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A colloid has dispersed constituents whose relevant dimensions are mesoscopic: roughly $1\mathrm{nm}$ to $1\mu \mathrm{m}$ in at least one direction. Interfaces and Brownian motion are consequently important. The size range is a useful convention, not a sharp boundary separating all physical behaviour. A sol of fine silicon dioxide particles in water illustrates solid in liquid; an oil-in-water emulsion illustrates liquid in another immiscible liquid; and a foam illustrates gas dispersed in a liquid, with thin intervening films supplying the small structural dimension. A colloid need not be a thermodynamic equilibrium state.^[ref]

Unprotected particles attract through Van der Waals forces. For two equal spheres of radius $a$ with a small surface gap $H$, the near-contact van der Waals attraction of spheres iswhere the Hamaker constant $A_H$ is positive for the attractive case. The large number of interacting constituent molecules can give a well much deeper than $k_{B}T$, even though individual molecular attractions are weak. Brownian motion brings particles together, and a sufficiently deep attractive well makes separation unlikely. This gives particle aggregation; solid particles can remain distinct within a cluster rather than merging into one continuous solid. Emulsion droplets can additionally coalesce if the separating film ruptures.

Charge stabilization of colloids supplies a repulsive barrier. Like-charged surfaces carry compensating counterions and electric double layers; bringing the layers together costs free energy. DLVO theory combines this repulsion with Van der Waals forces. A barrier large compared with $k_{B}T$ can make aggregation slow, even when sticking would lower the final free energy. Adding salt generally shortens the Debye–Hückel screening length and can lower the barrier; surface charge regulation and specific ion interactions also matter. A negatively charged particle and a positively charged particle need not repel.

Steric stabilization instead coats each particle with adsorbed or grafted polymer chains. In a good solvent, overlap and compression of these layers raise the osmotic pressure and reduce chain entropy, opposing close contact. Electrosteric stabilization combines the two mechanisms. Dense, securely bound layers matter: incomplete adsorption can allow polymer bridging, while nonadsorbing dissolved polymers can create an attractive depletion force. Poor solvent quality can collapse a protective layer. Reducing the density mismatch slows sedimentation but does not remove the interparticle attraction, so uniform suspension and resistance to aggregation are different requirements.

The left panel is an original qualitative potential with arbitrary coefficients: it illustrates a barrier and its reduction by shorter-ranged repulsion, without fitting an experimental DLVO theory curve or extrapolating the continuum attraction to molecular contact. The right panel plots the derived quasi-steady concentration profile in dimensionless variables.

In diffusion-controlled growth, attachment is rapid and delivery through the surrounding medium is the bottleneck. The growing surface depletes the neighboring solute, so a substantial concentration gradient develops. In reaction-controlled growth, arrival is fast compared with incorporation; many arrivals fail to attach, and the surface concentration stays near the reservoir value. The analogous distinction for particle aggregation is between near-unit sticking probability in diffusion-limited cluster aggregation and barrier-suppressed sticking in reaction-limited cluster aggregation. Here the subsequent calculation concerns accretion onto one growing particle, rather than a collision kernel for two mobile clusters.

