The Henderson-Hasselbalch Equation

A drop of blood at pH 7.35 instead of 7.40 can signal an emergency. Normally it means routine physiology. Yet blood is constantly assaulted by acid. Sources include lactic acid from exercising muscle. Carbonic acid comes from respiring mitochondria. Ketoacids arise from fasting metabolism. The pH barely moves, and this is not accidental dilution. It is the direct consequence of a buffering equilibrium. This equilibrium follows one deceptively simple relationship. That relationship is the Henderson-Hasselbalch equation. The equation looks almost too easy. It is a logarithm of a ratio, plus a constant. Its power lies in transforming an equilibrium constant into something usable. Equilibrium constants are notoriously abstract quantities. But this equation makes them a working recipe. Give a biochemist the pKa of a weak acid. Add the ratio of its two forms. The equation returns the pH directly. There are no simultaneous equations needed. There are no assumptions about "x is small." For a buffered system, x is never small. This is why the equation is the most-used formula in biochemistry labs. It helps in making buffers and reading titration curves. It explains how hemoglobin senses CO2. It also predicts whether a drug will cross a lipid membrane. The big picture is important to understand. Acid-base chemistry in biology is almost never about strong acids and bases. Instead, it involves weak acids and weak bases. These sit in mixtures with their conjugate partners. Examples include phosphate, bicarbonate, and histidine side chains. The amino and carboxyl termini of every protein also count. Each such pair is governed by exactly one number: the pKa. The Henderson-Hasselbalch equation bridges that number and solution pH. It also connects to the protonation state of the molecule. Protonation state, in turn, controls charge. Charge controls everything from enzyme catalysis to protein folding. It even affects drug solubility. This article derives the equation rigorously from the acid dissociation equilibrium. It works through the assumptions that make it valid. It applies the equation to three problems of increasing depth. These are a classroom acetate buffer, a clinical bicarbonate calculation, and a multi-protonation-state amino acid titration. Finally, it connects forward to why Buffers Resist pH Change. It also shows how Physiological Buffer Systems keep organisms alive.

Key takeaways

The Henderson-Hasselbalch equation, pH = pKa + log₁₀([A⁻]/[HA]), converts an equilibrium constant into a direct pH-composition relationship.
It is derived from the acid dissociation constant Ka by simple algebra and a base-10 logarithm — no approximation of "x is small" is needed because [HA] and [A⁻] are the actual equilibrium concentrations.
When [A⁻] = [HA], pH = pKa exactly — this is the buffer's point of maximum resistance to pH change.
The equation applies equally to conjugate acid–base pairs of weak bases, written as pH = pKa + log₁₀([B]/[BH⁺]).
Its main failure mode is at extreme ratios (>100:1) or very dilute/very concentrated solutions where activity coefficients and water autoionization become significant.
Key idea: pH, pKa, and the log of the conjugate base/acid ratio are locked together by one algebraic identity derived directly from the acid dissociation equilibrium.

Where this lives today

In the laboratory, the equation is the everyday recipe for making buffers of a target pH: a biochemist choosing Tris (pKa ≈ 8.1), phosphate (pKa2 ≈ 7.2), or HEPES (pKa ≈ 7.5) selects the buffering agent whose pKa lies closest to the desired pH, then uses Henderson-Hasselbalch to calculate exactly how many moles of acid and conjugate base form to weigh out — this single calculation underlies virtually every buffer recipe in molecular biology, protein purification, and cell culture media. In clinical medicine, arterial blood gas interpretation is built entirely on the bicarbonate form of the equation. Physicians read off [HCO3⁻] (metabolic component) and PCO2 (respiratory component, converted to dissolved CO2 via the solubility coefficient) and use pH = 6.1 + log([HCO3⁻]/0.03·PCO2) to diagnose respiratory versus metabolic acidosis or alkalosis, and to assess whether a compensatory response (renal or respiratory) is appropriate for the primary disturbance. In pharmacology, the equation predicts what fraction of a weak-acid or weak-base drug is in its neutral, membrane-permeable form at a given pH — critical for absorption across the stomach (pH ≈ 1.5-3.5), small intestine (pH ≈ 6-7), and renal tubules, where "ion trapping" (a drug becoming charged and unable to diffuse back across a membrane once it crosses into a compartment of different pH) is a direct, quantitative consequence of the equation and is exploited in aspirin overdose treatment (urinary alkalinization to trap salicylate as its anion and promote excretion). In protein biochemistry, the equation, applied residue-by-residue with residue-specific pKa values, generates the theoretical isoelectric point (pI) of a protein and predicts the protonation state of every titratable side chain (Asp, Glu, His, Cys, Tyr, Lys, Arg, and the N- and C-termini) at any solution pH — calculations used routinely in structural biology, in designing ion-exchange chromatography purification steps, and in computational pKa prediction software (e.g., PROPKA) used to model enzyme active-site chemistry. In environmental and marine chemistry, the same bicarbonate-carbonate framework (extended with a second pKa around 10.3 for HCO3⁻/CO3²⁻) governs ocean pH buffering and is central to modeling ocean acidification as atmospheric CO2 rises.

