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For an ideal dilute chemical solution, moving one mole of a species with signed ion charge number $z$ from outside to inside changes its electrochemical potential byHere $R$ is the molar gas constant, $T$ the uniform absolute temperature, $F$ the Faraday constant, and $c$ the concentration. This assumes the same reference chemical potential on both sides. At thermodynamic equilibrium, $\Delta \tilde{\mu }=0$, henceThe Nernst equilibrium potential uses thermodynamic activities in the general case; replacing them by concentrations is an ideal-solution approximation.

Choose an original passive model: only ${\mathrm{K}}^{+}$ permeates ($z=+1$), other ions cannot cross, there are no pumps or imposed electric currents, and there is no net water transfer. Stipulate reservoir concentrations $c_{\mathrm{in}}=140\mathrm{mM}$ and $c_{\mathrm{out}}=5\mathrm{mM}$, with $T=310\mathrm{K}$, $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ and $F=96485\mathrm{C}\mathrm{mo}{\mathrm{l}}^{-1}$. Treat reservoir concentrations as effectively unchanged by charging. ThenThe concentration ratio is dimensionless; $J/C=\mathrm{V}$ supplies the voltage unit. At $V_m=0$, $\Delta \tilde{\mu }>0$ for inward transport, so outward transport is favored. At $V_m=E_{\mathrm{K}}$ it vanishes. A more negative $V_m$ favors inward transport. These sign checks explain the result physically.

Keeping the other chosen inputs fixed, doubling $c_{\mathrm{out}}$ to $10\mathrm{mM}$ raises $E_{\mathrm{K}}$ by $(RT/F)\ln 2\approx 18.52\mathrm{mV}$. Equal inside and outside concentrations give $E_{\mathrm{K}}=0$; reversing their ratio reverses its sign. In finite reservoirs, transport changes both concentrations; the exact final thermodynamic equilibrium must also satisfy material conservation. The fixed-reservoir calculation is useful when only a small fraction of the available ions moves.

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.

Signal amplification is an increase in a specified downstream response relative to an initiating signal. In a molecular-count description, one activated component can generate or activate many downstream molecules. Repeated catalysis by an enzyme supplies a mechanism: the enzyme remains available for another chemical reaction. Successive stages can multiply the response. For a biological example, Cooper describes the epinephrine-to-glycogen phosphorylase pathway, where production of many cAMP molecules and repeated downstream protein phosphorylation provide opportunities for signal amplification. This is an example, not a claim that every signal transduction pathway has the same stages or gain. ^[ref]

Here is a counting toy extending the example under signal transduction. Suppose $N_0$ activated receptors each activate $a=5$ distinct relay proteins during a first interval. During a subsequent interval, each relay activates $b=20$ distinct target proteins. Assume enough inactive relays and targets, no shared targets, and no loss of activity during the count. ThenFor $N_0=1$, this gives $N_1=5$ and $N_2=100$: the target-to-receptor count ratio is $100$. These are assumed integer counts, not measured rates or a universal physiological amplification factor. The intervals are sequential; the equations do not specify a time course.

Availability limits the multiplication. As a second toy assumption, let there be only $T=250$ initially inactive target proteins shared by all branches, and suppose the branches can activate distinct targets until this pool is exhausted. ThenTwo activated receptors give $200$ active targets, whereas three give $250$, not $300$. At three activated receptors, the total count ratio is $250/3\simeq 83.3$; adding a fourth activates no additional targets. Thus total output per input and the increment caused by an extra input are different notions of gain. This finite-pool rule is an illustrative bound, not an enzyme kinetics law.

Signal amplification makes a small initiating signal capable of a larger response, but it does not supply the material or energy for downstream chemical reactions. In the second messenger transfer experiment, both stages still required added ATP and magnesium ions. Counts, changes in concentration, and changes in enzyme activity are different readouts; an amplification claim should name its input, output and observation interval. ^[ref]

Signaling adaptation is a decline of a response toward its earlier level while the initiating signal remains present. Specify which response is measured and its baseline: complete signaling adaptation means eventual return to that baseline; partial signaling adaptation leaves a lasting displacement. The continuing signal distinguishes signaling adaptation from simply switching off the stimulus. Signaling adaptation describes behavior, while signaling feedback describes a causal arrangement, so neither implies the other.

