Buffers and the Henderson-Hasselbalch Equation

A buffer is a solution containing significant amounts of a conjugate pair, such as CH₃COOH with CH₃COO⁻, or NH₄⁺ with NH₃. Added OH⁻ is consumed by the acid component and added H₃O⁺ is consumed by the base component, so neither accumulates.

A weak acid alone has too little conjugate base to absorb a useful added acid dose. A strong acid with its spectator salt, such as HCl/NaCl, also lacks a meaningful base component. A buffer can be prepared by mixing HA with a soluble A⁻ salt, or by partially neutralizing HA with strong base. For 0.100 mol HA and 0.0400 mol OH⁻, the reaction leaves 0.0600 mol HA and makes 0.0400 mol A⁻: both survive, so the product mixture is a buffer.

Rearranging K_a = [H₃O⁺][A-]/[HA] and taking negative logs gives the working equation. Because it involves a ratio, moles can be used instead of concentrations whenever both species share the same volume.

From K_a ≈ [H₃O⁺][A⁻]/[HA], rearrangement gives pH ≈ pK_a + log₁₀([A⁻]/[HA]). The equilibrium-activity form is the underlying relationship; substituting prepared component moles assumes further ionization and water contributions are small compared with those amounts. Test that assumption for very dilute mixtures or very unequal ratios. Equal prepared amounts give pH approximately pK_a, rather than an exact identity for every nonideal solution.

Treat the addition as a reaction that runs to completion: added OH⁻ converts HA into A-, and added H₃O⁺ converts A- into HA. Update the moles, then use Henderson-Hasselbalch again. If a component is exhausted, identify what remains: at exact exhaustion use the remaining weak-species equilibrium; beyond exhaustion excess strong acid or base often controls pH.

Use two stages: reaction stoichiometry first, acid-base equilibrium second. Starting with 0.200 mol HA and 0.100 mol A⁻, add 0.0300 mol HCl. H⁺ + A⁻ → HA leaves 0.0700 mol A⁻ and 0.230 mol HA. If pK_a = 4.76, pH ≈ 4.76 + log(0.0700/0.230) = 4.24. The common final volume cancels in this ratio, but it would not cancel in a later excess-strong-acid concentration.

If 0.100 mol HCl is added, the base component is exhausted exactly: the remaining solution is a weak-acid problem, not automatically an excess-strong-acid problem. Beyond that point, excess HCl usually controls pH, subject to equilibrium and dilution. Do not put a zero or negative component into a logarithm.

A beaker buffer has a finite inventory of acid and base. An open system can exchange a volatile component with its surroundings, so its response also depends on gas exchange and material balance. Carbon dioxide/bicarbonate buffering is an example: dissolved CO₂, carbonic acid and bicarbonate are linked equilibria. A pK_a referring to molecular H₂CO₃, one referring to combined dissolved CO₂ plus H₂CO₃, and a temperature-specific effective physiological value are different conventions and must not be mixed in one numerical Henderson–Hasselbalch ratio.

For ordinary lab design, choose a pair with pK_a near the target pH, calculate the required ratio, then choose total amounts large enough for the anticipated additions. Diluting both equally approximately preserves their ratio but lowers capacity per volume; severe dilution also invalidates the prepared-mole approximation.