Ask the tutor: How does the Henderson-Hasselbalch equation combine with mass balance to derive the full quantitative buffer capacity curve as a function of pH?

The physical picture — thorough prose

Consider a weak acid HA dissolved in water. It does not dissociate completely; instead it establishes a dynamic equilibrium,

\text{HA} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{A}^-

in which molecules of HA are continuously donating protons to water while A⁻ ions are continuously reclaiming them. At equilibrium the rates of these two processes are equal, but the concentrations of HA and A⁻ need not be equal — their ratio is fixed by how readily HA gives up its proton, a property quantified by the acid dissociation constant Ka. A small Ka means HA clings tightly to its proton (weak acid, dissociation equilibrium lies left); a large Ka means it lets go easily. The essential insight is that this equilibrium is a competition for the proton between two bases: A⁻ and H2O (as OH⁻, effectively). The position of equilibrium — how the total acid "pool" is split between HA and A⁻ — depends only on Ka and on the ambient concentration of H⁺, i.e., on pH. Raise the pH (remove H⁺, e.g., by adding NaOH) and the equilibrium in the dissociation reaction shifts right, converting HA into A⁻. Lower the pH (add H⁺) and it shifts left, converting A⁻ back to HA. The Henderson-Hasselbalch equation is simply this qualitative statement turned into precise algebra: it tells you exactly what fraction is HA and what fraction is A⁻ for any given pH, or conversely, what the pH must be for any given HA/A⁻ ratio. Because pH is a logarithmic quantity (pH = −log₁₀[H⁺]) and Ka is also usually reported logarithmically (pKa = −log₁₀ Ka), the natural language of this relationship is logarithms of ratios, not raw concentrations. This is why the equation has the elegant three-term structure pH = pKa + log(ratio): the pKa acts as an anchor point (the pH at which HA and A⁻ are equal), and the logarithmic term is a correction that shifts the pH up or down depending on which species dominates.

Figure

Titration curve and the Henderson-Hasselbalch relationship for acetic acid, pKa = 4.76

The core ideas — governing equations

The dissociation equilibrium of a general weak acid HA is described by its acid dissociation constant:

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

where:

$K_a$ — the acid dissociation constant (units of concentration, mol/L), a fixed number at a given temperature
$[\text{H}^+]$ — the equilibrium concentration of hydrogen ion (mol/L)
$[\text{A}^-]$ — the equilibrium concentration of the conjugate base
$[\text{HA}]$ — the equilibrium concentration of the undissociated acid

Rearranged and converted to logarithms, this becomes the Henderson-Hasselbalch equation:

\text{pH} = \text{p}K_a + \log_{10}\!\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)

where:

$\text{pH} = -\log_{10}[\text{H}^+]$ — the solution's acidity, dimensionless
$\text{p}K_a = -\log_{10}K_a$ — the characteristic constant of the acid, dimensionless
$[\text{A}^-]/[\text{HA}]$ — the ratio of conjugate base to conjugate acid at equilibrium, dimensionless

Reading it: the pH equals the intrinsic pKa of the acid plus a correction term that is positive when base dominates (pushing pH above pKa) and negative when acid dominates (pulling pH below pKa). When the two forms are present in equal concentration, the log term is log(1) = 0, and pH = pKa exactly — this is why pKa is operationally defined as the pH of half-neutralization. For a weak base B in equilibrium with its conjugate acid BH⁺,

\text{BH}^+ \rightleftharpoons \text{B} + \text{H}^+, \qquad K_a = \frac{[\text{H}^+][\text{B}]}{[\text{BH}^+]}

the identical algebra gives

\text{pH} = \text{p}K_a + \log_{10}\!\left(\frac{[\text{B}]}{[\text{BH}^+]}\right)

where $K_a$ here refers to the acid dissociation constant of the conjugate acid BH⁺ (not the base-association constant Kb — biochemistry convention always reports pKa of the acidic species, even for amines, to keep one unified scale).