The printed time-dependent equation is the diffusion equation, not the usual Poisson equation. For constant diffusion coefficient $D$, no bulk sources and negligible advection, Fick's laws give the number flux $\mathbf{j}=-D\nabla c$. Local conservation of mass, expressed here as conservation of the accreting constituent, then givesAn arbitrary time-independent exterior profile does not solve the full moving-boundary problem exactly. The required elementary result uses a quasi-steady approximation: diffusion establishes the exterior profile much faster than the particle radius changes. At each instantaneous radius $a$, set $\partial c/\partial t\simeq 0$. Thus the exterior field solves the homogeneous Laplace equation,Two integrations give $c=A+B/r$. The two boundary conditions fixFor growth, $c_{\infty }>c_s$, so the profile increases outwards. Its radial flux is negative, and the positive inward arrival rate isA material density is needed to turn an arrival rate into a radial velocity. Introduce the constant constituent number density $c_p$ inside the particle, using the same units as $c_s,c_{\infty }$. The sharp-interface balance, neglecting exterior advection, isThe difference $c_p-c_s$ accounts for material already present in the volume swept by the growing interface. In the usual dilute-feed approximation $c_s\ll c_p$, it can be replaced by $c_p$. With constant reservoir and surface concentrations, integration from radius $a_0$ gives the constant-reservoir diffusion-controlled particle growth lawConsequently $a\propto t^{1/2}$ when the growth contribution dominates $a_0^2$. For a finite seed at the literal limit $t\to 0$, the increment $a-a_0$ is instead initially proportional to $t$. Here “early time” means before appreciable reservoir depletion or interactions with other particles, while already beyond the rapid initial establishment of the diffusion field. Indeed, the ratio of diffusion-adjustment time $a^2/D$ to radial-growth time $a/\dot{a}$ is $(c_{\infty }-c_s)/(c_p-c_s)$; small supersaturation relative to the particle density makes the quasi-steady approximation self-consistent. No initial concentration field was specified, so an exact transient solution is not uniquely determined.

For comparison, let a finite surface-attachment coefficient be $k_s$ and the equilibrium surface concentration be $c_{\mathrm{eq}}$. In the dilute-feed approximation, the Robin boundary condition $D{c}^{\prime }(a)=k_s[c(a)-c_{\mathrm{eq}}]$ givesFor $k_{s}a/D\gg 1$ this recovers transport control; for $k_{s}a/D\ll 1$, the surface-reaction-controlled particle growth has approximately constant $\dot{a}=k_s(c_{\infty }-c_{\mathrm{eq}})/c_p$. Thus reaction control gives $a-a_0\propto t$, in contrast to the diffusion-controlled square-radius law. No numerical $c_p$ or $k_s$ is supplied or needed for these scaling results.

A permeating solute can cross the specified membrane appreciably during the observation interval. This is a property of a solute-membrane-time comparison, not an immutable label attached to a molecule. Passage can reduce a concentration difference through passive transport while water is also moving; it does not require water to wait.

For an original dilute ideal example, stipulate a flexible model biological cell with $20\mathrm{mmol}{\mathrm{L}}^{-1}$ retained particles and an external reservoir with $10\mathrm{mmol}{\mathrm{L}}^{-1}$ retained particles plus $10\mathrm{mmol}{\mathrm{L}}^{-1}$ freely permeating neutral solute. Initially the total osmolarities are equal. Suppose the permeating solute eventually has equal internal and external concentrations, there is negligible pressure difference, no active transport, and the internal retained particle amount is conserved. The permeating contributions cancel from the final osmotic balance; the internal retained concentration must fall from $20$ to $10\mathrm{mmol}{\mathrm{L}}^{-1}$. The required volume is twice its initial value. This illustrates an initially equal-osmolarity but hypotonic solution; it assumes the boundary can survive that expansion and predicts no universal time course.

Plasmolysis is withdrawal of the membrane-bounded contents from parts of a cell wall after water loss. A sufficiently concentrated external hypertonic solution, with solutes effectively excluded by the cell membrane over the observation time, can cause this. The cell wall remains; the enclosed material contracts. Plasmolysis is therefore distinct from osmotic lysis, which breaks the cellular boundary during excessive swelling. A return to a less concentrated surrounding chemical solution can reverse plasmolysis if the biological cell remains functional.^[ref]

A chemical solution with a polymer as a solute. The polymer conformation and interactions depend on the solvent and concentration.

Explanation of bulk assimilate transport from source to sink through hydrostatic pressure differences associated with solute loading and unloading.

A detector that measures changes of refractive index between a flowing sample and a reference stream.

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]

Positive hydrostatic pressure associated with roots, capable under suitable conditions of producing sap exudation.

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.

In ideal SEC pore access, geometric exclusion changes the fraction of pore liquid a solute can occupy. Diffusion permits entry into and exit from accessible pores; the through-flow chiefly transports the sample between packing particles. A hydrodynamic radius is a useful size descriptor, but actual pore entry also depends on shape and pore geometry. A flexible polymer coil is not a rigid ball with a perfectly sharp entry diameter.

