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Assume ideal, nondegenerate particles in local thermodynamic equilibrium, ground-state atomic hydrogen, no molecules, and a common temperature. Let ${\lambda }_j=h/\sqrt{2\pi m_j{k}_{B}T}$ be the thermal de Broglie wavelength. Choose zero energy for a separated stationary proton and electron, so a neutral atom has ground-state energy $-R$. The classical chemical potentials areChemical equilibrium for $H\rightleftharpoons p+e$ requires ${\mu }_H={\mu }_p+{\mu }_e$. Rearranging givesIncluding electron and proton spin degeneracy, $g_e=2$, $g_p=2$ and $g_H=4$ when hyperfine splittings are neglected. Hence the statistical factor is one. Also $m_H\simeq m_p$, so ${\lambda }_H^3/{\lambda }_p^3\simeq 1$. Charge neutrality gives $n_p=n_e$, and thereforeThis is the specified ground-state Saha equation. Keeping excited atomic partition functions or a different spin-counting convention can change a prefactor, so its convention must be stated rather than silently adding a factor of two. The left side has dimensions of a number density, not a dimensionless electron-to-neutral ratio.

Assume dilute, nondegenerate ideal electrons, protons and neutral atoms in thermal and chemical equilibrium, with charge neutrality $n_p=n_e$. Set ${\mu }_H={\mu }_p+{\mu }_e$, using each ideal species' translational factor $n{\lambda }_{\mathrm{th}}^3/g$ and hydrogen ground-state binding energy $-R$. Neglect excited states and take the heavy proton/atom thermal-wavelength ratio as one. With electron spin and neutral-atom degeneracies treated in the convention of the source, the remaining factor isDifferent partition-degeneracy conventions must be propagated consistently. Screening, dense-plasma level dissolution and departures from local thermodynamic equilibrium lie outside this derivation.

Assume a dilute nondegenerate hydrogen gas in thermal and chemical equilibrium, with Maxwell-Boltzmann translational partition functions, ground-state neutral atoms, single ionization and charge neutrality $n_p=n_e$. For $\mathrm{H}\rightleftharpoons p+e$, chemical equilibrium gives ${\mu }_H={\mu }_p+{\mu }_e$. The ideal-gas relation ${\mu }_i=E_{i0}+k_{B}T\log (n_i{\lambda }_i^3/g_i)$, with $E_{H0}=-R$, givesNeglecting the electron mass in $m_H/m_p$ and using the ground-state electronic statistical weights, $g_e=2$ and $g_H=2$ with cancelling nuclear-spin factors, this becomesExcited neutral states have an energy gap of at least $3R/4$ above the ground state. At the temperatures relevant to the plotted maximum their Boltzmann factors are very small despite the finite degeneracy increase. Highly excited levels also require a physical plasma cutoff; the untruncated infinite isolated-atom partition sum is not a valid finite-density model. Ground-state dominance is an approximation in this regime, not a theorem at all temperatures.

Take the heating process to keep the total hydrogen-nucleus density fixed, so $n_H+n_e=n_0$. This fixed-density assumption is needed to determine $n_0$ from the curve; freely expanding gas would require its density history. The excited-to-ground ratio is $n_2/n_H\simeq 4e^{-3R/(4k_{B}T)}$. ThereforeAt low temperature excitation is suppressed. At higher temperature excitation grows, but eventually ionization removes neutral atoms and the excited-neutral abundance decreases. To locate the maximum quantitatively, write $u=R/(k_{B}T)$. From $S/n_0=x^2/(1-x)$,Setting $d\log f/d\log T=0$ givesReading the original graph gives $T_{\max }\simeq (7.9\text{–}8.0)\times 10^3$ K and $f_{\max }\simeq 2.3\times 10^{-7}$. With $R=13.6$ eV, this implies $u\simeq 20$ and $\boxed{x_{\max }\simeq 0.82}$. Substitution intogives approximately $10^{18}{\mathrm{m}}^{-3}$. The result is particularly temperature-sensitive because of the Saha exponential; using $7900$ K gives about $9\times 10^{17}{\mathrm{m}}^{-3}$, while the coarse $8000$ K reading gives about $1.2\times 10^{18}{\mathrm{m}}^{-3}$. The plotted height is consistent with this range and should not be used to claim more precision than the source graph provides.

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.

For symmetric touching parabolic dispersions with constant density of states $D$ per energy band, including spin degeneracy, charge neutrality fixes $\mu =0$. The Fermi-Dirac distribution gives $N_c=D{k}_{B}T\ln 2$ and the thermal internal energy, including electrons and holes, is $\Delta U=({\pi }^2/6)D(k_{B}T{)}^2$. Thus $C=({\pi }^2/3)D{k}_B^{2}T$. Its large ratio to $N_c$ is compatible with the equipartition theorem because this excitation count grows with temperature and the heat capacity includes both species.