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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.

Potassium-40 is a long-lived radioactive isotope with 19 protons and 21 neutrons. It can decay to calcium-40 through beta-minus decay, or to argon-40 through electron capture and a rare beta-plus decay. The argon-producing branch supports potassium-argon dating.

Potassium hexamethyldisilazide is the potassium salt of the anion $\mathrm{N}(\mathrm{SiM}{\mathrm{e}}_3{)}_2^{-}$. The bulky silyl groups make it useful as a strong, sterically hindered base.

It can remove an acidic proton to form an enolate while competing less readily in some addition reactions than a smaller nucleophile. Counterion, solvent, and aggregation still influence the observed selectivity.

Let $m$ be the transferred sample mass, $p$ the known mass fraction of the intended chemical compound, and $M$ its molar mass. In a model with complete dissolution and chemically inactive impurities,The purity fraction is dimensionless: $99.95\%$ would mean $p=0.9995$, not $99.95$. These equations assume that the impurity contributes no titrant demand; an impurity that also reacts needs its own inventory equation. The value of $p$ must refer to the same preparation state as $m$, for example the specified dry material. A mass fraction assigned before unaccounted uptake of water cannot simply be used after it.

For an original worked example, stipulate pure potassium hydrogen phthalate, abbreviated KHP, so $p=1$, a selected mass $m=0.51055\mathrm{g}$, and rounded molar mass $M=204.22\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$. The NIST Chemistry WebBook lists formula ${\mathrm{C}}_8{\mathrm{H}}_5\mathrm{K}{\mathrm{O}}_4$ and molecular weight $204.2212$, consistent with the rounded molar mass used here.^[ref] The selected mass, purity assumption and volumes in this example are chosen inputs, not measured results or a claim about a real batch's purity.

The acid-bearing anion of potassium hydrogen phthalate loses one proton to hydroxide. The net chemical equation isThus sodium hydroxide has one-to-one stoichiometry with potassium hydrogen phthalate in this neutralization, givingChoose a delivered endpoint titre of $24.85\mathrm{mL}$ and matched titration blank of $0.05\mathrm{mL}$. ThenThe extra displayed digits show the arithmetic of the stipulated model; they do not establish experimental accuracy. A compact rounded result is $0.1008\mathrm{mol}{\mathrm{L}}^{-1}$.

When the entire weighed portion is titrated, added water changes concentration and the pH path but preserves the initial acid inventory. The dissolution volume therefore does not appear in $n_s=mp/M$. If instead a volumetric flask prepares a stock and a volumetric pipette selects an aliquot, the selected inventory is $n_s{V}_a/V_f$ for homogeneous stock, with delivered aliquot volume $V_a$ and final stock volume $V_f$. The same bookkeeping explains why complete transfer matters even when the amount of dissolution water is approximate.

A solvent containing an appropriate proton-donating group, often an O–H or N–H group. It can hydrogen bond, shuttle protons and quench strongly basic organometallic reagents.

The proton is a positively charged spin-$1/2$ baryon with ordinary valence quark content $uud$. It carries electric charge $+e$ and baryon number one. A hydrogen atomic nucleus is a single proton, while other ordinary atomic nuclei contain protons together with neutrons.

A proton is composite even when an atomic calculation treats it as a pointlike source. Electron scattering at increasing momentum transfer probes its spatial and magnetic response, whereas low-energy atomic behavior often needs only its charge, mass and magnetic moment. Its mass is about $1836$ electron masses; that ratio does not make it a bound collection of electrons.

Protonation adds a proton to a species. For a base $B$, the acid-base relation is $\mathrm{B}+{\mathrm{H}}^{+}\rightleftharpoons \mathrm{B}{\mathrm{H}}^{+}$, with the proton actually supplied by a donor in the surrounding medium.

Protonating ammonia gives an ammonium ion. Protonation can change charge, molecular geometry, solubility, and reactivity; protonating a carbonyl oxygen often increases the electrophilicity of its carbon. Which atom is protonated and how much of the sample is protonated depend on competing basic sites, solvent, and pH.

The root mean square radius of the proton charge distribution. Finite nuclear size changes hydrogen atom energy levels and enters precision predictions for the Lamb shift.^[ref]

Irradiation that removes resolved scalar spin-spin coupling to protons from another nucleus’s spectrum. Other couplings, such as carbon–fluorine coupling in a proton-decoupled carbon-13 NMR spectrum, can remain.

For a concrete independent example, stipulate perpendicular motion, $|q|=e$ with $e$ the elementary charge, uniform magnetic field $B=1.00\mathrm{T}$, incoming gyroradius $r_{\mathrm{in}}=0.100\mathrm{m}$ and outgoing gyroradius $r_{\mathrm{out}}=0.0500\mathrm{m}$. These are invented teaching inputs. Converting SI momentum to electronvolt units givesso $p_{\mathrm{in}}=29.9792458\mathrm{MeV}/c$ and $p_{\mathrm{out}}=14.9896229\mathrm{MeV}/c$. Both a Positron and a proton with these momenta trace the same ideal arcs. Their kinetic energies differ because the relativistic energy-momentum relation givesThe second expression avoids subtracting nearly equal quantities when $pc\ll m{c}^2$. For this calculation stipulate rounded rest energies $m_e{c}^2=0.511\mathrm{MeV}$ and $m_p{c}^2=938.272\mathrm{MeV}$. The Positron interpretation gives $K_{\mathrm{in}}\simeq 29.473\mathrm{MeV}$ and $K_{\mathrm{out}}\simeq 14.487\mathrm{MeV}$; the proton interpretation gives only $0.479\mathrm{MeV}$ and $0.120\mathrm{MeV}$. The corresponding proposed losses of kinetic energy are about $14.985\mathrm{MeV}$ and $0.359\mathrm{MeV}$.