The extra delay is best understood as extra accessible storage, rather than a prescribed longer streamline. Pore visits are reversible and random. An entirely excluded species samples the SEC void volume; a sufficiently small nonbinding species samples the SEC total accessible liquid volume. Intermediate species lie between these limits. Larger hydrodynamic size generally means earlier elution within the column's SEC fractionation range.

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]

Serial dilution repeatedly transfers a measured portion of one well-mixed chemical solution and dilutes that portion to a new final volume. With transferred volume $v_j$ and final volume $V_j$ at step $j$, conserved solute givesThis product follows by repeated substitution. Each step conserves the solute in the portion transferred into its receiving flask, not the entire inventory in the preceding flask. Uniform mixing makes the selected fraction representative.

Continue the original solution mass and volume bookkeeping example. A $10.00\mathrm{mL}$ portion of the $0.2000\mathrm{mol}{\mathrm{L}}^{-1}$ stock contains $0.002000\mathrm{mol}$ of analytical sodium chloride, corresponding to $0.11688\mathrm{g}$ before rounding. After dilution to $100.0\mathrm{mL}$, its analytical concentration is $0.02000\mathrm{mol}{\mathrm{L}}^{-1}$. A new $5.000\mathrm{mL}$ portion of that mixed chemical solution contains $0.0001000\mathrm{mol}$; dilution to $50.00\mathrm{mL}$ gives $0.002000\mathrm{mol}{\mathrm{L}}^{-1}$. The two factors $10.00/100.0$ and $5.000/50.00$ multiply to $1/100$.

The remaining stock holds $0.05000-0.002000=0.04800\mathrm{mol}$; the remaining first dilution holds $0.002000-0.0001000=0.001900\mathrm{mol}$. Including the final flask restores the starting inventory:This is a direct check that making a dilute final sample does not destroy the solute. Most of it remains in earlier vessels.

Serial dilution makes a small final concentration accessible using measurable portions instead of a single extremely small transfer. Its limitation is that systematic errors propagate: if each transferred volume is $1\%$ below its intended value while final volumes are correct, two steps give $0.99^2=0.9801$ times the intended concentration, a $1.99\%$ deficit. Repeated readings alone do not remove that bias. The gain in manageable volume therefore depends on calibrated transfer and representative mixing.

A solute is a component of a chemical solution distinguished from its solvent, typically present in a smaller proportion. There can be several solutes. A concentration must name both what is counted and the denominator: molar concentration uses amount of substance per final chemical solution volume; a solute-to-solvent mass ratio has another denominator.^[ref]

“Added sodium chloride” and “dissolved sodium cations” describe different levels of the same inventory. Dissociation of an ionic compound changes the grouping of the particles; it preserves their chemical elements. A dissolved ion remains an ion, and dissolved sucrose remains molecular unless a separate chemical reaction occurs. Thus the name of the weighed solute need not be the name of every dissolved chemical species.

A chemical solution is a homogeneous mixture in which designated solutes are distributed in a solvent. Representative equal-volume samples have the same composition after mixing. Individual molecules and ions remain distinguishable. The solute/solvent labels describe their roles in this mixture.^[ref]

A dissolved solute is dispersed as molecules, chemical atoms, or ions, rather than as the larger solid particles of a particle suspension. For example, sucrose retains its molecular formula during dissolution in water; a chemical reaction is not required. Dissolved sodium chloride supplies sodium cations and chloride anions, surrounded by water molecules through solvation. Dissolved does not mean chemically destroyed or converted into water.^[ref]

This makes a useful connection between microscopic identity and laboratory sampling. If a chemical solution is well mixed, the amount of substance in a measured portion follows from its concentration and volume. A scoop from a poorly mixed particle suspension need not represent the whole vessel. Visual clarity alone cannot establish composition or exclude a fine dispersion; the quantitative claim needs representative sampling.

