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Chemical equilibrium for $\mathrm{M}\rightleftharpoons {\mathrm{M}}^{+}+{\mathrm{e}}^{-}$ requiresInsert the given chemical potential form and define the positive ionization energyAfter collecting logarithms,soDefine a dimensionless concentration equilibrium constant using a standard number density $c^{\circ }$:Each $f_i$ is a molecular partition function density, with units of inverse volume. Consequently $f_e{f}_{{\mathrm{M}}^{+}}/f_{\mathrm{M}}$ has density units and division by $c^{\circ }$ makes $K_c$ dimensionless. All ground-state energies must use the same energy reference.

For an ideal gas, translation makes the molecular partition function proportional to volume. Define the molecular partition function density and number concentration byThenBoth $f_i$ and $c_i$ have units of inverse volume, so their ratio is dimensionless. If concentrations are specified in moles per volume instead, replace $c_i$ in this microscopic expression by $N_A$ times that molar concentration, with consistent volume units.

Let ${\rho }_i=N_i/V$ be a molecular number density and define the volume-independent molecular partition function density $f_i=q_i/V$. Then $q_i/N_i=f_i/{\rho }_i$. Also define the reaction reference-energy differenceSubstitution into the equilibrium chemical-potential relation givesAn equilibrium constant expressed using concentration activities $a_i={\rho }_i/c^{\circ }$ is dimensionless:Here every $f_i$ has units ${\mathrm{m}}^{-3}$, and $c^{\circ }$ is the chosen standard concentration, initially one molecule per cubic metre. The factor $c^{\circ }$ is essential: without it the association quotient has units of volume.