The Henderson-Hasselbalch equation

Result

The Henderson-Hasselbalch equation

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Statement

Buffer pH from the acid-to-base ratio.

Why it matters

bronsted-lowry and water-autoionisation-ph established what acids, bases, and pH fundamentally are; the Henderson–Hasselbalch equation is the single most-used working formula built on that foundation, giving the pH of a buffered solution — a mixture of a weak acid and its conjugate base — directly from their concentration ratio, without needing to solve an equilibrium expression from scratch each time. Buffers resist pH change on addition of small amounts of strong acid or base, a property essential to everything from laboratory reaction control to blood pH regulation, and this equation is the standard quantitative tool for both preparing a buffer at a target pH and predicting how far it can be pushed before it fails.

Hypotheses

Both the weak acid \(\text{HA}\) and its conjugate base \(\text{A}^-\) are present in bulk, comparable concentrations, well above the concentration of \(\text{H}^+\) or \(\text{OH}^-\) contributed by water's own autoionisation or by the acid's own dissociation.This "buffer approximation" lets the equilibrium concentrations of \(\text{HA}\) and \(\text{A}^-\) be replaced directly by their nominal (initially prepared, or formal) concentrations without solving the full equilibrium expression exactly; it breaks down for very dilute buffers, or when the desired pH is far from the acid's \(\text{p}K_a\) (Fails without). Only a single acid–base equilibrium (one \(K_a\)) is relevant across the pH range of interest.For a polyprotic acid (polyprotic-acids), several distinct equilibria with different \(K_a\) values may be simultaneously relevant, and the simple single-\(K_a\) form of the equation below applies cleanly only when the pH of interest is close to one specific \(\text{p}K_a\) and well separated from the others.

Proof

K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}

The equilibrium constant expression for the weak acid's dissociation, \(\text{HA}\rightleftharpoons\text{H}^++\text{A}^-\), is the starting point; this is simply the law of mass action applied to a specific weak-acid equilibrium. A

[\text{H}^+] = K_a\cdot\frac{[\text{HA}]}{[\text{A}^-]}

Rearranging Step 1 algebraically to isolate \([\text{H}^+]\) gives it directly in terms of the equilibrium constant and the ratio of acid to conjugate-base concentrations. A

-\log[\text{H}^+] = -\log K_a - \log\frac{[\text{HA}]}{[\text{A}^-]}

Taking the negative base-10 logarithm of both sides of Step 2 converts the multiplicative relationship into an additive one, matching the definitions \(\text{pH}=-\log[\text{H}^+]\) and \(\text{p}K_a=-\log K_a\). A

\text{pH} = \text{p}K_a + \log\frac{[\text{A}^-]}{[\text{HA}]}

Since \(-\log(x/y)=\log(y/x)\), flipping the ratio inside the logarithm in Step 3 gives the standard Henderson–Hasselbalch form, with the conjugate base over the acid; substituting the nominal, prepared concentrations for the true equilibrium concentrations (Hypotheses) gives the equation its familiar, directly usable working form. A

Result

\text{pH} = \text{p}K_a + \log\frac{[\text{A}^-]}{[\text{HA}]}

Reading. A buffer's pH is set by its acid's intrinsic \(\text{p}K_a\), shifted by the logarithm of the ratio of conjugate base to acid present — when the two are equal, pH equals \(\text{p}K_a\) exactly.

Scope. Reliable when both \(\text{HA}\) and \(\text{A}^-\) are present at bulk concentrations well above the buffer's own dissociation contribution (Hypotheses), and generally most accurate, and most resistant to added acid or base, when the ratio \([\text{A}^-]/[\text{HA}]\) is within roughly a factor of ten of unity (i.e. pH within about one unit of \(\text{p}K_a\)).

Corollaries & converses

At the half-equivalence point of a titration (titration-curves), exactly half the original acid has been converted to its conjugate base, so \([\text{A}^-]=[\text{HA}]\) and, by the Result, \(\text{pH}=\text{p}K_a\) exactly — the standard method for reading a weak acid's \(\text{p}K_a\) directly off a titration curve.
Biological buffer systems (such as the bicarbonate buffer regulating blood pH) rely on exactly this equation to maintain pH within a narrow physiological range despite continual metabolic production of acid or base.
Converse: given a target pH and a chosen weak acid's known \(\text{p}K_a\), the Result can be solved directly for the required ratio \([\text{A}^-]/[\text{HA}]\), the standard calculation used when preparing a buffer at a specified pH.

Fails without

Apply the equation to a very dilute buffer, where the buffer approximation of the Hypotheses breaks down: if the concentrations of \(\text{HA}\) and \(\text{A}^-\) are not large compared to the \([\text{H}^+]\) or \([\text{OH}^-]\) generated by the acid's own dissociation or water's autoionisation, the nominal concentrations substituted in Step 4 no longer closely approximate the true equilibrium concentrations, and the predicted pH becomes noticeably inaccurate.
Use the equation far outside the buffer's effective range (pH far from \(\text{p}K_a\)): when the ratio \([\text{A}^-]/[\text{HA}]\) becomes very large or very small, the solution contains predominantly only one of the two species and no longer resists pH change effectively on addition of acid or base, even though the Result's arithmetic still technically returns a number.

