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Buffer exchange changes the small components surrounding a retained macromolecule toward those of a desired buffer solution. Components absent from fresh dialysate tend to leave; supplied components can enter. Each permeant solute has its own exchange problem.^[ref]

For a dilute, nonbinding permeant solute, an ideal equilibrium calculation approaches the bath's specified concentration. That approximation need not describe a solute bound to a protein or a charged mixture with additional electrical constraints. Retaining a protein also does not by itself establish that its protein structure or catalytic function survives the change of buffer solution.

Dialysis exchanges small solutes between a sample and an external bath through a semipermeable membrane, while retaining suitably large macromolecules, such as proteins. The transport is passive diffusion: random motion carries permeant molecules both ways, with net transport down their concentration gradient. The bath is the dialysate. At equal permeant solute concentrations, microscopic exchange continues but net exchange stops.^[ref]

For example, a protein sample containing excess sodium chloride can lose the small ions to a low-salt dialysate without deliberately transferring the protein. Conversely, components of a new buffer solution can enter from the bath. This is useful before studying enzyme kinetics: changing the surrounding chemical solution can remove an interfering small solute while keeping the enzyme available for the next experiment.

The striking feature is that the retained protein does not have to be captured and released again to change its surroundings. However, retention is conditional on the semipermeable membrane and the particular macromolecule, and dialysis is poorly suited to separating proteins of nearly the same size. The molecular weight cut-off is a nominal retention rating, not a sharp geometrical boundary.

The molecular weight cut-off rates retention of reference molecules. One manufacturer's definition uses the mass at which approximately $90\%$ remains after seventeen-hour dialysis, tested with globular proteins. A $10\mathrm{kDa}$ rating uses dalton units: a mass, not a length or an absolute passage threshold.^[ref]

A semipermeable membrane has a distribution of passage sizes, and a molecule's shape affects passage as well as its mass. Near the nominal MWCO, slow passage can make substantial retention compatible with eventual leakage. The protein-retention experiment below makes retention an observable quantity rather than a universal yes-or-no property inferred from the label.

Consider an original ideal model with sample volume $V_s$, bath volume $V_b$, and initial sample solute concentration $c_0$. Assume each volume is fixed, a freely permeant, nonbinding solute, no chemical reactions, complete mixing on each side, and equal concentrations at equilibrium. Each fresh bath initially contains none of this solute. The protein is perfectly retained in this model.

The first equilibration redistributes the initial amount of substance $c_0{V}_s$ through both compartments. Conservation givesThis denominator includes the sample itself. For the chosen $V_s=1\mathrm{mL}$ and $V_b=9\mathrm{mL}$, $c_1/c_0=1/10$, rather than $1/9$. Nine-tenths of the initial solute is in the bath and one-tenth remains in the sample; nothing has vanished during equilibration.

Discarding the equilibrated bath removes its solute. A second fresh bath starts with only $c_1{V}_s$ of solute in the entire system, so the same argument applies again. Induction gives the geometric progressionHere $n$ counts completed equilibrations with fresh baths: $n=1$ is the first bath, $n=2$ follows one replacement, and $n=3$ follows two replacements. The normalized sample concentrations are respectively $0.1$, $0.01$ and $0.001$. Leaving the first bath in place for longer cannot lower the ideal equilibrium value below $0.1$.

The same conservation argument handles a fresh bath already containing a permeant solute at fixed concentration $b$:Thus the sample approaches $b$, not necessarily zero. For example, if a hypothetical new buffer component starts at zero in the sample and at $b$ in every bath, its sample concentration reaches $0.9b$, $0.99b$ and $0.999b$ in the three equilibrations. Outward removal and inward replacement follow the same mathematics.

Equilibrium fractions do not specify elapsed time. An illustrative well-mixed linear ordinary differential equation model within one bath isHere $K$ is an effective transfer coefficient with units of volume per time. Adding the two equations conserves $V_s{c}_s+V_b{c}_b$; subtracting their normalized forms givesThe difference undergoes exponential decay. A bath replacement resets $c_b$, renewing the driving difference without making $c_s$ jump instantly. The plotted points are equilibrium endpoints, not a measured time course or a timetable for an arbitrary protein.

Real sample volume can change through osmosis; proteins can be lost through leakage or adsorption to the semipermeable membrane. Consequently, a measured decrease in protein concentration alone cannot distinguish dilution from loss of protein. Measuring recovered volume as well as protein concentration permits an estimate of recovered protein mass.^[ref]