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For the conjugate base ${\mathrm{A}}^{-}$ accepting a proton from water,This base dissociation constant uses the same water thermodynamic activity approximation and standard state conventions as the acid dissociation constant. Multiplying the two relations cancels the partner thermodynamic activities:Here $K_w$ is the aqueous ion product associated with self-ionization of water. At fixed temperature, a larger acid dissociation constant gives a smaller base dissociation constant for its own conjugate base; this reciprocal rule compares matching conjugate acid-base pairs. A weak acid does not necessarily have a conjugate base that reacts almost completely with water. For example, stipulated $K_a=10^{-5}$ and $K_w=10^{-14}$ give $K_b=10^{-9}$.^[ref]

The value of $K_w$ changes with temperature; $10^{-14}$ is the familiar approximate value near $25{}^{\circ }\mathrm{C}$, not a universal constant. If water is retained explicitly in both full equilibrium constants, their product instead equals $a_{{\mathrm{H}}_3{\mathrm{O}}^{+}}{a}_{{\mathrm{OH}}^{-}}/a_{{\mathrm{H}}_2\mathrm{O}}^2$, the full equilibrium constant for $2{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{H}}_3{\mathrm{O}}^{+}+{\mathrm{OH}}^{-}$. The acid dissociation constant, base dissociation constant and self-ionization of water convention must agree.

Stipulate pKa $5$ and $K_w=10^{-14}$; these are model inputs, not measurements of an identified substance. At pH $5$ the pair contribution is $0.575646\dots C$. For analytical concentrations $C=0.010$ and $0.100\mathrm{mol}{\mathrm{L}}^{-1}$, respectively, it isThe free hydrogen ion and hydroxide contribution is the same at this pH, ${\beta }_{\mathrm{water}}=(\ln 10)(10^{-5}+10^{-9})\mathrm{mol}{\mathrm{L}}^{-1}=0.0000230282\dots \mathrm{mol}{\mathrm{L}}^{-1}$ per pH. The full buffer capacities are therefore approximately $0.00577949$ and $0.0575877\mathrm{mol}{\mathrm{L}}^{-1}$ per pH. Identical pH fixes the free hydrogen ion thermodynamic activity; it does not fix the inventory of weak acid and conjugate base available for neutralization.

At one unit of pH on either side of pKa, $r=10$ or $0.1$, giving ${\beta }_{\mathrm{pair}}/{\beta }_{\mathrm{pair},\max }=40/121\approx 0.331$. The partners still form a buffer solution, but its local resistance is only about one third of the maximum for that analytical concentration. At fixed pH, the pair contribution scales exactly with $C$; the full buffer capacity contains the additional $C$-independent water contribution. Consequently, extreme dilution does not justify dropping self-ionization of water.

Assume an ideal dilute aqueous chemical solution with just one monoprotic acid pair $\mathrm{HA}/{\mathrm{A}}^{-}$, self-ionization of water, and other ions that balance electric charge without undergoing further chemical reactions. Hold the analytical concentration $C=[\mathrm{HA}]+[{\mathrm{A}}^{-}]$ constant. Keep the acid dissociation constant $K_a$ and the self-ionization of water constant $K_w$ dimensionless, consistently with pH:Thus $H$ denotes normalized hydrogen ion concentration, not a dimensional concentration. The chemical equilibrium and conserved analytical concentration giveAdding chemical base converts weak acid to conjugate base, increases free hydroxide, and decreases free hydrogen ion concentration. Balancing electric charge, including the added reagent's accompanying ions, gives the signed dose relative to any initial chemical equilibrium at $p_0$:Differentiate with respect to pH, using $\mathrm{d}H/\mathrm{d}p=-(\ln 10)H$. The result isThe first term is the monoprotic acid pair's contribution, ${\beta }_{\mathrm{pair}}$; the remaining terms are the free hydrogen ion and hydroxide contributions, ${\beta }_{\mathrm{water}}$. Each has units $\mathrm{mol}{\mathrm{L}}^{-1}$ per pH. Equivalently, with dimensional $H_c=[{\mathrm{H}}^{+}]$, $K_{a,c}=c^{\circ }{K}_a$ and $K_{w,c}=(c^{\circ }{)}^2{K}_w$, the expression is $(\ln 10)(C{K}_{a,c}{H}_c/(K_{a,c}+H_c{)}^2+H_c+K_{w,c}/H_c)$. Mixing the dimensional and dimensionless conventions would give incorrect units.