Derivation AFrom the equilibrium constant to the log-ratio form

Goal: starting only from the mass-action expression for Ka, derive pH = pKa + log([A⁻]/[HA]) without any approximation.

Step 1 — Write the equilibrium expression. For HA ⇌ H⁺ + A⁻ in dilute aqueous solution (activities ≈ concentrations), the equilibrium constant is

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

This is an exact thermodynamic statement at equilibrium; it carries no assumption about how much HA dissociated, because $[\text{H}^+]$ , $[\text{A}^-]$ , and $[\text{HA}]$ are whatever they equilibrate to — not initial concentrations.

Step 2 — Isolate the hydrogen ion concentration. Multiply both sides by $[\text{HA}]/[\text{A}^-]$ :

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

This says the free proton concentration is set by the acid's intrinsic Ka, scaled by how much HA remains relative to A⁻ — more undissociated acid means more protons available to be released, hence a higher $[\text{H}^+]$ .

Step 3 — Take the base-10 logarithm of both sides.

\log_{10}[\text{H}^+] = \log_{10}K_a + \log_{10}\!\left(\frac{[\text{HA}]}{[\text{A}^-]}\right)

Logarithm of a product is the sum of logarithms — this is the only mathematical operation used, applied validly because both sides are strictly positive quantities.

Step 4 — Multiply through by −1 to introduce pH and pKa. By definition $\text{pH} \equiv -\log_{10}[\text{H}^+]$ and $\text{p}K_a \equiv -\log_{10}K_a$ . Negating both sides of Step 3:

-\log_{10}[\text{H}^+] = -\log_{10}K_a - \log_{10}\!\left(\frac{[\text{HA}]}{[\text{A}^-]}\right)

\text{pH} = \text{p}K_a - \log_{10}\!\left(\frac{[\text{HA}]}{[\text{A}^-]}\right)

Step 5 — Invert the ratio to flip the sign of the logarithm. Since $\log_{10}(x^{-1}) = -\log_{10}(x)$ ,

-\log_{10}\!\left(\frac{[\text{HA}]}{[\text{A}^-]}\right) = \log_{10}\!\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)

Substituting gives the final closed form:

\text{pH} = \text{p}K_a + \log_{10}\!\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)

Consistency check. No step assumed anything about initial concentrations, degree of dissociation, or "x is small." The entire derivation used only the mass-action law and the definitions of pH and pKa. This is precisely why the equation remains valid even when [A⁻] and [HA] are comparable in magnitude — the very regime (buffers) where na?ve "assume x is small" approximations from introductory weak-acid problems would fail.

Go deeper · Derivation B · 1 minFraction protonated as a function of pH — the sigmoidal titration curve

Derivation BFraction protonated as a function of pH — the sigmoidal titration curve

Goal: derive an explicit formula for the fraction of acid present as HA (call it $f_{HA}$ ) as a function of pH − pKa, explaining the shape of the titration curve.

Step 1 — Define the fraction protonated. Let the total acid concentration be $C_T = [\text{HA}] + [\text{A}^-]$ (mass balance — nothing is created or destroyed, only interconverted). Define

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

Step 2 — Express the ratio [A⁻]/[HA] via Henderson-Hasselbalch. Rearranging pH = pKa + log([A⁻]/[HA]) gives

\frac{[\text{A}^-]}{[\text{HA}]} = 10^{(\text{pH}-\text{p}K_a)}

This single exponential relation is the algebraic heart of the sigmoidal titration curve: pH − pKa is the "driving force" and the ratio responds exponentially (in log space, linearly) to it.

