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The external chemical solution on the other side of a dialysis semipermeable membrane. Its composition supplies the boundary condition for exchange: it can receive unwanted small solutes or supply desired small solutes. A bath replacement discards the old dialysate and introduces a new reservoir with a specified composition.

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.

A molecule with a high relative molecular mass, typically built from many smaller units joined by covalent bonds. Proteins and nucleic acids are examples. In dialysis, a semipermeable membrane can retain these large molecules while smaller solutes diffuse through.

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.

Osmosis is net movement of water across a boundary whose permeability differs for water and relevant solutes. The direction depends on the chemical potential of water, including both dissolved material and pressure. A cell wall can let water reach the cell membrane, yet oppose expansion through turgor pressure. Zero net movement does not mean individual water molecules have stopped crossing.

Imagine pure water on side A and a chemical solution on side B of a semipermeable membrane. At equal temperature and pressure, dissolved particles lower the chemical potential of water in B, and net water transfer initially favors A to B. Retaining the solutes lets water redistribute without simply mixing everything. Increasing B's pressure opposes entry; enough pressure gives thermodynamic equilibrium with unequal solute concentrations. This is why “water moves toward more solute” requires a pressure condition.^[ref]

At thermodynamic equilibrium, the chemical potentials of exchangeable water are equal. Individual water molecules still move and cross in both directions, with equal mean opposing transfer rates and zero net flux. A stationary liquid level alone does not prove this: a blocked route can also give no visible transfer. Selective permeability provides a route; water potential supplies the direction, and neither alone determines the speed. The cell wall, turgor pressure and plasmolysis articles show why a living boundary's mechanical support and its selective permeability must be distinguished.

Take two reservoirs separated by a semipermeable membrane that passes water and retains one neutral solute. Stipulate $T=300\mathrm{K}$, $c_A=10\mathrm{mol}{\mathrm{m}}^{-3}$ and $c_B=30\mathrm{mol}{\mathrm{m}}^{-3}$, equivalent to $10$ and $30\mathrm{mmol}{\mathrm{L}}^{-1}$. Treat both chemical solutions as dilute and ideal. These are invented model conditions, not measurements or recommended biological conditions.

For pressures measured at the same height beside the membrane, define $\Delta P=P_B-P_A$. The model predictsAt $\Delta P=0$, net water transfer favors A to B. At $\Delta P\simeq 49.89\mathrm{kPa}$, the opposing pressure balances the osmotic pressure difference, giving zero net transfer with unequal concentrations. At $\Delta P=75\mathrm{kPa}$, the sign reverses and net water transfer favors B to A, toward the more dilute reservoir. Raising pressure changes the total water potential, rather than changing the membrane into a one-way gate.

Fixed concentrations require large maintained reservoirs or an instantaneous comparison. In a closed membrane osmometer, entering water causes dilution of the receiving chemical solution and changes its volume; the final balance must use final concentrations. A supported cell wall provides another way to develop the opposing pressure, as explained under turgor pressure.

Osmotic pressure is the excess pressure needed on a chemical solution to maintain osmotic thermodynamic equilibrium with pure solvent across a semipermeable membrane. For a nearly incompressible solvent, equality of chemical potential gives $\Pi \simeq -(RT/{\bar{V}}_w)\ln a_w$. It is an equilibrium property; an observed initial swelling rate is a different measurement.^[ref]

For a dilute ideal chemical solution of retained particles,Here $n$ is the amount of substance of independently dissolved solute particles, measured in moles; $V$ is the final volume of chemical solution, not the starting volume of pure solvent. Count the resulting ions separately if a solute dissociates; do not multiply twice if $n$ already counts them. With $c_{\mathrm{osm}}$ in $\mathrm{mol}{\mathrm{m}}^{-3}$ and $R$ in $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$, $RT{c}_{\mathrm{osm}}$ has units $\mathrm{J}{\mathrm{m}}^{-3}=\mathrm{Pa}$. This resembles the ideal gas law, although the particles here are dissolved in a liquid.

The assumptions are uniform temperature, a dilute ideal chemical solution, a nearly incompressible solvent, and a semipermeable membrane that retains the counted solutes while passing solvent. Association, incomplete dissociation and interactions change the relation between added material and independent particles; nonideal thermodynamic activities invalidate simple particle counting. The formula does not describe a concentrated brine merely because the apparatus contains a membrane.

A short derivation explains why particles are counted. In an ideal mixture, $a_w=x_w=n_w/(n_w+n)$, with $n_w$ the amount of substance of water. If $n\ll n_w$, then $-\ln x_w=\ln (1+n/n_w)\simeq n/n_w$ and $V\simeq n_w{\bar{V}}_w$. Substitution gives $\Pi \simeq RTn/V$. This is a dilute-limit approximation, not a claim that water molecules stop moving or that dissolved particles are a gas.

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]

Selective permeability means that a cell membrane permits different substances to pass at different rates under comparable conditions. Both the lipid bilayer and the available membrane proteins contribute to this selection.

Artificial boundaries can also have selective permeability. For osmosis, specify which solutes cross and over what observation interval: a boundary that retains one solute need not retain another. A semipermeable membrane is the ideal solvent-only boundary used in the osmotic pressure calculation.

A semipermeable membrane ideally passes the solvent while excluding the specified solutes. In the aqueous osmosis model below, water can cross but the counted dissolved particles cannot. Real selective permeability is often a matter of different rates, so this idealization needs checking over the measurement interval. Neither an impermeable sheet nor a freely mixing opening realizes this model.^[ref]