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Buffer capacity measures resistance to a change in pH: how much additional chemical acid or chemical base a chemical solution accommodates per small change in pH. Its differential form is a local derivative, rather than the total inventory available before a buffer solution is exhausted. Two buffer solutions can have identical pH and very different buffer capacities.

Choose a constant volume $V$ and fixed temperature. Let $n_B$ count added moles of fully dissociated, monovalent chemical base supplying hydroxide, and $n_A$ added moles of fully dissociated, monovalent strong acid supplying hydrogen ions. Neither addition supplies the weak acid component itself. The signed addition is $b=(n_B-n_A)/V$, with units $\mathrm{mol}{\mathrm{L}}^{-1}$, andThe last two forms describe separate base-only and acid-only additions. The buffer capacity is positive for the stable chemical equilibrium model below and is reported in $\mathrm{mol}{\mathrm{L}}^{-1}$ per unit of pH. Since pH is dimensionless, the dimensional unit is just $\mathrm{mol}{\mathrm{L}}^{-1}$; stating “per pH” makes the measurement convention clear. Here constant volume is an idealization, realized approximately by small additions of concentrated reagent.

For a sufficiently small addition, $\Delta \mathrm{pH}\approx \Delta b/\beta$. A large buffer capacity means a small local change of pH for a given addition. Over a finite interval, use the changing buffer capacity instead:This integral gives a finite dose in $\mathrm{mol}{\mathrm{L}}^{-1}$; it need not equal the initial buffer capacity times the eventual change in pH.

A buffer solution contains appreciable quantities of a weak acid and its conjugate base, and resists changes in pH caused by small quantities of added chemical acid or chemical base. The conjugate base accepts added protons; the weak acid supplies protons when a chemical base consumes them. Consuming most of either partner undermines the buffering action. See Flowers et al., Chemistry 2e, buffers.

For a separate toy calculation, stipulate pKa $5$ for a weak acid $\mathrm{HA}$; this is an input, not measured material data. Initially $1\mathrm{L}$ of buffer solution contains $0.10\mathrm{mol}$ of $\mathrm{HA}$ and $0.10\mathrm{mol}$ of its conjugate base ${\mathrm{A}}^{-}$. In the ideal concentration approximation, the Henderson-Hasselbalch equation gives initial pH $5$. Add $0.010\mathrm{mol}$ of hydroxide with negligible volume change. The dominant neutralization has one-to-one stoichiometry:Each mole of consumed hydroxide removes one mole of weak acid and forms one mole of conjugate base and water. Thus the inventories after neutralization areThe shared volume cancels from the concentration ratio. The Henderson-Hasselbalch equation therefore predictsThis small increase illustrates how conversion between abundant partners absorbs added chemical base. The calculation neglects the tiny free hydronium and hydroxide inventories, changes during re-establishment of chemical equilibrium, self-ionization of water, and activity coefficient corrections. The displayed inventories approximate the final chemical species distribution; they are not an exact solution of all chemical equilibrium conditions.

A buffer solution has a finite buffer capacity: its weak acid inventory supplies protection against added chemical base, and its conjugate base inventory supplies protection against added chemical acid. Here an addition approaching $0.10\mathrm{mol}$ of hydroxide uses up the available $\mathrm{HA}$ in the dominant stoichiometry calculation. Close to that limit, the neglected chemical equilibrium contributions matter; once $\mathrm{HA}$ is exhausted, inserting a zero denominator into the Henderson-Hasselbalch equation cannot predict the actual pH. Excess hydroxide and the remaining chemical species require a fresh chemical equilibrium calculation. This is why a buffer solution stabilizes conditions for chemical reactions over a finite range of additions. See Flowers et al., Chemistry 2e, buffer capacity and added base.

For the pair $\mathrm{HA}/{\mathrm{A}}^{-}$, rearranging the acid dissociation constant gives the activity relationWith thermodynamic activity of water approximated by one and ideal dilute thermodynamic activities for both partners, this yields the familiar Henderson-Hasselbalch equation:The concentrations are at chemical equilibrium; using initial values requires negligible equilibration changes. Below, also neglect self-ionization of water and competing chemical reactions.

Stipulate $\mathrm{p}{K}_a=4.76$ for an illustrative weak acid, without identifying it with a measured substance. A conjugate base to weak acid concentration ratio of two givesEqual concentrations give a ratio of one, whose logarithm is zero, hence pH $4.76$. Multiplying the ratio by ten raises pH by one: ratios $0.1$, $1$, and $10$ give pH values $3.76$, $4.76$, and $5.76$. The ratio controls the ideal prediction, while the absolute quantities control how much added chemical acid or chemical base the buffer solution can accommodate.

The Henderson-Hasselbalch equation estimates buffer solution pH using a weak acid's acid dissociation constant and the conjugate base to weak acid concentration ratio.

Its activity form is an exact rearrangement of the chosen chemical equilibrium relation. Write the acid dissociation step as $\mathrm{HA}\rightleftharpoons {\mathrm{H}}^{+}+{\mathrm{A}}^{-}$, with hydrated hydrogen ions understood in water and the solvent convention incorporated into the acid dissociation constant. For consistent standard states,Here each $a_i$ is a dimensionless thermodynamic activity at chemical equilibrium. Exactness refers to this specified acid dissociation constant relation and the chosen single-ion thermodynamic activity convention, not to identifying the ratio with quantities initially mixed. If water is instead written explicitly as a reactant, its thermodynamic activity must be treated consistently. See Hippler and Metcalfe, Using activities to correct the Henderson-Hasselbalch equation (2020), equations 3–4.

For a toy weak acid with stipulated pKa $5$, use the ideal dilute concentration approximation and put $r=[{\mathrm{A}}^{-}]/[\mathrm{HA}]$. The Henderson-Hasselbalch equation becomes $\mathrm{pH}\approx 5+{\log }_{10}r$:More conjugate base relative to weak acid means higher pH. Each tenfold increase in the ratio contributes exactly one to the calculated pH. The stipulated pKa is not a tabulated value for an identified substance.

The distinction between thermodynamic activity and concentration is explicit when $a_i={\gamma }_i[i]/c^{\circ }$, where ${\gamma }_i$ is an activity coefficient and $c^{\circ }$ is the reference concentration:Neglecting the final term gives the usual approximate Henderson-Hasselbalch equation. It is justified when the ratio of activity coefficients is sufficiently close to one, as in a suitable dilute ideal approximation. Ionic strength, including contributions from other dissolved ions, affects activity coefficients; a nominally dilute weak acid alone does not establish that all corrections are small. The ratio must use chemical equilibrium concentrations. Replacing these by prepared amounts or post-neutralization inventories is a further approximation, suitable only when subsequent redistribution is small. See Hippler and Metcalfe, concentration, activity, and analytical-concentration approximations.

The short derivation is to solve the acid dissociation constant relation for $a_{{\mathrm{H}}^{+}}$, take a negative base-ten logarithm, and use the definitions of pH and pKa. Its useful beauty is that a chemical equilibrium becomes a simple rule about a ratio: select a weak acid with pKa near the desired pH, then adjust the partner ratio. Scaling both partner concentrations equally leaves the ideal pH prediction unchanged, while changing buffer capacity. In practical buffer solutions, temperature, ionic strength, and the available inventories still matter.