For $r=K_a/H=10^{\mathrm{pH}-\mathrm{p}{K}_a}$,because $(1+r{)}^2-4r=(1-r{)}^2\ge 0$. Equality occurs at $r=1$: equal weak acid and conjugate base concentrations, or pH equal to pKa. This is the exact maximum of the pair contribution in this model. The full buffer capacity need not have its maximum there: free hydrogen ions and hydroxide make it grow toward the acidic and alkaline extremes. When these contributions are small, the central peak lies close to pKa.

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.

A chemical equation represents a chemical reaction using chemical formulas, with reactants on the left and reaction products on the right. A balanced chemical equation has the same number of chemical atoms of each chemical element on both sides. For molecular hydrogen and molecular oxygen forming water,The coefficients $2,1,2$ give the relative numbers of molecules; the coefficient $1$ is normally omitted. Count hydrogen as $2\times 2=4$ on the left and $2\times 2=4$ on the right, and oxygen as $1\times 2=2$ on the left and $2\times 1=2$ on the right. See Flowers et al., Chemistry 2e, §4.1: writing and balancing chemical equations.

To balance the initial expression ${\mathrm{H}}_2+{\mathrm{O}}_2⟶{\mathrm{H}}_2\mathrm{O}$, first put $2$ before water to match the two oxygen chemical atoms. This creates four hydrogen chemical atoms on the right, so put $2$ before molecular hydrogen. Changing a coefficient multiplies the entire chemical formula. Changing a subscript changes composition and usually the chemical substance: $2{\mathrm{H}}_2\mathrm{O}$ is two water molecules, whereas ${\mathrm{H}}_2{\mathrm{O}}_2$ is one hydrogen peroxide molecule. Never balance a chemical equation by changing its subscripts.

The bookkeeping follows from conservation of the atomic nuclei in ordinary, non-nuclear chemical reactions; chemical bonds and the distribution of electrons can change. For a chemical equation involving ions, total electric charge must also match. For example, the existing self-ionization of water equation has zero total electric charge on each side, since the $+1$ of hydronium cancels the $-1$ of hydroxide. See 2.6 Ionic and Molecular Compounds (§2.6: nuclei and ions).

The same $2:1:2$ ratio relates changes in amount of substance: two moles of molecular hydrogen consumed per mole of molecular oxygen correspond to two moles of water formed. A mole is a defined unit, not one molecule; the little whole-number picture is an abstract counting model, not a claim that a macroscopic sample contains just five molecules. See molecular count and amount of substance for the exact conversion and SI base unit: mole by Bureau International des Poids et Mesures for the definition.

This is why a balanced chemical equation is useful: stoichiometry turns the same atom accounting into relations between measurable quantities. It does not determine reaction rate, the final conversion set by chemical equilibrium, or the actual yield of recovered reaction products. The limiting reagent gives an upper bound under the specified chemical reaction.

The equivalence point is the composition at which added titrant and the initial sample inventory match the specified chemical reaction's stoichiometry. It is a theoretical inventory condition; the titration endpoint is an observed stopping criterion. For a monoprotic acid consuming one hydroxide per acid unit, initial acid molar concentration $c_A$, sample volume $V_A$, and base titrant molar concentration $c_B$ giveThe proof is amount of substance conservation with one-to-one neutralization; a different stoichiometry requires its coefficients. All volumes must use compatible units. Since $1\mathrm{mL}=10^{-3}\mathrm{L}$, $0.100\mathrm{mol}{\mathrm{L}}^{-1}\times 10.00\mathrm{mL}=0.001000\mathrm{mol}$, not $1\mathrm{mol}$.