Step 3 — Substitute into the fraction formula. Divide numerator and denominator of $f_{HA}$ by $[\text{HA}]$ :

f_{HA} = \frac{1}{1 + [\text{A}^-]/[\text{HA}]} = \frac{1}{1 + 10^{(\text{pH}-\text{p}K_a)}}

Step 4 — Similarly derive the fraction deprotonated.

f_{A^-} = 1 - f_{HA} = \frac{10^{(\text{pH}-\text{p}K_a)}}{1+10^{(\text{pH}-\text{p}K_a)}}

Step 5 — Interpret the limits. As $\text{pH}-\text{p}K_a \to -\infty$ (strongly acidic relative to pKa), $10^{(\text{pH}-\text{p}K_a)}\to 0$ , so $f_{HA}\to 1$ : essentially all acid form, as expected. As $\text{pH}-\text{p}K_a \to +\infty$ , $10^{(\text{pH}-\text{p}K_a)}\to\infty$ , so $f_{HA}\to 0$ : essentially all base form. At $\text{pH}=\text{p}K_a$ exactly, $f_{HA}=f_{A^-}=1/2$ — the inflection point of the sigmoid, matching the titration curve's half-equivalence point shown graphically above.

Step 6 — Quantify the "useful buffer range." At pH = pKa + 1, the ratio [A⁻]/[HA] = 10¹ = 10, so $f_{HA} = 1/11 \approx 9.1\%$ . At pH = pKa − 1, [A⁻]/[HA] = 10⁻¹ = 0.1, so $f_{HA} = 10/11 \approx 90.9\%$ . This is the origin of the standard rule that a buffer is effective only within pH = pKa ± 1, since outside that window one form is depleted below about 9% and can no longer supply significant buffering capacity.

Figure

Fraction protonated versus pH − pKa: the logistic titration curve

Worked examples

Example 1 — Foundational: an acetate buffer with an asymmetric ratio. A buffer is prepared by dissolving 0.30 mol of acetic acid (CH3COOH) and 0.70 mol of sodium acetate (CH3COO⁻Na⁺) in enough water to make 1.00 L. The pKa of acetic acid is 4.76. Find the pH. Because the mixture is prepared directly from the acid and its conjugate base salt, and no reaction with water or acid/base has yet occurred, the equilibrium concentrations are (to excellent approximation) the nominal concentrations:

[\text{HA}] = 0.30 \text{ M}, \qquad [\text{A}^-] = 0.70\text{ M}

Apply Henderson-Hasselbalch directly:

\text{pH} = 4.76 + \log_{10}\!\left(\frac{0.70}{0.30}\right)

\text{pH} = 4.76 + \log_{10}(2.333)

\text{pH} = 4.76 + 0.368 = 5.13

The asymmetric ratio (0.70:0.30, not 1:1) correctly pushes the pH above pKa by an amount reflecting the excess of base over acid — exactly as physically expected, since the solution contains more of the deprotonated (basic) form. Limiting-case check: if the ratio were instead 1:1 (0.50 mol each), the log term would be log(1) = 0, giving pH = 4.76 = pKa exactly, matching the defining property of pKa as the half-equivalence pH. This confirms the equation reduces correctly to the known special case.

Example 2 — Applied: bicarbonate buffer in blood plasma. Blood pH is maintained near 7.40 by the CO2/bicarbonate buffer system,

\text{CO}_2 (\text{dissolved}) + \text{H}_2\text{O} \rightleftharpoons \text{H}_2\text{CO}_3 \rightleftharpoons \text{H}^+ + \text{HCO}_3^-

For this system biochemists use an apparent pKa (pKa′ = 6.1) that already folds in the CO2 hydration equilibrium, and write concentrations in terms of dissolved CO2 rather than carbonic acid directly, since [H2CO3] is tiny and hard to measure but is in fixed proportion to dissolved CO2:

\text{pH} = 6.1 + \log_{10}\!\left(\frac{[\text{HCO}_3^-]}{[\text{CO}_2]_{\text{dissolved}}}\right)

Normal plasma has $[\text{HCO}_3^-] = 24$ mM and $[\text{CO}_2]_{\text{dissolved}} = 1.2$ mM (this comes from $P_{CO_2}=40$ mmHg times the solubility coefficient 0.03 mM/mmHg). Then:

\text{pH} = 6.1 + \log_{10}\!\left(\frac{24}{1.2}\right) = 6.1 + \log_{10}(20)

