The Henderson-Hasselbalch Equation
HL Extension
Henderson-Hasselbalch Equation
The Henderson-Hasselbalch Equation:
\[\text{pH} = \text{p}K_{\text{a}} + \log_{10}\left(\frac{[\text{conjugate base}]}{[\text{weak acid}]}\right) = \text{p}K_{\text{a}} + \log_{10}\left(\frac{n(\text{A}^-)}{n(\text{HA})}\right)\]Equimolar Buffer: When \([\text{A}^-] = [\text{HA}]\), \(\log(1) = 0 \implies \textbf{pH = pK}_\text{a}\) (maximum buffering capacity).
Worked Example
Worked Example
Buffer pH Calculation
Problem: Calculate the pH of a buffer prepared by mixing \(0.20\text{ mol dm}^{-3}\text{ CH}_3\text{COOH}\) (\(\text{p}K_{\text{a}} = 4.76\)) and \(0.30\text{ mol dm}^{-3}\text{ CH}_3\text{COONa}\).
\[\text{pH} = 4.76 + \log_{10}\left(\frac{0.30}{0.20}\right) = 4.76 + \log_{10}(1.50) = 4.76 + 0.18 = 4.94\]
Examiner Trap
Partial Neutralisation and Mole Substitution
- Use mole ratios: Because both components share the same total volume, you can substitute moles directly: \(\frac{n(\text{A}^-)}{n(\text{HA})}\).
- Partial neutralisation: Adding strong base to excess weak acid forms a buffer: moles of \(\text{A}^-\) formed = moles of strong base added.
AQA GCSE & IB Chemistry
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