An equivalence point does not prescribe pH $7$. A weak acid titrated with a strong base leaves its conjugate base, which reacts with water; the corresponding chemical equilibrium can make the chemical solution alkaline. The neutral strong acid/strong base case additionally depends on temperature, through self-ionization of water.^[ref]

Adding sodium hydroxide to acidified glycine successively removes its two titratable protons. Two broad buffer solution regions appear around the two pKa values, separated by a steep pH rise near the isoelectric point. This is useful evidence for two separated acid-base chemistry steps even though the changing chemical species are invisible. In the approximation that added hydroxide only changes glycine protonation, $b=1-\bar{z}$ is the added base amount of substance per initial amount of substance of glycine. Then $b\approx 0.5$, $1$ and $1.5$ correspond to $\mathrm{p}{K}_{a1}$, $\mathrm{pI}$ and $\mathrm{p}{K}_{a2}$; endpoints are not exactly pure chemical species. ^[ref]

For the solid-curve model, start with $n_G=0.00200\mathrm{mol}$ of glycine in $V_0=0.0200\mathrm{L}$ and exactly $n_G$ of fully dissociated hydrochloric acid; this fixes the chloride inventory, but does not force every glycine molecule to remain $G^{+}$. Let the sodium hydroxide concentration be $C_B=0.100\mathrm{mol}{\mathrm{L}}^{-1}$ and $b=n_{\mathrm{NaOH}}/n_G$. Assume volume is additive and no other ions are present; stipulate $K_w=10^{-14}$ for self-ionization of water at a temperature of $25{}^{\circ }\mathrm{C}$. ThenBalancing all positive and negative electric charges requiresSolving this equation gives the solid curve. It remains finite at $b=0$ and beyond $b=2$, where ignoring free hydronium and hydroxide would fail. At the first pKa, the full model gives $b\approx 0.436$, rather than $0.5$: appreciable free hydronium makes that half-equivalent shortcut inaccurate. At the second pKa, $b\approx 1.502$. Known acid inventory, dilution, and thermodynamic activity assumptions are therefore part of any quantitative interpretation, not optional corrections inferred from the curve's shape.

Consider a neutral monoprotic acid $\mathrm{HA}$ dissolved in water, with no added salts or other acids. Assume dilute ideal thermodynamic activities, fixed temperature, water thermodynamic activity one, and negligible self-ionization of water. Write $C_T$ for its analytical concentration, $c=C_T/c^{\circ }$ with $c^{\circ }=1\mathrm{mol}{\mathrm{L}}^{-1}$, and $\alpha$ for its weak acid dissociation fraction. Conservation of acid units and electric charge givesHere ${\mathrm{A}}^{-}$ is the conjugate base, and ${\mathrm{H}}^{+}$ is conventional shorthand for hydrated hydrogen ions, represented by hydronium in this model. Substitution into the dimensionless acid dissociation constant givesSolving this quadratic equation and rejecting its negative root of a polynomial gives an exact result within the stated model:The second form avoids subtracting nearly equal numbers. Unlike the approximation $\alpha \approx \sqrt{K_a/c}$, it retains depletion of neutral weak acid and always satisfies $0<\alpha <1$ for positive $K_a,c$. The broader weak acid mass and charge balances restore self-ionization of water when that omitted contribution matters.

For an original worked example, stipulate $K_a=10^{-5}$ without identifying a real substance. At $C_T=10^{-2}\mathrm{mol}{\mathrm{L}}^{-1}$, the weak acid dissociation fraction is approximately $0.0311267$, or $3.11267\%$, and $[{\mathrm{H}}_3{\mathrm{O}}^{+}]=3.11267\times 10^{-4}\mathrm{mol}{\mathrm{L}}^{-1}$. A hundredfold dilution to $C_T=10^{-4}\mathrm{mol}{\mathrm{L}}^{-1}$ raises the fraction to $0.270156$, or $27.0156\%$, but lowers $[{\mathrm{H}}_3{\mathrm{O}}^{+}]$ to $2.70156\times 10^{-5}\mathrm{mol}{\mathrm{L}}^{-1}$. A completely ionized strong acid at $C_T=10^{-4}\mathrm{mol}{\mathrm{L}}^{-1}$ would have $[{\mathrm{H}}_3{\mathrm{O}}^{+}]\approx 10^{-4}\mathrm{mol}{\mathrm{L}}^{-1}$: less than the more concentrated weak acid in the first calculation, despite its greater converted fraction.

The directions of both changes follow without relying on the plot. Put $x=[{\mathrm{H}}_3{\mathrm{O}}^{+}]/c^{\circ }=c\alpha$. Rearranging the chemical equilibrium relation givesThus smaller $c$ forces smaller $x$, which in turn makes $\alpha$ larger. The derivative proves that total hydronium concentration and converted fraction move in opposite directions during dilution. In this model, adding solvent also makes additional acid units ionize at fixed total acid amount of substance, even though their ions are spread over more volume. That distinction explains why “more ionized” alone does not predict greater electrical conductivity.