\text{pH} = 6.1 + 1.301 = 7.40

This matches physiological blood pH precisely, and the large bicarbonate-to-CO2 ratio (20:1, not 1:1) shows this buffer operates well outside its "ideal" pH = pKa ± 1 window centered on 6.1 — it is not the most efficient conceivable buffer by that narrow criterion, but its effectiveness in the body comes from a different mechanism: the lungs and kidneys actively regulate the numerator and denominator independently (respiration controls CO2, kidneys control HCO3⁻), turning it into an open buffering system rather than a simple closed mixture. This is the physiological subtlety explored further under Physiological Buffer Systems. Consistency check via reverse calculation: if a patient's ratio drops to 15:1 (respiratory or metabolic acidosis), pH = 6.1 + log(15) = 6.1 + 1.176 = 7.28, correctly predicting acidemia — illustrating how clinicians use this equation directly at the bedside to interpret arterial blood gas results.

Example 3 — Exam/competition level: histidine, a three-pKa amino acid, and its fractional charge at physiological pH. Free histidine has three ionizable groups: the α-carboxyl (pK1 = 1.8), the imidazole side chain (pK2 = 6.0), and the α-amino group (pK3 = 9.2). Find the fraction of histidine molecules with a protonated (positively charged) imidazole side chain at pH 7.4, and comment on why this matters for enzyme catalysis. Because the three ionizations are well separated (each pKa differs by more than 3 units from its neighbors: 1.8, 6.0, 9.2), they can be treated independently — the protonation state of the imidazole ring is governed by its own Henderson-Hasselbalch equilibrium without significant coupling to the carboxyl or amino groups' states at pH 7.4 (both of those are already fully resolved: carboxyl fully deprotonated since pH ≫ pK1, amino group fully protonated since pH ≪ pK3). For the imidazole equilibrium alone (ImH⁺ ⇌ Im + H⁺, pK2 = 6.0):

\text{pH} = \text{p}K_2 + \log_{10}\!\left(\frac{[\text{Im}]}{[\text{ImH}^+]}\right)

Solve for the ratio:

\log_{10}\!\left(\frac{[\text{Im}]}{[\text{ImH}^+]}\right) = \text{pH} - \text{p}K_2 = 7.4 - 6.0 = 1.4

\frac{[\text{Im}]}{[\text{ImH}^+]} = 10^{1.4} = 25.1

Using the fraction-protonated formula derived above with $f_{\text{ImH}^+}$ playing the role of $f_{HA}$ :

f_{\text{ImH}^+} = \frac{1}{1+10^{(\text{pH}-\text{p}K_2)}} = \frac{1}{1+25.1} = \frac{1}{26.1} = 0.0383

So only about 3.8% of histidine side chains are protonated (positively charged) at physiological pH, while 96.2% are neutral. This small but non-negligible protonated fraction is exactly why histidine is the most common catalytic residue at enzyme active sites (e.g., the catalytic triad of chymotrypsin, the distal histidine of hemoglobin, ribonuclease A's His12/His119 pair): with pKa near 6–7 for the imidazole in many protein microenvironments, histidine sits in rapid equilibrium between protonated and neutral forms at physiological pH, allowing it to act alternately as a general acid (donating the ~4% protonated proton) and a general base (accepting a proton via the 96% neutral form) within a single catalytic cycle — a chemical versatility no other proteinogenic amino acid side chain offers so precisely tuned to pH 7.4. Limiting-case check: if the local protein environment shifted this histidine's effective pKa up to 7.4 (matching solution pH exactly, as can happen when the imidazole is buried near a negative charge), the ratio would become 10⁰=1 and exactly 50% would be protonated — the maximum possible buffering/catalytic flexibility for that residue, illustrating why enzymes often tune active-site pKa values toward the operating pH.

