= Buffer solution
{wiki}
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 https://openstax.org/books/chemistry-2e/pages/14-6-buffers[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>:
$$
\mathrm{HA}+\mathrm{OH}^-\longrightarrow\mathrm A^-+\mathrm H_2\mathrm O.
$$
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> are
$$
n_{\mathrm{HA}}\approx(0.10-0.01)\,\mathrm{mol}=0.09\,\mathrm{mol},\qquad
n_{\mathrm A^-}\approx(0.10+0.01)\,\mathrm{mol}=0.11\,\mathrm{mol}.
$$
The shared volume cancels from the <concentration> ratio. The <Henderson-Hasselbalch equation> therefore predicts
$$
\mathrm{pH}\approx5+\log_{10}\!\left(\frac{0.11}{0.09}\right)=5+\log_{10}\!\left(\frac{11}{9}\right)=5.0871501757\ldots.
$$
This 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.
\Image[/buffer-neutralization-inventory.png]
{title=Added <hydroxide> transfers a finite inventory from <weak acid> to <conjugate base>}
{description=Toy <pKa> $5$, initial volume $1\,\mathrm L$, and negligible added volume. Bars show the dominant <neutralization> bookkeeping before and after adding $0.010\,\mathrm{mol}$ of <hydroxide>; the <pH> values use the ideal <concentration> approximation. The acid-plus-base inventory stays $0.20\,\mathrm{mol}$. Neither the bars nor the <pH> values are measured material data or exact speciation.}
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 https://openstax.org/books/chemistry-2e/pages/14-6-buffers[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 relation
$$
\mathrm{pH}=\mathrm pK_a+\log_{10}\!\left(\frac{a_{\mathrm A^-}}{a_{\mathrm{HA}}}\right).
$$
With <thermodynamic activity> of <water> approximated by one and ideal dilute <thermodynamic activities> for both partners, this yields the familiar <Henderson-Hasselbalch equation>:
$$
\mathrm{pH}\approx\mathrm pK_a+\log_{10}\!\left(\frac{[\mathrm A^-]}{[\mathrm{HA}]}\right).
$$
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 $\mathrm pK_a=4.76$ for an illustrative <weak acid>, without identifying it with a measured substance. A <conjugate base> to <weak acid> <concentration> ratio of two gives
$$
\mathrm{pH}\approx4.76+\log_{10}2=5.061029995\ldots\approx5.06103.
$$
Equal <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.
\Image[/buffer-ratio-ph.png]
{title=The <Henderson-Hasselbalch equation> turns a tenfold ratio change into one unit of <pH>}
{description=The stipulated <weak acid> has $\mathrm pK_a=4.76$. Points mark <conjugate base> to <weak acid> <concentration> ratios $0.1$, $1$, $2$, and $10$. The curve uses ideal dilute <thermodynamic activities>, <chemical equilibrium>, and negligible <self-ionization of water>; it is a calculation, rather than measured data.}
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