Do not extend the no-background model to unlimited dilution: its prediction $[{\mathrm{H}}_3{\mathrm{O}}^{+}]\to 0$ omits water's own ions. The weak acid mass and charge balances include that background. Added salts, other acids or bases, and nonideal activity coefficients also change the relation between analytical concentration, converted fraction and hydronium concentration.

The ionic product of water describes the balance between hydronium and hydroxide produced by self-ionization of water:For a dilute water-based chemical solution whose water activity is approximately one,The ionic thermodynamic activities are dimensionless. In a dilute molal convention, $a_i\simeq b_i/b^{\circ }$, where $b_i$ are molalities and $b^{\circ }=1\mathrm{mol}\mathrm{k}{\mathrm{g}}^{-1}$. The full equilibrium constant for the reaction includes water activity squared in the denominator; concentrated solutions need that solvent contribution as well.

Near ambient pressure at $25{}^{\circ }\mathrm{C}$, $K_w$ is approximately $10^{-14}$. Equal ionic activities in neutral dilute water are therefore approximately $10^{-7}$, giving pH near seven. The value changes with temperature and pressure, so neutrality is not universally the same as pH seven. ^[ref]

A nonelectrolyte supplies neutral molecules rather than ions on dissolving in the specified solvent, as illustrated by sucrose in water.

For example, dissolving an idealized $1\mathrm{mmol}$ sucrose inventory gives $1\mathrm{mmol}$ dissolved sucrose molecules, rather than separate cations and anions. Dissolving the same amount of substance of a fully dissociated one-to-one strong electrolyte instead gives $1\mathrm{mmol}$ of each ion. Both can disappear visibly into the solvent; the resulting chemical species distinguish them.^[ref]

A nonelectrolyte is not necessarily insoluble: sucrose provides a soluble counterexample. Nor does the label assert exactly zero measured electrical conductivity. Self-ionization of water, impurities or residual electrolyte from a probe can supply background ions. Comparing with a solvent blank and rinsing between samples, as in the strong electrolyte experiment, tests whether the dissolved substance itself produces appreciable additional mobile ions. This separates visible dissolution from ion production without treating lamp darkness as an absolute absence of electric charge.

The pH of a chemical solution is the negative base-ten logarithm of its hydrogen ion thermodynamic activity:The thermodynamic activity is a dimensionless quantity, so the logarithm has a dimensionless argument. The definition uses thermodynamic activity, rather than total chemical acid concentration. Single-ion thermodynamic activities require a convention; practical pH measurements are tied to calibrated standards. See IUPAC Gold Book, pH.

In a dilute ideal chemical solution in water, a useful approximation isSquare brackets denote concentration in moles per litre. In water, $[{\mathrm{H}}^{+}]$ is shorthand for the hydronium concentration, often written $[{\mathrm{H}}_3{\mathrm{O}}^{+}]$; it does not describe isolated bare protons. Departures from ideal behavior need activity coefficients. The concentration formula approximates the thermodynamic activity definition.

For an ideal strong acid supplying one proton per initial unit, stipulate a total concentration of $0.001\mathrm{mol}{\mathrm{L}}^{-1}$ and negligible self-ionization of water. Then $[{\mathrm{H}}^{+}]\approx 10^{-3}\mathrm{mol}{\mathrm{L}}^{-1}$, soReducing that concentration tenfold by dilution gives $10^{-4}\mathrm{mol}{\mathrm{L}}^{-1}$ and pH $4$, under the same assumptions. More generally, dividing the hydrogen ion thermodynamic activity by ten raises pH by one. This is why the scale compresses an enormous range: equal steps of pH represent multiplicative changes, rather than equal differences in concentration. These are model calculations, rather than measured samples.