Limiting cases, scope & common traps

Regime	Behavior	Why
[A⁻] = [HA]	pH = pKa exactly	log(1) = 0
[A⁻]/[HA] = 10	pH = pKa + 1	log(10) = 1
[A⁻]/[HA] = 0.1	pH = pKa − 1	log(0.1) = −1
[A⁻]/[HA] → ∞ or → 0	equation still holds mathematically but buffer capacity → 0	one reservoir is nearly exhausted; small further addition of acid/base causes large pH swing
Very dilute solutions (< 10⁻⁴ M total buffer)	equation becomes unreliable	water's autoionization ([H⁺] and [OH⁻] from water itself) is no longer negligible compared to buffer species
High ionic strength / concentrated solutions	equation becomes approximate	activities deviate from concentrations; true relation uses activities, $K_a = a_{H^+}a_{A^-}/a_{HA}$
Polyprotic acids with close pKa values (e.g., some dicarboxylic acids)	simple single-pKa treatment fails	overlapping equilibria require the full simultaneous system, not independent treatment

Common trapMisconception: Henderson-Hasselbalch is only an approximation for buffers

Students often carry over the "x is small, so ignore it" approximation from simple weak-acid pH problems and assume Henderson-Hasselbalch is a similar shortcut valid only when a buffer's components are in large excess. This is incorrect. The equation is an exact rearrangement of the Ka expression using the actual equilibrium concentrations of HA and A⁻ — it makes no approximation whatsoever. The only assumptions are (1) concentrations closely approximate activities (fails only at high ionic strength) and (2) water's autoionization contributes negligibly to [H⁺] and [A⁻]/[HA] (fails only in extremely dilute solutions). For any solution where a buffer is actually doing its job — concentrations in the millimolar-to-molar range — the equation is exact for all practical purposes.

The chemist behind it

The equation bears two names for good reason: it was assembled in two independent steps, a decade apart, by researchers working in very different contexts — one a physiologist studying blood chemistry, the other a chemist formalizing acid-base notation. Lawrence Joseph Henderson (1878–1942) was an American physiologist and biochemist at Harvard, deeply influenced by his reading of the physical chemistry of ionic equilibria being developed in Europe at the turn of the twentieth century. In 1908, working on the chemistry of blood, Henderson derived a relationship between the concentrations of carbonic acid and bicarbonate and the hydrogen ion concentration of blood, essentially expressing what we would now call the mass-action equilibrium in a form suited to physiological calculation — but he wrote it in terms of raw concentrations and Ka, not logarithms, because the pH scale itself had not yet been popularized for biological use. Karl Albert Hasselbalch (1874–1962), a Danish physician and chemist working at the Finsen Institute in Copenhagen, took Henderson's relationship in 1916 and reformulated it using the newly introduced pH notation (Sørensen had defined pH in 1909, also in Copenhagen, while working on enzyme kinetics at the Carlsberg Laboratory). Hasselbalch's logarithmic reformulation — converting Henderson's equilibrium expression into the pH = pKa + log(ratio) form — is precisely what made the relationship practically usable for clinical blood-gas interpretation, and it is this form that all of biochemistry now uses. The equation thus represents a beautiful case of applied science: Henderson supplied the physiological chemistry (why the equilibrium matters for blood), and Hasselbalch supplied the mathematical packaging (logarithms) that turned an obscure equilibrium expression into a formula any physician or student could use directly at the bedside or bench.

Origins · 1 minLawrence Joseph Henderson and how this was discovered

Lawrence Joseph Henderson

1878–1942 · Harvard physiologist and biochemist who first derived the blood buffer relationship in 1908, later a pioneer of physiological chemistry and systems thinking in biology

History

- 1909 — Søren Sørensen introduces the pH scale at the Carlsberg Laboratory, Copenhagen, while studying enzyme activity as a function of hydrogen ion concentration.

- 1908 — Lawrence Henderson derives the blood buffer equilibrium relationship connecting CO2, carbonic acid, bicarbonate, and hydrogen ion concentration.

- 1916 — Karl Hasselbalch reformulates Henderson's relationship using Sørensen's pH notation, producing the logarithmic pH = pKa + log(ratio) form still used today.

- 1920s–1930s — the equation becomes standard in clinical blood-gas analysis and physiological chemistry textbooks.

- Present — the equation is foundational across biochemistry, pharmacology, analytical chemistry, and clinical medicine for buffer design, drug ionization prediction, and acid-base diagnosis.

Does the Henderson-Hasselbalch equation work for strong acids like HCl?

