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The conjugate base of acetic acid. It is a reaction product in the conductometric saponification assay.

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 carboxylate is the conjugate base of a carboxylic acid; an acetate ion is one example.

A conjugate acid-base pair consists of two chemical species related by transfer of one proton. The conjugate acid has one more proton and one more positive unit of electric charge than the conjugate base.

A conjugate base is what remains after a chemical acid donates one proton. In the conjugate acid-base pair $\mathrm{HA}/{\mathrm{A}}^{-}$, $\mathrm{HA}$ is the conjugate acid of ${\mathrm{A}}^{-}$; accepting a proton reverses the conversion. For example, acetic acid $\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}$ loses a proton to give its conjugate base $\mathrm{C}{\mathrm{H}}_3\mathrm{CO}{\mathrm{O}}^{-}$. Partners differ by one proton and one unit of electric charge; they need not always be neutral and negative. Hydronium/water and water/hydroxide are also conjugate acid-base pairs.

At a prescribed pH, the ideal dilute acid dissociation constant determines the partition of a monoprotic acid inventory between its chemical species:These dimensionless fractions sum to one. They are the weak acid dissociation fraction and its complement when the only acid-bearing chemical species are $\mathrm{HA}$ and ${\mathrm{A}}^{-}$. Equal fractions occur at $\mathrm{pH}=\mathrm{p}{K}_a$ in this ideal approximation. For nonideal chemical solutions, equality of thermodynamic activities need not mean equality of concentrations.

For stipulated pKa $5$, the conjugate base fractions at pH $4$, $5$ and $6$ are $1/11$, $1/2$ and $10/11$. Here an external reservoir sets pH; its ions must supply the additional electric charge balance. The diagram omits those ions and water populations. It does not represent three pure $\mathrm{HA}$ samples at identical analytical concentration with freely chosen pH. In the latter experiment, the weak acid mass and charge balances determine pH as well as the partition.

A disulfide bond is the single covalent bond between the two adjacent sulfur chemical atoms in a disulfide, conventionally written $\mathrm{R}-\mathrm{S}-\mathrm{S}-{\mathrm{R}}^{\prime }$. Here $R$ and $R^{\prime }$ stand for the attached organic groups; they may belong to the same molecule or to initially separate molecules. A disulfide bond is a chemical bridge, rather than a hydrogen bond or a mere contact between nearby groups. Two thiols, $\mathrm{R}-\mathrm{SH}$ and ${\mathrm{R}}^{\prime }-\mathrm{SH}$, can supply its two sulfur endpoints. ^[ref]

The stoichiometry of forming a disulfide bond is especially revealing. Written as an oxidation half-reaction, it isThe reverse reduction half-reaction consumes two electrons and two hydrogen ions to restore two thiols. Both sides contain two sulfur chemical atoms; the two hydrogen chemical atoms have moved into the two hydrogen ions, and the net electric charge remains zero because $+2-2=0$. Thus one disulfide bond corresponds to two thiol groups and two transferred electrons. These half-reactions describe redox bookkeeping, not a complete standalone chemical reaction: an accompanying reactant must accept or supply the electrons, and pH affects the actual protonation of the groups. Mercaptoethanol can promote reduction through exchange of disulfide bonds, but adequate excess and appropriate conditions are needed; adding it does not by itself prove complete reduction. ^[ref]

In a protein disulfide bond, the endpoints are normally the amino acid side chains of two cysteine amino acid residues. An intrachain protein disulfide bond closes an intramolecular bridge within one polypeptide chain; an interchain protein disulfide bond connects different polypeptide chains, and can arise by intermolecular crosslinking of initially separate molecules. Forming or reducing these bridges changes covalent bond connectivity and can change protein structure, while preserving the order of amino acid residues and the peptide bonds in the protein backbone. A complete chemical account of protein primary structure includes crosslinks as well as sequence, so preservation of sequence does not imply preservation of every covalent bond. Mature insulin has two interchain protein disulfide bonds and one intrachain protein disulfide bond. ^[ref]

A bridge is useful because it constrains which parts of a polypeptide chain can move apart without breaking a covalent bond. Its placement therefore matters as well as its presence. The ribonuclease A refolding experiment provides the existing experimental example: breaking and reforming four protein disulfide bonds can accompany loss and recovery of function. Counting four bridges alone cannot tell whether their endpoints have the native pairing. This is an attractive connection between simple chemical counting and the richer problem of protein folding.