Neutrality depends on temperature. In the dilute ideal water model, define the dimensionless ion product $K_w=([{\mathrm{H}}^{+}]/c^{\circ })([{\mathrm{OH}}^{-}]/c^{\circ })$ and $\mathrm{p}{K}_w=-{\log }_{10}{K}_w$. Equal hydronium and hydroxide concentrations implyAt $25{}^{\circ }\mathrm{C}$, the textbook approximation $K_w\approx 10^{-14}$ gives neutral pH $7$. Since $K_w$ varies with temperature, neutrality need not mean pH $7$. Nor is pH restricted to $0$–$14$: the activity definition permits negative pH when $a_{{\mathrm{H}}^{+}}>1$, and pH above $14$ when $a_{{\mathrm{H}}^{+}}<10^{-14}$. High concentrations can invalidate the dilute ideal approximation. See Flowers et al., Chemistry 2e, pH and pOH.

S. P. L. Sørensen introduced pH at the Carlsberg Laboratory in 1909, as recorded in the Carlsberg Foundation historical timeline.

A solubility product constrains a product of thermodynamic activities; the available inventories determine which combinations of concentrations are possible. Their intersection determines the molar solubility. Consider an original, abstract AB solid, with the preceding one-to-one dissolution chemical equation. Stipulate $c^{\circ }=1\mathrm{M}$, ${\gamma }_{\mathrm{A}}={\gamma }_{\mathrm{B}}=1$ and $K_{\mathrm{sp},c}=10^{-4}{\mathrm{M}}^2$ at fixed temperature and pressure. Thus $K_{\mathrm{sp}}=10^{-4}$ is dimensionless. These are chosen model parameters, not data for an identified material.

Initially, a completely dissociated soluble XB salt supplies B anions at molar concentration $c_0\ge 0$ and X cations at the same molar concentration; A is absent. All three ions have unit electric charge magnitude. X remains a spectator ion, and only AB can form a solid. Assume constant volume, no additional chemical reactions, and negligible contributions from self-ionization of water. Dissolving AB raises both reacting-ion concentrations by the same $s$:Charge neutrality is satisfied: $c_{\mathrm{A}}+c_{\mathrm{X}}-c_{\mathrm{B}}=s+c_0-(s+c_0)=0$. Adding B alone without its accompanying cation would omit the initial electric charge balance.

In the plane of the two reacting-ion concentrations, this inventory constraint is the line $c_{\mathrm{B}}=c_{\mathrm{A}}+c_0$. Saturation supplies the hyperbola $c_{\mathrm{B}}=K_{\mathrm{sp},c}/c_{\mathrm{A}}$ for $c_{\mathrm{A}}>0$. Their positive intersection solves a quadratic equation:The quadratic formula supplies two roots of a polynomial; only the positive one is a physical concentration. Uniqueness also follows because $s(s+c_0)$ strictly increases for $s\ge 0$, starting at zero and growing without bound. The second expression, obtained by multiplying by the conjugate, avoids subtraction of nearly equal numbers at large $c_0$.

For pure water, $c_0=0$ gives $s_{\ast }=\sqrt{K_{\mathrm{sp},c}}=0.010\mathrm{M}$. For $c_0=0.030\mathrm{M}$,Their product is $10^{-4}{\mathrm{M}}^2$, although neither concentration equals the pure-water value. This is the useful geometrical distinction: an inventory line and an equilibrium constant specify different constraints.

Check the solid inventory before imposing saturation. If the initially added AB has amount of substance $n_0$ in volume $V$, the saturated branch requires $n_0>V{s}_{\ast }$ for excess solid to remain. For example, choose $n_0=0.0020\mathrm{mol}$ and $V=0.100\mathrm{L}$. In pure water, $0.0010\mathrm{mol}$ dissolves and the same amount of substance remains as solid. At $c_0=0.030\mathrm{M}$, approximately $0.0003027756\mathrm{mol}$ dissolves, leaving $0.0016972244\mathrm{mol}$ as solid. In this model, an insufficient supply instead givesIf $n_0/V<s_{\ast }$, all AB dissolves and the ion product is below the solubility product. At $n_0/V=s_{\ast }$, the last solid disappears exactly at saturation.

