To calculate the pH of a 0.02 M solution of the weak acid, HA, we need to follow these steps:
1. Write the dissociation equation:
[tex]\[
HA \rightleftharpoons H^+ + A^-
\][/tex]
2. Set up the expression for the acid dissociation constant (Ka):
[tex]\[
Ka = \frac{[H^+][A^-]}{[HA]}
\][/tex]
Given:
[tex]\[
Ka = 2.46 \times 10^{-4}
\][/tex]
Initial concentration of HA:
[tex]\[
[HA] = 0.02 \, \text{M}
\][/tex]
3. Define the equilibrium concentrations:
Assume that the initial concentration of HA is roughly equal to its concentration at equilibrium, and let [tex]\( x \)[/tex] represent the concentration of [tex]\( H^+ \)[/tex] and [tex]\( A^- \)[/tex] at equilibrium. Therefore:
[tex]\[
[H^+] = x
\][/tex]
[tex]\[
[A^-] = x
\][/tex]
[tex]\[
[HA] \approx 0.02 - x \approx 0.02 \, (\text{since } x \text{ will be very small})
\][/tex]
4. Substitute the values into the Ka expression:
[tex]\[
Ka = \frac{x \cdot x}{0.02} = \frac{x^2}{0.02}
\][/tex]
5. Solve for [tex]\( x \)[/tex]:
[tex]\[
2.46 \times 10^{-4} = \frac{x^2}{0.02}
\][/tex]
[tex]\[
x^2 = 2.46 \times 10^{-4} \times 0.02
\][/tex]
[tex]\[
x^2 = 4.92 \times 10^{-6}
\][/tex]
[tex]\[
x = \sqrt{4.92 \times 10^{-6}}
\][/tex]
[tex]\[
x \approx 0.00222
\][/tex]
This value of [tex]\( x \)[/tex] represents the concentration of [tex]\( H^+ \)[/tex] ions in the solution.
6. Calculate the pH:
[tex]\[
\text{pH} = -\log [H^+]
\][/tex]
[tex]\[
\text{pH} = -\log (0.00222)
\][/tex]
[tex]\[
\text{pH} \approx 2.65
\][/tex]
So, the pH of a 0.02 M solution of the weak acid HA is approximately 2.65.
