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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 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.