The denominator of molar concentration is the final volume of the chemical solution, including the dissolved solute. It is not the volume of solvent initially poured in. A volumetric flask fixes the final total volume; weighing the solute supplies its mass, and a specified molar mass converts that mass to amount of substance.^[ref]

Consider an original illustrative account, not reported measurements. Suppose $2.922\mathrm{g}$ of pure dry sodium chloride, with stipulated molar mass $58.44\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$, is completely dissolved and quantitatively transferred into a chemical solution whose final volume is $250.0\mathrm{mL}$. Assume no losses, no additional chemical reactions, and complete dissociation of an ionic compound. ThenHere analytical concentration counts the sodium chloride inventory, not intact dissolved $\mathrm{NaCl}$ molecules. Under these assumptions, the amount of substance of sodium cations and of chloride anions is separately $0.05000\mathrm{mol}$; each has molar concentration $0.2000\mathrm{mol}{\mathrm{L}}^{-1}$. Their combined ionic amount of substance is $0.1000\mathrm{mol}$, but this does not double either the salt inventory or the mass. Each sodium cation still contains a sodium atomic nucleus, and each chloride anion a chlorine atomic nucleus. Counting chemical elements survives dissociation of an ionic compound.

The mass ledger is different from the volume ledger. If the final chemical solution had an illustrative net mass of $252.922\mathrm{g}$, the water contribution would be $252.922-2.922=250.000\mathrm{g}$, assuming nothing entered or escaped except those two components. The ratio $2.922/252.922\approx 0.011553$ expresses the solute share of total mass; the separate mass density would be $252.922/250.0=1.011688\mathrm{g}\mathrm{m}{\mathrm{L}}^{-1}$ before rounding. These stipulated quantities are mutually consistent bookkeeping inputs, not a measured mass density for saltwater. Neither establishes that $250.0\mathrm{mL}$ of water was added.

More generally, with full transfer and conserved solute, $n=m/M$ and $c=m/(MV)$. No additive volume or ideal chemical solution assumption is needed when $V$ is the measured final volume. If only a fraction $f$ of the weighed solute reaches the flask, the actual result is $c=fm/(MV)$. This short calculation explains why an undissolved residue or an incomplete rinse creates a systematic deficit even when the final calibration mark is read correctly.

A calibration mark can be tested against an independent mass measurement. The Buffalo State-hosted Laboratory Experiment 1. Calibration of Volumetric Glassware uses a $25\mathrm{mL}$ pipette, filler, conical flask, weighing bottle, balance, and thermometer. It specifies clean, freely draining glassware, bubble-free filling, an eye-level liquid meniscus reading, gravity drainage, and retaining the residual liquid allowed for by the pipette's calibration. Degassed water is delivered to a tared bottle; ten weighings and initial/final temperature readings give repeated delivered volumes using the water mass density. Their mean and standard deviation assess delivery and repeatability. The balance and thermometer must themselves be checked before claiming accuracy. The same handout describes weighing an empty volumetric flask and then weighing it filled to its mark.^[ref]

The independent inference is $V=m_{\mathrm{water}}/{\rho }_{\mathrm{water}}(T)$. Agreement between repeats constrains scatter; agreement with the nominal delivery additionally needs trustworthy mass density, temperature, and balance readings. This experiment tests delivered or contained volume under its conditions. It does not prove that a subsequent solute transfer is complete or a prepared chemical solution is homogeneous.

The University of Wisconsin–Madison teaching module supplies the complementary preparation demonstration. Dissolved solute and vessel rinsings reach the volumetric flask; the contents reach room temperature before a dropper adjusts the eye-level liquid meniscus to the calibration mark. The module excludes a stirring bar during filling because its displaced volume is absent from the calibration, and requires mixing the neck liquid with the bulk chemical solution. Its self-check identifies incomplete dissolution and transfer as a cause of lower-than-intended molar concentration.^{[ref][ref][ref]}

The observable endpoints are a liquid boundary at a mark and a delivered mass, not a direct count of solute particles. An overfilled flask has excessive final volume; removing some well-mixed chemical solution also removes solute, so it cannot recover the intended inventory merely by restoring the mark. Measurement uncertainty, contamination, incomplete transfer, and sampling before full mixing limit the inferred molar concentration. The apparatus demonstrates why the words “dilute to” refer to final chemical solution volume.