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Acid-base chemistry studies how chemical acids and chemical bases exchange protons or electron pairs. Here the focus is transfer of protons in water: a chemical acid donates a proton, and a chemical base accepts it. The resulting chemical equilibrium determines pH, while a buffer solution can resist changes in pH. This connects microscopic chemical reactions with a compact, measurable description of a chemical solution.

The important distinction is between how much chemical acid is present and how readily it transfers a proton. A weak acid and a strong acid at the same total concentration need not have the same pH. Conversely, two buffer solutions can have the same pH while containing very different total quantities of their chemical acids and chemical bases.

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.

In acid-base chemistry, hydrogen ion usually refers to positively charged hydrogen, represented by ${\mathrm{H}}^{+}$. In water, this notation describes solvated species, commonly represented by hydronium, rather than an isolated bare proton. The hydrogen ion thermodynamic activity defines pH.

Before calculating electronic energy, ask a simpler question: where can the available valence electrons go? A Lewis structure makes that budget visible, and nitrate shows why a correct budget can still require several resonance structures.

A Lewis structure records which chemical atoms are connected in a molecule or ion and how its valence electrons are allocated. One line represents one bond pair, two lines two bond pairs, and three lines three bond pairs; two dots beside a chemical atom represent one lone pair. The marks are a counting language, rather than measured electron positions. See the IUPAC Gold Book, Lewis formula.

For water, ${\mathrm{H}}_2\mathrm{O}$, the budget is $6+2\times 1=8$ valence electrons. Two O–H lines use four electrons; the other four form two lone pairs on oxygen. For Ammonia, ${\mathrm{NH}}_3$, $5+3\times 1=8$ valence electrons give three N–H lines and one lone pair on nitrogen. Both molecules contain four electron pairs, but their allocations differ. Each shared bond pair is counted once in the total budget, even though both endpoints count it when checking the octet rule.

A practical construction method is:

For carbon dioxide, $4+2\times 6=16$ valence electrons. Starting with two single C–O connections leaves twelve electrons, enough for three lone pairs on each oxygen, but only four electrons around carbon. Converting one lone pair from each oxygen into a second bond pair gives $\mathrm{O}=\mathrm{C}=\mathrm{O}$. Eight electrons are in the four bond pairs, and eight are in four lone pairs, two on each oxygen. Carbon has no lone pair; all three chemical atoms satisfy the octet rule, and all three formal charges vanish.

For the nitrate ion, ${\mathrm{NO}}_3^{-}$, the valence electron budget isNitrogen contributes five, the three oxygen chemical atoms contribute eighteen, and the negative electric charge adds one electron. Start with nitrogen connected to each oxygen by one bond pair: those three lines use six electrons. Put the remaining eighteen electrons into three lone pairs on each oxygen. Each oxygen now satisfies the octet rule, but nitrogen counts only six electrons. Reassign one oxygen lone pair as a second bond pair on its N–O connection. This changes the allocation without adding an electron; it applies the construction procedure in Flowers et al., Chemistry 2e, 7.3: drawing Lewis structures.

The resulting nitrate Lewis structure has one N=O connection and two N–O connections. Its four bond pairs use eight electrons, and its eight lone pairs use sixteen: $8+16=24$. The doubly connected oxygen has two lone pairs; each singly connected oxygen has three; nitrogen has none. All four chemical atoms satisfy the octet rule. Any of the three oxygen chemical atoms can supply the additional bond pair, giving the three equivalent resonance structures below.

Why not draw three N=O connections to make the ion look symmetric? With two lone pairs on each oxygen, that drawing still spends $12+12=24$ electrons and has the right total formal charge, but places twelve electrons around nitrogen. It fails the octet rule for this second-period chemical element. Three equivalent resonance structures express the symmetry without that invalid local count.

The economy is beautiful: a handful of dots and lines rules out many impossible counts and identifies lone pairs that can participate in acid-base chemistry. A Lewis structure helps choose rational candidate connections before a detailed quantum chemistry calculation. It does not specify a complete many-electron wavefunction, actual electron density, or molecular geometry. Adding the separate VSEPR model, described in Chemistry 2e, 7.6: molecular structure and polarity, predicts bent water and pyramidal Ammonia; the layouts above alone make no such prediction. Molecular orbital theory supplies a richer electronic description, including the unpaired electrons of molecular oxygen that an all-paired Lewis structure misses.