Common errors

Inverting the ratio, writing \(\log([\text{HA}]/[\text{A}^-])\) instead of \(\log([\text{A}^-]/[\text{HA}])\) (Step 4), which flips the sign of the correction term relative to \(\text{p}K_a\).
Using initial (pre-equilibrium) concentrations that do not actually correspond to the buffer's true acid-to-base ratio after any prior reaction (for instance, forgetting to account for acid or base already consumed by a partial neutralisation) before substituting into the Result.
Applying the equation confidently far from \(\text{p}K_a\), where the Hypotheses' approximations weaken and the buffer's actual capacity to resist pH change is small (Fails without).
Confusing \(K_a\) and \(\text{p}K_a\) directly, or forgetting the negative sign relating them, when converting between a tabulated \(K_a\) value and the \(\text{p}K_a\) used in the Result.

Discussion

Lawrence Henderson derived the underlying equilibrium relationship in 1908, originally in the context of blood's buffering capacity as a physiological chemist studying acid–base balance; Karl Albert Hasselbalch reformulated it explicitly in logarithmic, pH form in 1917, in the same physiological context, giving the equation its now-standard name and its most immediately usable working form.

The equation's continued central role in physiology is not accidental: the bicarbonate buffer system, \(\text{H}_2\text{CO}_3/\text{HCO}_3^-\), keeps human blood pH within an extremely narrow range (approximately 7.35–7.45) despite continuous metabolic acid production, and clinical interpretation of blood gas measurements is built directly on Henderson–Hasselbalch reasoning applied to this specific buffer pair.

Common misconception: that a buffer maintains pH at a perfectly fixed value regardless of how much strong acid or base is added. A buffer only resists pH change within its finite capacity (set by how much \(\text{HA}\) and \(\text{A}^-\) are actually present); once enough strong acid or base is added to consume nearly all of one component, the pH begins to change rapidly, exactly as an unbuffered solution would — the Result's Scope note on effective range describes this limitation directly.

Worked examples

\text{Acetate buffer: } \text{p}K_a(\text{CH}_3\text{COOH})=4.76,\quad [\text{CH}_3\text{COO}^-]=0.30\,\text{M},\quad [\text{CH}_3\text{COOH}]=0.20\,\text{M}

Substituting directly into the Result: \(\text{pH}=4.76+\log(0.30/0.20)=4.76+\log(1.5)=4.76+0.18=4.94\). A

\text{Target: prepare a buffer at pH } 5.00 \text{ using the same acid, } \text{p}K_a=4.76

Solving the Result for the ratio: \(\log([\text{A}^-]/[\text{HA}]) = \text{pH}-\text{p}K_a = 5.00-4.76=0.24\), so \([\text{A}^-]/[\text{HA}]=10^{0.24}=1.74\) — the practical prescription for how much sodium acetate to add per mole of acetic acid to hit the target pH. A

\text{pH}=4.94\ (\text{Example 1}); \qquad [\text{A}^-]/[\text{HA}]=1.74\ \text{needed for pH }5.00

Reading. The same equation runs equally well forward (concentrations to pH) or backward (target pH to required concentration ratio), making it equally useful for predicting a buffer's pH and for designing one.

Scope. Both calculations remain reliable provided the buffer stays within roughly a factor of ten of equal acid and base concentrations, consistent with the Result's stated scope.

Problems

A buffer is prepared with \(0.15\,\text{M}\) \(\text{NH}_3\) and \(0.25\,\text{M}\) \(\text{NH}_4^+\), where \(\text{p}K_a(\text{NH}_4^+)=9.25\). Find the buffer's pH.
Solution
Here \(\text{NH}_4^+\) is the acid (\(\text{HA}\)) and \(\text{NH}_3\) is the conjugate base (\(\text{A}^-\)). \(\text{pH}=9.25+\log(0.15/0.25)=9.25+\log(0.60)=9.25+(-0.22)=9.03\).
What ratio of \([\text{A}^-]/[\text{HA}]\) is required to prepare a buffer exactly one pH unit below the acid's \(\text{p}K_a\)? State the general rule this illustrates.
Solution
\(\text{pH}-\text{p}K_a = -1 = \log([\text{A}^-]/[\text{HA}])\), so \([\text{A}^-]/[\text{HA}]=10^{-1}=0.10\), i.e. a \(1{:}10\) ratio of base to acid. This illustrates the general rule that each whole pH unit away from \(\text{p}K_a\) corresponds to exactly a factor-of-ten change in the \([\text{A}^-]/[\text{HA}]\) ratio, directly from the logarithmic form of the Result.
Explain why a buffer prepared with \([\text{A}^-]/[\text{HA}]=100\) (i.e. pH two units above \(\text{p}K_a\)) is a poor buffer against added acid, even though the Result still returns a valid pH value for it.
Solution
At this ratio, the solution contains roughly 100 times as much conjugate base as undissociated acid; while the Result's arithmetic still gives a definite pH, the buffer's capacity to neutralise added acid specifically depends on having a substantial reservoir of \(\text{A}^-\) to consume it, which is present here, but its capacity to neutralise added base depends on having a substantial reservoir of \(\text{HA}\), which is now comparatively very small. The buffer is therefore strongly asymmetric and only weakly resists further base addition, illustrating why the Result's Scope note recommends staying within about a factor of ten of equal concentrations for balanced buffering capacity in both directions.