📘 IB Understanding
Because weak acids only partially dissociate, we can approximate [HA]eqm ≈ [HA]initial. The IB does not expect use of the quadratic formula.
The Approximation
\(K_a = \frac{[\text{H}^+]^2}{[\text{HA}]_{\text{initial}}}\)
\(\boxed{[\text{H}^+] = \sqrt{K_a \times [\text{HA}]_{\text{initial}}}}\)
Worked Example
Q: Calculate the pH of 0.10 M ethanoic acid (Ka = 1.76 × 10⁻⁵).
Step 1: \([H^+] = \sqrt{1.76 \times 10^{-5} \times 0.10} = \sqrt{1.76 \times 10^{-6}}\)
Step 2: \([H^+] = 1.33 \times 10^{-3} \text{ mol dm}^{-3}\)
Step 3: \(\text{pH} = -\log(1.33 \times 10^{-3}) = \textbf{2.88}\)
📋 Exam Tips
- Sig figs: Decimal places in pH = sig figs in [H⁺]. E.g., [H⁺] = 5.3 × 10⁻⁵ (2 s.f.) gives pH = 4.28 (2 d.p.)
- Diluting a weak acid increases pH but increases % ionisation (Le Chatelier's)