In 1916 Gilbert N. Lewis proposed two-dot notation for the shared electrons of a chemical bond and fuller dot formulae for the outer electrons, explicitly relating a shared pair to the existing single-line notation. See Lewis, The Atom and the Molecule, JACS 38, 762–785 (1916), pp. 777–778, available as an ^[ref]; the relevant passage is also readable in a ^[ref]. The modern quantum mechanics interpretation should be distinguished from his historical model of fixed outer electrons.

A lone pair is an electron pair assigned to one chemical atom without a connecting line to another chemical atom in a Lewis structure. It is usually shown as two dots. Water has two lone pairs on oxygen, accounting for four valence electrons; Ammonia has one lone pair on nitrogen, accounting for two. In $\mathrm{O}=\mathrm{C}=\mathrm{O}$, each oxygen has two lone pairs, and carbon has none.

A lone pair can be reactive. In protonation of Ammonia, the nitrogen lone pair becomes a new bond pair to a proton:The proton supplies no electron, so the eight valence electrons are retained: three bond pairs plus one lone pair become four bond pairs. The resulting nitrogen has formal charge $5-0-8/2=+1$; each hydrogen has formal charge zero. This is why identifying an available lone pair helps explain the chemical base behaviour of Ammonia, rather than merely decorating its formula.

Lewis already used the Ammonia to ${\mathrm{NH}}_4^{+}$ example in The Atom and the Molecule (1916), pp. 777–778. In modern terms the dots identify an allocation of electrons, not two stationary particles or an isolated patch unaffected by the rest of the molecule. A molecular orbital description can spread electronic amplitude over several chemical atoms while a localized Lewis structure remains useful for acid-base chemistry.

A monoprotic acid supplies one proton per acid unit in the acid-base chemistry model under consideration. Acetic acid is an example. Several successive proton transfers require several acid dissociation constants and a larger set of chemical species.

A formal two-electron reductive elimination couples two anionic, mutually cis donors and reduces Ti(IV) to Ti(II). There are two chemically distinct couplings: Cl–Cl elimination gives $\mathrm{C}{\mathrm{l}}_2$, leaving a Ti(II) complex with the intact anionic O,N,P ligand and one chloride; O–Cl elimination gives an aryl hypochlorite group, $\mathrm{ArOCl}$, breaking the original Ti–O attachment and leaving a Ti(II) center coordinated by N,P and two chlorides. The latter product can remain tethered through the N,P part of the same ligand; it is not necessarily a free small molecule. The starting facial arrangement supplies cis pairs of both kinds.

Cl–Cl coupling is the more plausible of these formal alternatives, because it retains the particularly strong early-metal Ti–O interaction and makes the stronger Cl–Cl rather than relatively weak O–Cl bond. Both alternatives reduce the valence count by two, from 12 to 10, so the 18-electron rule does not choose between them. Neither should be asserted to occur readily: reduction of a stable, oxophilic $d^0$ Ti(IV) center is intrinsically unfavorable. Coupling a chloride to the neutral amine or phosphine is not the ordinary elimination of two X-type ligands. proton transfer between N–H and oxygen, if it occurred, would instead be acid-base chemistry, not automatically a two-electron reductive elimination.

A redox chemical reaction couples oxidation and reduction. In the electrode examples here, oxidation releases electrons and reduction accepts them. ^[ref]

Redox can be recognized by changes in oxidation state: an increase records oxidation, a decrease reduction. These are formal electron assignments, distinct from measured partial atomic charges. Half-reactions expose the matching loss and gain, making electric charge conservation visible.^[ref]

The electron accounting need not describe free electrons travelling between spatially separated chemical species. For example, in $2{\mathrm{H}}_2+{\mathrm{O}}_2⟶2{\mathrm{H}}_2\mathrm{O}$, hydrogen changes from oxidation state $0$ to $+1$ and oxygen from $0$ to $-2$. Four hydrogen chemical atoms lose four formally assigned electrons in total; two oxygen chemical atoms gain four. The water product has covalent bonds, and these assignments do not turn it into separated ions. See ionic approximation of bonds and electron transfer and ionic character.

By contrast, neutralization can rearrange chemical bonds without changing oxidation states:Hydrogen remains $+1$ and oxygen remains $-2$. The aqueous proton notation abbreviates its interaction with water. This acid-base chemistry example conserves electric charge but has no matching increase and decrease of oxidation state; movement of ions or a change of chemical bonds alone does not establish redox.