] concentration):
[tex]\[
x^2 \approx 6.3 \times 10^{-5} \times 0.050
\][/tex]
[tex]\[
x^2 \approx 3.15 \times 10^{-6}
\][/tex]
[tex]\[
x \approx \sqrt{3.15 \times 10^{-6}}
\][/tex]
[tex]\[
x \approx 0.0017748
\][/tex]
Therefore, the concentration of [tex]\(\text{H}^+\)[/tex] ions ([tex]\(x\)[/tex]) at equilibrium is approximately [tex]\(0.0017748\)[/tex] M.
6. Calculating pH:
The pH of the solution is calculated using the formula:
[tex]\[
\text{pH} = -\log [\text{H}^+]
\][/tex]
Substituting the value of [tex]\(x\)[/tex]:
[tex]\[
\text{pH} = -\log (0.0017748)
\][/tex]
[tex]\[
\text{pH} \approx 2.75
\][/tex]
Therefore, the pH of the 0.050 M benzoic acid solution is approximately [tex]\(2.75\)[/tex].