The small proton kinetic energy follows from $K\simeq p^2/(2m_p)$, whereas here the Positron has $pc\gg m_e{c}^2$ and $K\simeq pc-m_e{c}^2$. Consequently magnetic bending alone cannot reject the proton. One needs an independent material constraint: could a proton at the proposed kinetic energy penetrate the plate and then travel at least the observed visible distance in the specified gas? A particle range shorter than that visible distance contradicts the hypothesis. The visible track need not include the stopping point: it already supplies a lower bound on particle range. No numerical particle range is assigned to this teaching example; that would require stopping data for the actual material and gas conditions. The experimental insight is to combine a geometric momentum measurement with an independent test of penetration.

A missing proton relative to a filled nuclear shell. In a closed-core approximation a single hole in a $j$ orbital has the same total angular momentum $j$ and orbital parity as the removed particle. For example, removing a $p_{1/2}$ proton from a spin-zero closed core gives $J^P=1/2^{-}$.

A proton-motive force is a proton electrochemical potential difference per unit electric charge. It combines the electric potential difference and the chemical potential difference across a cell membrane; the direction and sign convention must be specified.

A substance facilitating proton transfer across a membrane and dissipating the proton electrochemical gradient. It can uncouple respiration from ATP synthesis without directly inhibiting every electron-transfer reaction.

An absolute scalar magnetometer that infers $|\mathbf{B}|$ from the free Larmor precession frequency of polarized protons. It measures field strength rather than field direction.

The proton–proton chain first produces deuterium by a weak reaction and then helium-3 by proton capture. pp I ends with ${}^3\mathrm{He}+{}^3\mathrm{He}\to {}^4\mathrm{He}+2p$. pp II passes through beryllium-7 electron capture to lithium-7 and then proton capture. pp III passes through boron-8 beta-plus decay and beryllium-8 breakup. Each branch produces one net helium-4 nucleus from four protons, but the neutrino spectrum and deposited heat differ. In pp II the net weak reactions involve one emitted positron and one captured electron.

A membrane transport system that uses an energy source to move protons against their electrochemical gradient.

With bare nuclear masses, the proton separation energy is the energy needed to remove a proton. A negative value means the daughter plus proton has lower rest energy. Proton emission can still have a finite lifetime because of a Coulomb or centrifugal barrier.

A quark is a spin one-half fermion carrying color charge and participating in quantum chromodynamics. There are six quark flavors: up quark, down quark, strange quark, charm quark, bottom quark and top quark. Up-type flavors have electric charge $+2e/3$ and down-type flavors $-e/3$; every quark has baryon number $1/3$.^[ref]

Confinement prevents ordinary isolated quark final states. A proton has net valence content $uud$, and a neutron $udd$; their quark and gluon fields also contain additional fluctuations. A meson has minimal valence content consisting of one quark and one antiquark. Deep inelastic scattering tests quark structure through short-distance interactions, while hadron spectroscopy tests the allowed collective quantum states.

For a three-light-flavor example, take up quarks, down quarks and strange quarks with the following additive quantum numbers. Here $Q$ denotes electric charge in units of the positive elementary charge $e$, $B$ is baryon number, $S$ is strangeness, $I_3$ is the third component of isospin, and $Y=B+S$ is flavor hypercharge. An antiquark reverses all five displayed values. These are flavor coordinates, not spatial coordinates or electroweak hypercharge.

Adding these columns immediately explains familiar hadrons. A proton with valence quarks $uud$ has $Q=2/3+2/3-1/3=1$ and $B=1$; a neutron with $udd$ has $Q=0$ and $B=1$. A positive kaon with $u\bar{s}$ has $Q=2/3+1/3=1$, $B=0$ and $S=+1$. Thus the same constituent rules describe both baryons and mesons.

Two hadrons can have identical additive labels while differing in their internal state. The Lambda baryon and neutral Sigma baryon both have valence content $uds$, yet their total isospin differs. Likewise, the neutral pion flavor direction $(u\bar{u}-d\bar{d})/\sqrt{2}$ and the Eta octet state $(u\bar{u}+d\bar{d}-2s\bar{s})/\sqrt{6}$ both have $Q=Y=0$, but their inner product vanishes. Expanding that inner product gives $(1-1+0)/\sqrt{12}=0$. Constituent counting must therefore be supplemented by quantum superposition and the symmetry of the state.

The parton distribution function is a number density. Integrating the quark distribution minus the antiquark distribution counts the net flavour charge of the hadron, since every sea quark-antiquark pair contributes zero. The corresponding flavour-vector conserved current counts particles minus antiparticles. For a proton, the net up/down numbers are two and one; for a neutron they are one and two. The rule fixes the difference of distributions, not their sum, and does not imply absence of a sea.