How can an invisible dissolved inventory be measured? The unsigned laboratory handout Solubility Product Constant, Ksp, for Calcium Hydroxide by Acid-Base Titration, Procedure, starts with saturated calcium hydroxide, supplied by the teacher or prepared from calcium and water. Its apparatus includes filter paper, a funnel, stoppered flasks, a volumetric pipette, a volumetric flask and a burette. Two successive rounds of filtration remove particles. The filtrate is stoppered to reduce carbon dioxide exposure. A $10.00\mathrm{mL}$ aliquot is diluted with deionized water and titrated with standardized hydrochloric acid; bromothymol blue gives a pale-green titration endpoint. The readout is the difference between final and initial burette readings; four consistent trials are required.^[ref]

Unlike AB, calcium hydroxide has one-to-two dissolution stoichiometry:For molar solubility $s$ in the ideal model with no background calcium or hydroxide, no additional chemical reactions, and negligible self-ionization of water, $[{\mathrm{Ca}}^{2+}]=s$ and $[{\mathrm{OH}}^{-}]=2s$. Hence $K_{\mathrm{sp},c}=4s^3$, with units ${\mathrm{M}}^3$; it is not the AB constant used in the figures. The neutralization chemical equation is ${\mathrm{H}}^{+}+{\mathrm{OH}}^{-}⟶{\mathrm{H}}_2\mathrm{O}$. Let $C_H$ be the known titrant molar concentration, $v$ its delivered volume at the equivalence point, and $V_a$ the original aliquot volume. Then the inventory calculation isDilution of the aliquot changes its concentration during the acid-base titration, but preserves its initial amount of substance of hydroxide. Therefore $V_a$, rather than the diluted flask volume, belongs in the denominator.

The handout identifies two biases: carried-over calcium hydroxide solid dissolves during acid-base titration, consuming extra titrant; carbon dioxide consumes hydroxide and forms calcium carbonate. Filtration and stoppering address these effects, while repeat trials test consistency. Its review questions also ask about dissolved ion pairs, which filtration cannot remove.^[ref]

Further controls would compare a water blank, independently verify titrant standardization, and record the saturation temperature; these are proposed checks, not reported measurements. A titration endpoint estimates the equivalence point and may introduce a systematic offset. Even perfect inventory measurement does not determine activity coefficients. Recovering the dimensionless thermodynamic solubility product additionally requiresThis explains both the power and the limit of the experiment: acid-base titration measures a dissolved inventory, while the link from that inventory to a thermodynamic equilibrium constant still depends on a chemical model.

Consider an original ideal model: $V_A=10.00\mathrm{mL}$ of hydrochloric acid at $c_A=0.100\mathrm{mol}{\mathrm{L}}^{-1}$, titrated with sodium hydroxide at $c_B=0.100\mathrm{mol}{\mathrm{L}}^{-1}$. Treat these numbers as exact model inputs. Assume complete dissociation of an ionic compound, complete strong acid ionization, additive volumes, uniform mixing, and activity coefficients equal to one. Use stipulated $K_w=10^{-14}$, approximately appropriate near $25{}^{\circ }\mathrm{C}$. These are calculated pH values, not measurements of these materials.

The chemical equation for neutralization isInitial acid inventory is $n_A=1.000\mathrm{mmol}$, so the equivalence point is $v=10.00\mathrm{mL}$. Added water does not change this inventory. But the final volume $V_A+v$, including the titrant, determines all concentrations.

Keep self-ionization of water even at the equivalence point. With $c^{\circ }=1\mathrm{mol}{\mathrm{L}}^{-1}$, define the dimensionless acid excess $d$, hydronium level $h$, and hydroxide level $o$ byThe sodium cation and chloride anion inventories fix the electric charge balance $h-o=d$, while chemical equilibrium gives $ho=K_w$. Eliminating $o$ yields a quadratic equation $h^2-dh-K_w=0$ with exactly one positive root. For stable calculation on either side, writeThe second expression follows by multiplying by $s-d$; it avoids subtracting two nearly equal large quantities on the alkaline side. A logarithm acts on the dimensionless $h$, as required by the pH definition.

At $v=5.00\mathrm{mL}$, the excess is $0.500\mathrm{mmol}$ in $15.00\mathrm{mL}$, giving $d=1/30$ and pH $1.477121\dots$. At $v=15.00\mathrm{mL}$, excess hydroxide is $0.500\mathrm{mmol}$ in $25.00\mathrm{mL}$, giving $d=-1/50$ and pH $12.301030\dots$. Neglecting dilution would give the wrong values even far from the equivalence point.

