= Repeated-bath dialysis model
{title2=$c_n/c_0=(V_s/(V_s+V_b))^n$}
Consider an original ideal model with sample <volume (thermodynamics)> $V_s$, bath <volume (thermodynamics)> $V_b$, and initial sample <solute> <concentration> $c_0$. Assume each <volume (thermodynamics)> 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> $c_0V_s$ through both compartments. Conservation gives
$$
c_0V_s=c_1(V_s+V_b),\qquad
c_1=c_0\frac{V_s}{V_s+V_b}.
$$
This 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 $c_1V_s$ of <solute> in the entire system, so the same argument applies again. Induction gives the <geometric progression>
$$
q=\frac{V_s}{V_s+V_b},\qquad
c_n=q c_{n-1}=c_0q^n,\qquad
\frac{c_n}{c_0}=10^{-n}\quad(V_s=1,\ V_b=9).
$$
Here $n$ 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 $0.1$.
\Image[/biochemical-dialysis-repeated-bath-model.png]
{title=Repeated fresh baths reduce normalized <solute> <concentration> geometrically}
{description=Original fixed-<volume (thermodynamics)> model, not experimental data. The left panel normalizes the first bath's conserved <amount of substance> by $c_0V_s$, distinguishing the sample's $0.1$ share from the bath's $0.9$ share. The right panel plots sample <solute> <concentration> divided by $c_0$ after each completed fresh-bath equilibration on a logarithmic vertical axis. The <protein> curve is normalized independently by its own initial <concentration>. The perfectly retained <protein> stays at normalized <concentration> one only because its <amount of substance> and the sample <volume (thermodynamics)> are fixed.}
The same conservation argument handles a fresh bath already containing a permeant <solute> at fixed <concentration> $b$:
$$
c_n=\frac{V_sc_{n-1}+V_bb}{V_s+V_b},\qquad
c_n-b=q(c_{n-1}-b),\qquad
c_n=b+(c_0-b)q^n.
$$
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 is
$$
V_s\frac{dc_s}{dt}=-K(c_s-c_b),\qquad
V_b\frac{dc_b}{dt}=K(c_s-c_b),\qquad K>0.
$$
Here $K$ is an effective transfer coefficient with units of <volume (thermodynamics)> per time. Adding the two equations conserves $V_sc_s+V_bc_b$; subtracting their normalized forms gives
$$
\frac{d(c_s-c_b)}{dt}=-K\left(\frac1{V_s}+\frac1{V_b}\right)(c_s-c_b).
$$
The 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 (thermodynamics)> 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 (thermodynamics)> as well as <protein> <concentration> permits an estimate of recovered <protein> <mass>.<Dialysis: an overview, Pierce Biotechnology technical note>[Tech Tip 20, p. 2, Other Factors in Dialysis: water transport, volume change and nonspecific membrane binding]{ref}
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