Question

Unit 6 Activity 2: Weak Acid/Weak Base Calculations [9 marks]
4. A weak acid solution with a concentration of 0.0107 M has a pH of 4.17. What is its KA value? [8]

Answer 1

To find the \( K_a \) value of the weak acid, we can use the relationship between the pH, the concentration of the weak acid, and the \( K_a \).

1. **Calculate the \([H^+]\) from pH**:
\[
[H^+] = 10^{-\text{pH}} = 10^{-4.17} \approx 6.76 \times 10^{-5} \, \text{M}
\]

2. **Set up the equilibrium expression**:
Let's denote the weak acid as HA. We know that it dissociates in water according to the following equation:
\[
HA \rightleftharpoons H^+ + A^-
\]
For a solution of concentration \( [HA]_0 = 0.0107 \, \text{M} \):
- The change in concentration when dissociation occurs can be represented as \( -x \) for \( HA \) and \( +x \) for both \( H^+ \) and \( A^- \).

3. **Establish the concentrations at equilibrium**:
- At equilibrium:
- \([HA] \approx [HA]_0 - x \approx 0.0107 - 6.76 \times 10^{-5} \approx 0.0106 \, \text{M}\) (since \( x \) is very small compared to \( [HA]_0 \))
- \([H^+] = [A^-] = x \approx 6.76 \times 10^{-5} \, \text{M}\)

4. **Write the expression for \( K_a \)**:
\[
K_a = \frac{[H^+][A^-]}{[HA]} = \frac{(6.76 \times 10^{-5})(6.76 \times 10^{-5})}{0.0106}
\]

5. **Calculate \( K_a \)**:
First calculate \((6.76 \times 10^{-5})^2\):
\[
(6.76 \times 10^{-5})^2 \approx 4.57676 \times 10^{-9}
\]

Now substitute it into the \( K_a \) formula:
\[
K_a = \frac{4.57676 \times 10^{-9}}{0.0106} \approx 4.32 \times 10^{-7}
\]

Thus, the \( K_a \) value for the weak acid is approximately:
\[
\boxed{4.32 \times 10^{-7}}
\]