Exactly at $v=10.00\mathrm{mL}$, $d=0$ and $h=o=\sqrt{K_w}=10^{-7}$, giving pH $7$ in this model. At $9.999\mathrm{mL}$ and $10.001\mathrm{mL}$, the same smooth expression gives pH $5.300835\dots$ and $8.699121\dots$. The familiar excess-only formulas would predict zero hydronium or zero hydroxide at the equivalence point; retaining self-ionization of water removes that failure. A steep curve is still continuous.

The sensitivity is especially revealing. Differentiating $h-K_w/h=d$ giveswhen $c_A=c_B$. At this model's equivalence point, the derivative is about $1.086\times 10^4$ pH units per millilitre. This is a local derivative, not a linear prediction over a whole drop. Real endpoint accuracy also depends on dosing, mixing, measurement uncertainty, temperature, extra reacting chemical species, and nonideal thermodynamic activities.

For a monoprotic acid with analytical concentration $C_T=[\mathrm{HA}]+[{\mathrm{A}}^{-}]$, define the dimensionless weak acid dissociation fractionThis counts acid units in the conjugate base form. Identifying it with $[{\mathrm{H}}_3{\mathrm{O}}^{+}]/C_T$ additionally requires negligible self-ionization of water and no other sources of electric charge. In the ideal dilute chemical solution model, with $h=[{\mathrm{H}}_3{\mathrm{O}}^{+}]/c^{\circ }$, the acid dissociation constant gives $\alpha =K_a/(K_a+h)$. Thus lowering hydronium thermodynamic activity favors the conjugate base.

The diagram uses stipulated dimensionless acid dissociation constant $K_a=10^{-5}$ and self-ionization of water ion product $K_w=10^{-14}$ at fixed temperature; these are model inputs, not measurements of a named substance. It solves the weak acid mass and charge balances. At $C_T=0.010\mathrm{mol}{\mathrm{L}}^{-1}$, the calculated weak acid dissociation fraction is about $3.11\%$ and pH about $3.51$; at $C_T=10^{-4}\mathrm{mol}{\mathrm{L}}^{-1}$, they are about $27.0\%$ and $4.57$. More acid units are dissociated as a fraction of the total, yet fewer hydronium ions occur per volume.

At extreme dilution, self-ionization of water makes $h$ approach $\sqrt{K_w}$, so pH approaches $7$ for the chosen model. The limiting weak acid dissociation fraction is $K_a/(K_a+\sqrt{K_w})\approx 0.9901$, rather than exactly one. The calculation assumes pure added water, a neutral monoprotic acid, and no competing chemical reactions; actual contamination becomes increasingly important as $C_T$ falls.

Consider only neutral $\mathrm{HA}$ added to pure water, at fixed temperature, with ideal dilute thermodynamic activities, water thermodynamic activity approximated by one, and no added salts or other chemical species. Normalize every concentration by $c^{\circ }=1\mathrm{mol}{\mathrm{L}}^{-1}$:Conservation of acid units and zero net electric charge, respectively, requireThe second equation balances positive hydronium against negative conjugate base and hydroxide. It would need all additional ions if salts or a chemical base were added. Together with acid dissociation constant $K_a=hx/y$ and the self-ionization of water relation $K_w=ho$, these giveThis is exact within the stated ideal model, including depletion and self-ionization of water. Multiplication by $h(K_a+h)$ yields the cubic equationFor positive $K_a,K_w,c$, the unmultiplied left-hand side increases strictly from negative to positive values as $h$ increases from zero to infinity; its derivative is $1+K_w/h^2+K_{a}c/(K_a+h{)}^2>0$. There is therefore exactly one positive root of a polynomial. A bisection method can find it without choosing an unphysical root of a polynomial.

If $K_w/h^2\ll 1$, neglect hydroxide, so $h\approx x$ and the quadratic equation retaining acid depletion givesThe last form avoids subtracting nearly equal numbers. If also $x/c\ll 1$, neglect depletion to obtainCheck both conditions after computing: $x/c$ measures the fraction lost from the initial undissociated weak acid inventory, while $K_w/h^2$ measures the omitted hydroxide contribution to electric charge balance. A $5\%$ depletion criterion is a chosen tolerance, not a universal boundary. Keeping depletion does not cure the neglect of self-ionization of water. These weak acid mass and charge balances also explain why the square root estimate cannot be extended to arbitrarily dilute chemical solutions.