The Henderson–Hasselbalch Equation

Core Concept

The Henderson–Hasselbalch equation is a rearranged, log-form version of the Ka expression that makes buffer pH easy to read: pH = pKa + log([A⁻]/[HA]). It says buffer pH is set by the pKa of the weak acid plus a correction for the ratio of conjugate base to acid. When the two concentrations are equal, pH = pKa. The equation is a shortcut, not a new law — it assumes the approximations of the common-ion ICE table (that x is small), so it is accurate for buffers where both members are present at appreciable, comparable concentrations.

Key Ideas

• Origin: take −log of Ka = [H₃O⁺][A⁻]/[HA] and solve for pH.
• Form: pH = pKa + log([A⁻]/[HA]).
• Equal concentrations: [A⁻] = [HA] ⇒ log(1) = 0 ⇒ pH = pKa.
• Ratio matters, not total amount: diluting a buffer changes [A⁻] and [HA] by the same factor, so the ratio (and pH) stays nearly constant.
• Base version: pOH = pKb + log([BH⁺]/[B]).
• Assumptions: x is negligible compared to both [HA] and [A⁻]; activity effects ignored.

Equations and Variables

• Ka = [H₃O⁺][A⁻] / [HA]
• pH = pKa + log([A⁻]/[HA])
• pOH = pKb + log([BH⁺]/[B])
• Target ratio: [A⁻]/[HA] = 10^(pH − pKa)

How It Works

1. Start from the equilibrium expression Ka = [H₃O⁺][A⁻]/[HA].
2. Solve for [H₃O⁺]: [H₃O⁺] = Ka × [HA]/[A⁻].
3. Take the negative log of both sides: −log[H₃O⁺] = −log Ka − log([HA]/[A⁻]).
4. Apply the definitions pH = −log[H₃O⁺] and pKa = −log Ka, and flip the sign of the last log (since −log([HA]/[A⁻]) = +log([A⁻]/[HA])).
5. Result: pH = pKa + log([A⁻]/[HA]). To design a buffer, invert it: choose a weak acid with pKa near the target, then set the ratio [A⁻]/[HA] = 10^(pH − pKa).

Worked Example

You need a buffer at pH 5.00. Acetic acid has pKa = 4.74. What [CH₃COO⁻]/[CH₃COOH] ratio is required?

pH − pKa = 5.00 − 4.74 = 0.26

[A⁻]/[HA] = 10^0.26 = 1.8

So you need about 1.8 times as much acetate as acetic acid (e.g., 0.18 M CH₃COO⁻ with 0.10 M CH₃COOH). Check: pH = 4.74 + log(1.8) = 4.74 + 0.26 = 5.00 ✓.

How it works

1. Start from the equilibrium expression Ka = [H₃O⁺][A⁻]/[HA].
2. Solve for [H₃O⁺]: [H₃O⁺] = Ka × [HA]/[A⁻].
3. Take the negative log of both sides: −log[H₃O⁺] = −log Ka − log([HA]/[A⁻]).
4. Apply the definitions pH = −log[H₃O⁺] and pKa = −log Ka, and flip the sign of the last log (since −log([HA]/[A⁻]) = +log([A⁻]/[HA])).
5. Result: pH = pKa + log([A⁻]/[HA]). To design a buffer, invert it: choose a weak acid with pKa near the target, then set the ratio [A⁻]/[HA] = 10^(pH − pKa).

Common confusions

• "It's a new fundamental law." — It is just the Ka expression with logs; nothing new, only convenient.
• "Adding water changes buffer pH." — Dilution changes [A⁻] and [HA] by the same factor, so the ratio — and pH — barely changes (a tiny shift remains because the x-small assumption improves).
• "Use molarity of the salt and acid before mixing." — Use the equilibrium (post-stoichiometry) concentrations; if strong acid/base was added, do stoichiometry first.
• "It works for any acid." — It is meant for buffers (weak acid + conjugate base); it fails for strong acids or very dilute buffers.
• "log([A⁻]/[HA]) is always positive." — If [HA] > [A⁻], the ratio is < 1 and the log is negative, so pH < pKa.

Quick review

• pH = pKa + log([A⁻]/[HA]).
• Derived from Ka by taking −log and rearranging.
• Equal concentrations ⇒ pH = pKa.
• Target ratio: [A⁻]/[HA] = 10^(pH − pKa).
• Dilution preserves the ratio, so pH stays nearly constant.