Chemistry / Unit 10 Chem 1893 worked example, 60 to 90 seconds
Henderson-Hasselbalch Equation Application
The idea
Applying the Henderson-Hasselbalch equation means taking the negative log of Ka to get pKa, then adding the log of the ratio of conjugate base to acid concentrations directly: pH = pKa + log([base]/[acid]), a shortcut derived from the weak-acid equilibrium expression itself once both concentrations are already known from preparation, skipping a full ICE table.
In practice
A buffer of 0.30 M acetate and 0.20 M acetic acid, Ka = 1.8 × 10⁻⁵ (pKa ≈ 4.74): pH = 4.74 + log(0.30/0.20) = 4.74 + 0.18 ≈ 4.92, slightly above pKa since base exceeds acid here in this particular mixture.
Where people slip
Inverting the ratio, base over acid versus acid over base, flips the sign of the log term and gives the wrong pH entirely, not merely a small error.
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