How can an experiment reveal an initially hidden disulfide bond? Ellman's reagent, also called DTNB, measures free thiols by producing a yellow absorbing species. Its reactive disulfide bond undergoes exchange with the conjugate base $\mathrm{R}-{\mathrm{S}}^{-}$ of a thiol, producing a mixed disulfide attached to $R$ and one released 2-nitro-5-thiobenzoate species, abbreviated TNB. At the assay pH, the strongly colored form is the dianion $\mathrm{TN}{\mathrm{B}}^{2-}$: one negative charge is on sulfur and the other is on the carboxylic acid-derived $\mathrm{CO}{\mathrm{O}}^{-}$ group. With excess DTNB and complete reaction, one released TNB reports one reacted thiol. George L. Ellman's 1959 paper introduced the tissue assay; it was subsequently reexamined quantitatively. ^{[ref] [ref]}

The modern apparatus can be simple: tubes, pipettes, a buffer solution, cysteine standards, DTNB, and an instrument that measures absorbance through a cuvette at wavelength $412\mathrm{nm}$. The Pierce protocol uses $50\mu \mathrm{L}$ of a $4\mathrm{mg}\mathrm{m}{\mathrm{L}}^{-1}$ DTNB stock, $2.50\mathrm{mL}$ of buffer solution, and $250\mu \mathrm{L}$ of sample. Its buffer solution is $0.1\mathrm{mol}{\mathrm{L}}^{-1}$ sodium phosphate at pH 8.0 with a metal-binding additive. After 15 minutes at room temperature, measure absorbance against a buffer solution plus DTNB blank; run known cysteine concentrations in parallel. The observable is increased yellow color and increased blank-corrected absorbance as more accessible thiols react, rather than direct imaging of disulfide bonds. These are published protocol conditions, not measurements performed for this article. ^[ref]

For a before/after test, split a purified protein sample into matched aliquots. Measure accessible free thiols in one aliquot; subject the other to complete reduction of its disulfide bonds, remove the reducing reagent, then measure the resulting thiols with fresh DTNB. Process the unreduced aliquot through a matching cleanup. Normalize to recovered protein and known dilution, since cleanup can change both. To estimate all reducible protein disulfide bonds, expose buried groups using matching protein denaturation conditions in both aliquots and calibrate in that same chemical solution; a native-accessibility measurement answers a different question. A larger corrected reading after reduction supports conversion of previously unreactive groups into detectable thiols. Attribution to disulfide bonds requires excluding other reducible sulfur modifications. This is a proposed comparison based on the detection principles, not a reconstruction of Ellman's original tissue measurements. ^[ref]

Under the Beer-Lambert law, the blank-corrected absorbance is $A=\epsilon lc$, where $l$ is the cuvette path length, $c$ is the released TNB concentration, and $\epsilon$ is its molar absorption coefficient. Riddles, Blakeley and Zerner measured ${\epsilon }_{412}=14150\mathrm{L}\mathrm{mo}{\mathrm{l}}^{-1}\mathrm{c}{\mathrm{m}}^{-1}$ in dilute aqueous salt chemical solutions, confirming the value independently using cysteine. It must not be applied indiscriminately to other solvents or protein denaturation mixtures. ^[ref] For aliquots compared on the same final volume and protein concentration basis, the ideal estimate isThe factor of two follows from the reduction stoichiometry; ${\bar{n}}_{\mathrm{S}-\mathrm{S}}$ is the number of disulfide bonds per protein molecule, or an average for a mixed population. If the aliquots have different dilutions or recoveries, convert each reading to a common sample basis before subtracting.

As a hypothetical calculation, choose $c_{\mathrm{protein}}=10\mu \mathrm{mol}{\mathrm{L}}^{-1}$ in each final cuvette, two free thiols and two disulfide bonds per protein molecule, and $l=1\mathrm{cm}$. Before reduction, $c_{\mathrm{SH}}=20\mu \mathrm{mol}{\mathrm{L}}^{-1}$ gives $A=0.283$; complete reduction adds four thiols per protein molecule, so $c_{\mathrm{SH}}=60\mu \mathrm{mol}{\mathrm{L}}^{-1}$ gives $A=0.849$. The difference $0.566$ gives $c_{\mathrm{S}-\mathrm{S}}=20\mu \mathrm{mol}{\mathrm{L}}^{-1}$ and ${\bar{n}}_{\mathrm{S}-\mathrm{S}}=2$. The published tube volumes imply a dilution factor $2.80/0.250=11.2$; reversing that dilution multiplies both thiol and protein concentrations by 11.2 and leaves the bridge count per protein molecule unchanged.

