The Henderson-Hasselbalch Equation
\[\text{pH} = pK_a + \log\frac{[\text{A}^-]}{[\text{HA}]}\]
💡 Key Insight
When [A⁻] = [HA] (equal concentrations of acid and conjugate base), the log term = 0 and pH = pKa. This is also the half-equivalence point in a titration.
Worked Example
Q: Calculate the pH of a buffer made from 0.20 M CH₃COOH and 0.30 M CH₃COONa (pKa = 4.75).
\(\text{pH} = 4.75 + \log\frac{0.30}{0.20} = 4.75 + \log(1.5)\)
\(\text{pH} = 4.75 + 0.18 = \textbf{4.93}\)
📋 Exam Tips
- You can use moles instead of concentrations in the ratio if both components are in the same solution volume (the volumes cancel)
- A buffer has maximum capacity when [HA] = [A⁻]