 SEC partition coefficient

For ideal SEC, define the SEC partition coefficient byHere ${\overline{c}}_{\mathrm{pore}}$ averages solute concentration over all pore liquid, including inaccessible regions. $V_e$ is column elution volume. Pure steric exclusion gives $0\le K_d\le 1$; the endpoints mean exclusion and full access.^[ref]

A storage argument explains the relation. Assume a dilute stable solute, unchanged packing and solvent, constant volumetric flow rate $Q$, rapid local chemical equilibrium between pore liquid and interparticle liquid, negligible binding, and negligible transport through the pores by bulk flow. At interparticle concentration $c$, the column holds an amount of substance $n=c{V}_0+K_{d}c{V}_i$. Its exit transports $Qc$ per unit time. The ratio $n/(Qc)$ is the mean residence time in this equilibrium transport model; multiplying by $Q$ gives $V_e=V_0+K_d{V}_i$. For a narrow symmetric SEC chromatogram peak, the peak maximum approximates that mean; a strongly asymmetric peak needs an explicitly stated location convention.

The handbook also uses an operational normalizationThis denominator includes solid matrix, so full access gives $K_{av}<1$ when $V_{\mathrm{matrix}}>0$. It differs from the pore-liquid SEC partition coefficient. Read each source's volume convention before comparing coefficients.^[ref]

For an original numerical example, take $V_0=5\mathrm{mL}$, $V_i=8\mathrm{mL}$, and $V_{\mathrm{matrix}}=3\mathrm{mL}$. Then $V_t=13\mathrm{mL}$ and $V_c=16\mathrm{mL}$. Elution at $9\mathrm{mL}$ gives $K_d=1/2$ but $K_{av}=4/11$. Even a fully included tracer gives $K_{av}=8/11$, not $1$. These are stipulated volumes, not experimental measurements.