Henderson–Hasselbalch from Ka

Introduction

The Henderson–Hasselbalch equation is probably the most used formula in buffer chemistry, yet it is not a new law. It is simply the acid dissociation constant rearranged into logarithmic form. Deriving it carefully shows exactly which assumptions it relies on, when it can be trusted, and how to correct it for the non-ideal behaviour of ions in real solutions. It also gives a clear way to read a buffer ratio as a pH.

Core explanation

Derivation. For a weak acid HA + H₂O ⇌ H₃O⁺ + A⁻,

Ka = [H₃O⁺][A⁻] ÷ [HA]

Rearranging gives [H₃O⁺] = Ka × [HA] ÷ [A⁻]. Taking −log₁₀ of both sides:

−log[H₃O⁺] = −log Ka − log([HA] ÷ [A⁻])

pH = pKa + log([A⁻] ÷ [HA])

Up to this point the equation is exact (in terms of concentration-based constants), because it is just the equilibrium expression.

The key approximation. In practice we substitute the formal concentrations, c(HA) and c(A⁻), that were put into the solution. Strictly, the equilibrium concentrations are

[HA] = c(HA) − [H₃O⁺] + [OH⁻] and [A⁻] = c(A⁻) + [H₃O⁺] − [OH⁻]

These corrections are negligible when c(HA) and c(A⁻) are both much larger than [H₃O⁺] and [OH⁻]. For a typical buffer at 0.1 mol dm⁻³ and pH 4–10, the corrections are well under 1%. They become important for very dilute buffers, for acids with small pKa (below about 3), or for bases whose conjugate acid has large pKa (above about 11).

Ratio interpretation. Because pH depends on log of the ratio:

- ratio 1 : 1 gives pH = pKa - each tenfold increase in [A⁻]/[HA] raises pH by 1 unit - ratio 2 : 1 raises pH by log 2 = 0.30

So a buffer ratio can be read directly as a displacement from pKa. The ratio may be written using moles instead of concentrations because both species share one volume.

Activities. The true thermodynamic constant uses activities, a = γc. With activities,

pH = pKa + log([A⁻] ÷ [HA]) + log(γ(A⁻) ÷ γ(HA))

A neutral HA has γ ≈ 1, while an anion in a solution of ionic strength 0.1 mol dm⁻³ has γ ≈ 0.78. The correction log(0.78) ≈ −0.11 means the measured pH of such a buffer is about 0.1 lower than the ideal equation predicts. For a charged acid–base pair such as H₂PO₄⁻/HPO₄²⁻ the correction is larger, because the doubly charged ion has a much smaller activity coefficient. This is why tabulated buffer recipes are often calibrated by experiment rather than calculated from pKa alone.

Formulae

pH = pKa + log([A⁻] ÷ [HA]); equivalently pOH = pKb + log([BH⁺] ÷ [B]) for a weak-base buffer. Activity-corrected: pH = pKa + log(c(A⁻)γ(A⁻) ÷ c(HA)γ(HA)).

Step-by-step reasoning

To use the equation responsibly:

1. Identify the conjugate pair and the correct pKa for that pair. 2. Find amounts of HA and A⁻ after any stoichiometric reaction. 3. Check that both are much larger than the expected [H₃O⁺] and [OH⁻]. 4. Substitute the ratio and calculate pH. 5. If ionic strength is significant, add the activity correction.

Visual explanation

Plot pH against log([A⁻]/[HA]). The Henderson–Hasselbalch equation gives a straight line of gradient 1 passing through pH = pKa at log ratio 0. Real buffer data follow this line in the middle and bend away at the ends, where the concentration approximation breaks down.

Real-world analogy

A buffer ratio behaves like a gear setting on a bicycle. The pKa is the chosen gear (the base setting), and the ratio makes fine adjustments either side. A tenfold change in the ratio corresponds to one full "click" of pH.

Real-world example

Clinical blood-gas analysers report plasma pH using the bicarbonate/carbon dioxide version of this equation: pH = 6.1 + log([HCO₃⁻] ÷ (0.03 × pCO₂)), with pCO₂ in mmHg. The value 6.1 is an apparent pKa for plasma conditions, already absorbing activity effects and the hydration of CO₂.

Why?

Why do moles work as well as concentrations? Both species are in the same volume V, so [A⁻]/[HA] = (n(A⁻)/V) ÷ (n(HA)/V) = n(A⁻)/n(HA). The volume cancels, which is also why moderate dilution does not change buffer pH.

Common misconception

"The Henderson–Hasselbalch equation works for any acid solution." It applies to mixtures containing significant amounts of both conjugate forms. It gives nonsense for a pure weak acid solution, where [A⁻] comes only from dissociation.

Worked example

Question: A buffer contains 0.150 mol dm⁻³ sodium ethanoate and 0.100 mol dm⁻³ ethanoic acid (pKa = 4.76). Find the ideal pH and the activity-corrected pH if γ(CH₃COO⁻) = 0.77.

Reasoning: Ideal: pH = 4.76 + log(0.150 ÷ 0.100) = 4.76 + 0.176 = 4.94. Correction: log(0.77) = −0.11, taking γ(HA) = 1.

Answer: Ideal pH 4.94; corrected pH ≈ 4.83.

Quick check

1. By how much does the pH of a buffer change if the ratio [A⁻]/[HA] is increased by a factor of 100? Answer: It rises by 2 pH units, because log 100 = 2.

Exam focus

You may be asked to derive the equation from Ka, so practise the rearrangement and the logarithm step. State the assumption that equilibrium concentrations equal formal concentrations, and say when it fails: very dilute buffers or pKa values near the ends of the pH scale.

Advanced insight

When the approximation fails, the exact treatment combines Ka with mass and charge balances. The result, the Charlot equation, is [H₃O⁺] = Ka × (c(HA) − [H₃O⁺] + [OH⁻]) ÷ (c(A⁻) + [H₃O⁺] − [OH⁻]). It reduces to Henderson–Hasselbalch when the correction terms are small and can be solved iteratively otherwise.

Summary

Henderson–Hasselbalch is the Ka expression in logarithmic form: pH = pKa + log([A⁻]/[HA]). Using formal concentrations or moles is valid when both components greatly exceed [H₃O⁺] and [OH⁻]. Each tenfold change in the ratio shifts pH by one unit. Activity coefficients lower the measured pH of anion-based buffers, typically by about 0.1 at ionic strength 0.1.

Practice questions

1. Derive the Henderson–Hasselbalch equation starting from the Ka expression. Answer: Rearrange to [H₃O⁺] = Ka[HA]/[A⁻], take −log of both sides, and use pH = −log[H₃O⁺], pKa = −log Ka to get pH = pKa + log([A⁻]/[HA]). 2. What ratio [A⁻]/[HA] gives a pH 0.50 units below pKa? Answer: 10^(−0.50) ≈ 0.32, so about one part base to three parts acid. 3. Why is Henderson–Hasselbalch unreliable for a 1.0 × 10⁻⁴ mol dm⁻³ ethanoate buffer? Answer: At pH near 4.8, [H₃O⁺] is about 1.7 × 10⁻⁵ mol dm⁻³, which is not negligible compared with the component concentrations, so the formal concentrations differ from the equilibrium ones. 4. State the direction of the activity correction for an ethanoate buffer at high ionic strength and explain. Answer: The pH is lower than the ideal value, because γ(A⁻) is less than 1 while γ(HA) is close to 1, so log(γ(A⁻)/γ(HA)) is negative.