Controls determine whether that arithmetic has chemical meaning. Use fresh cysteine standards and a processed blank without protein to detect color generated by residual reducing reagent. Even a reducing reagent containing no thiols can interfere: Han and Han showed that tris(2-carboxyethyl)phosphine, abbreviated TCEP, reduces DTNB to two TNB equivalents per TCEP. ^[ref] In the hypothetical model, $5\mu \mathrm{mol}{\mathrm{L}}^{-1}$ residual TCEP in only the reduced cuvette would add $A=0.1415$ and falsely add $0.5$ bridge per protein molecule. A stable endpoint and agreement after a further reduction treatment help check completion. A sample without DTNB checks its own color; cloudy samples can also attenuate light without producing TNB. Match pH, temperature, accessibility and calibration chemical solution, and minimize delay that permits reoxidation. If two already-free but buried thiols become accessible only in the reduced aliquot, they mimic one new disulfide bond per protein molecule. Finally, bulk absorbance supplies a count under these assumptions: it does not locate the cysteine endpoints, distinguish intrachain from interchain bridges, or establish the native pairing of a protein disulfide bond.

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]

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.

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 concentration-based ionic strength is $I_c=\frac{1}{2}{\sum }_i{c}_i{z}_i^2$, where $c_i$ is each dissolved ion's concentration and $z_i$ its signed electric charge number. It has the same units as $c_i$; some formulas instead use the dimensionless ratio $I_c/c^{\circ }$. All dissolved ions contribute, including those that balance the electric charge of a buffer solution's conjugate base. Ionic strength helps describe the interactions represented by activity coefficients. See Hippler and Metcalfe, activity coefficients and dimensionless ionic strength.

Lactate is the conjugate base of lactic acid. Lactate dehydrogenase couples its interconversion with pyruvate to the NAD+/NADH redox pair.

A neutralization is a chemical reaction between a chemical acid and a chemical base. In the buffer solution example, hydroxide takes a proton from the weak acid to form water and the conjugate base. The stoichiometry updates the partner inventories before the approximate chemical equilibrium calculation. The name does not require the resulting pH to be $7$.

The all-carbonyl hydride, HCo(CO)${}_4$, is more acidic. Loss of ${\mathrm{H}}^{+}$ gives $[\mathrm{Co}(\mathrm{CO}{)}_4{]}^{-}$, whose negative charge is stabilized by pi backbonding into four carbonyl ligands. Replacing one CO by triphenyl phosphite reduces the net electron-withdrawing character of the ligand set relative to four carbonyls. The substituted hydride is more electron rich and its conjugate-base anion is less stabilized against protonation.Both conjugate bases have a conventional 18-electron count, so the difference is not a missing-electron argument. It is the relative stabilization of acid and base by the ligands in coordination chemistry. This concerns heterolytic deprotonation, not the homolytic Co–H bond-dissociation energy.

Low-spin Co(III) is substitution-inert. In acid, coordinated ammine deprotonation is suppressed, so the strongly donating amido conjugate base that assists ligand loss is not generated appreciably. Direct aquation/linkage change consequently remains very slow.

The persistence of the starting S-bound complex therefore supports a base-activated kinetic pathway. It is consistent with the rapid rearrangement in base and does not mean that the S-bound form must be the thermodynamically preferred linkage isomer under every condition.

A pH indicator changes color as its protonation distribution changes. In a simple weak acid/conjugate base model, its two forms have different colors and their ratio depends on pH through the acid dissociation constant. The visible transition spans a range of pH; it is not a direct signal that the sample has reached an equivalence point. An indicator is useful when its transition lies in a steep part of the titration curve, so a modest range of pH corresponds to a small added volume.^[ref]

The pKa is the negative base-ten logarithm of the dimensionless acid dissociation constant. A lower pKa means a larger acid dissociation constant and a stronger tendency to donate a proton under the same standard state, solvent, and temperature conventions. The Henderson-Hasselbalch equation places equal weak acid and conjugate base thermodynamic activities at pH equal to pKa. See Flowers et al., Chemistry 2e, the Henderson-Hasselbalch equation.

Pyruvate is the conjugate base of pyruvic acid.