No, not meaningfully. Strong acids dissociate essentially completely in water, so at equilibrium [HA] ≈ 0 and [A⁻] ≈ total concentration — the ratio [A⁻]/[HA] becomes enormous or undefined, and the "pKa" of a strong acid is a large negative number that carries little quantitative meaning in aqueous solution (water itself becomes the limiting base). The equation is specifically a tool for weak acids and bases, where a genuine equilibrium exists between meaningful, comparable concentrations of both protonated and deprotonated forms. For strong acids, pH is calculated directly from the (essentially complete) dissociation, i.e., pH ≈ −log[HCl]initial, without needing any equilibrium constant at all.

Why does pH equal pKa exactly at the half-equivalence point of a titration?

At half-equivalence, exactly half of the original weak acid has been converted to its conjugate base by the titrant, so [HA] = [A⁻] in solution. Substituting this equality into the Henderson-Hasselbalch equation gives a log term of log(1) = 0, leaving pH = pKa exactly. This is in fact the standard experimental method for measuring a weak acid's pKa: titrate it with base, plot pH versus volume added, and read the pH at the volume corresponding to half of the total equivalence volume.

Can the Henderson-Hasselbalch equation be used with concentrations expressed in any units, or must they be molarity?

Because the equation involves the logarithm of a ratio of concentrations, the units cancel as long as both [A⁻] and [HA] are expressed in the same units — molarity, millimolarity, mass per volume, or even moles (if the volume is identical for both, as in a single solution). What must not be mixed is inconsistent units between numerator and denominator, or comparing a concentration to an absolute amount (moles) in a different total volume. Clinically, [HCO3⁻] and dissolved CO2 are both routinely expressed in mM/L or mmol/L, and PCO2 in mmHg is converted to an equivalent concentration using the solubility coefficient (0.03 mM per mmHg) before use in the ratio.

Why is the "effective buffer range" said to be pKa ± 1, and not some other window?

This convention comes directly from the fraction-protonated formula, $f_{HA} = 1/(1+10^{\text{pH}-\text{p}K_a})$ . At pH = pKa ± 1, the minority species makes up about 1/11 ≈ 9% of the total buffer pool. Outside this window the minority species becomes so depleted that adding even a small further amount of strong acid or base can completely exhaust it, causing the pH to swing sharply — the buffer "runs out" of one of its two reservoirs. Within pKa ± 1, both reservoirs remain substantial (at least ~9% and at most ~91%), giving meaningful resistance to pH change in both directions, a property formalized quantitatively in the treatment of Buffers Resist pH Change.

How does the equation relate to the definition of buffer capacity?

Buffer capacity (β), defined as the amount of strong acid or base needed to change the pH by a small increment, can be derived by differentiating the fraction-protonated formula with respect to pH. This produces a bell-shaped function of pH that peaks exactly at pH = pKa (where the buffer has the steepest, most resistant response) and falls off symmetrically as pH moves away from pKa in either direction — mathematically confirming, via calculus, the same pKa ± 1 rule of thumb obtained by inspection of the fraction-protonated values above. The Henderson-Hasselbalch equation is thus the equilibrium backbone from which buffer capacity, a kinetic/quantitative-resistance concept, is derived.

Does the equation apply to polyprotic acids like phosphoric or citric acid?

Yes, but one pKa (and one HH equation) is used for each successive ionization step, treated independently as long as the pKa values are well separated (generally by more than about 3 pH units, as with phosphoric acid's pKa1 = 2.15, pKa2 = 7.21, pKa3 = 12.4). Within a couple of pH units of any single pKa, that particular equilibrium dominates and can be treated with a standalone Henderson-Hasselbalch equation, exactly as done for the phosphate buffer example above using only pKa2. When pKa values are close together (within about 2 units), the ionization steps overlap significantly and a full multi-equilibrium treatment (not simple independent Henderson-Hasselbalch steps) is required for high accuracy.

Why do biochemists prefer log base 10 rather than natural log in this equation?

Purely historical and practical convention: pH itself is defined using log base 10 (following Sørensen's original 1909 definition), so pKa is defined the same way for consistency, and the derivation naturally inherits base-10 logarithms throughout. There is nothing mathematically special about base 10 — the same relationship could be written with natural logarithms and a rescaled constant — but using base 10 keeps pH, pKa, and the Henderson-Hasselbalch correction term on the same familiar 0–14-ish numerical scale used throughout all of biochemistry and clinical medicine.