By reducing a persistent response to background ligand concentration, signaling adaptation can make a biological cell sensitive to changes rather than only to absolute levels. Resetting the response helps leave room for another change over a useful range of ligand concentrations. This is valuable in chemotaxis: swimming bacteria encounter changing chemical conditions, and their signal transduction influences whether a run continues or a tumble occurs. Signaling adaptation helps separate a change encountered during motion from a lasting background. This usefulness does not imply exact return or equal sensitivity at every possible concentration. ^{[ref] [ref]}

Here is a solvable incoherent feedforward loop toy that exhibits complete signaling adaptation without signaling feedback. The input $s(t)$ jumps from $0$ to $1$ at $t=0$ and remains $1$, a Heaviside step function. Let $x$ be a fast activating branch, $a$ a slower inhibitory branch, and $r=x-a$ a readout. The activities $s,x,a,r$ are dimensionless variables; physical time is in seconds. Choose time constants ${\tau }_x=1$ second and ${\tau }_a=5$ seconds, with initial conditions $x(0)=a(0)=0$:Both branches read $s$ directly. Neither branch depends on $r$, and $a$ does not change $x$: the subtraction only defines the readout. Thus the slow opposing influence is feedforward, with no closed feedback control loop.

For $t\ge 0$, solving the two ordinary differential equations givesWriting $t$ for the numerical time in seconds, these are $x=1-e^{-t}$, $a=1-e^{-t/5}$ and $r=e^{-t/5}-e^{-t}$. Initially $r=0$, then $x$ rises faster than $a$, producing a pulse. At long times $x$ and $a$ both approach $1$, so $r$ returns to $0$ despite sustained $s=1$. For this upward step $e^{-t/5}\ge e^{-t}$, hence $r(t)\ge 0$. The striking feature of this signaling adaptation toy is that both branches remain active while the readout becomes quiet: the continuing background and the response to its arrival are separable.

Find the pulse maximum by differentiating the exponential functions. With numerical time in seconds, the derivative isThe stationary condition is $e^{-4t/5}=1/5$, so the natural logarithm givesBefore this time the derivative is positive; afterwards it is negative. Together with $r(0)=0$ and ${\lim }_{t\to \infty }r(t)=0$, this proves that it is the unique pulse maximum. As another numerical check, at $t=10$ seconds $x\simeq 0.999955$, $a\simeq 0.864665$ and $r\simeq 0.135290$: the slow branch is still catching up.

The subtraction $r=x-a$ is a mathematical readout, not a claim that an actual chemical concentration equals this difference for arbitrary inputs. Nonnegativity was established only for the specified upward step; other input histories can give a negative signed readout. Complete signaling adaptation here also relies on equal eventual branch gains: changing the readout to $r=x-0.9a$ leaves $r_{\ast }=0.1$ for $s=1$. This toy illustrates cancellation and different time constants; it does not establish molecular implementation or robustness to arbitrary parameter changes. Ma and colleagues' 2009 study separately analyzed biochemical mathematical models with negative feedback and incoherent feedforward loops, identifying additional conditions for signaling adaptation in those models. ^[ref]

A historical connection to feedback control comes from Yi, Huang, Simon and Doyle's 2000 analysis of a bacterial chemotaxis model. They identified integral action in its regulation of receptor activity: accumulated deviations alter a slower adjustment, which in turn changes the measured activity. At a stationary state, an adjustment that integrates a nonzero error cannot remain constant, explaining how zero error can follow. Their conclusion depended on stated biochemical assumptions and stable behavior; it was not a claim that every biological cell uses that mechanism. A transient pulse alone, as the incoherent feedforward loop toy shows, cannot identify the underlying signaling feedback architecture. ^[ref]

For $r=-\lambda >0$ and vanishing solid-to-liquid diffusivity ratio, $H(r)=\sqrt{\pi }r{e}^{r^2}\mathrm{erfc}(-r)$ gives the interfacial liquid concentration. It increases from zero to $C_s$ as $r$ increases. Its derivative is $C_s{H}^{\prime }/(1+H{)}^2$, which decreases from $\sqrt{\pi }{C}_s$ to zero. For a direct proof, write $A=H/r=\sqrt{\pi }{e}^{r^2}\mathrm{erfc}(-r)$. Then $A^{\prime }=2rA+2$, so $A\ge \sqrt{\pi }+2r$. Differentiation gives $2H^{\prime 2}-(1+H)H^{\prime \prime }=2[(1+r^2+2r^4)A^2+r(4r^2-1)A+2r^2-2]>0$: the bracket increases with $A$, and substitution of its lower bound gives only positive polynomial coefficients, including constant $\pi -2$. Thus the concentration is strictly concave as a function of melting speed parameter $r$. Combining this with a linearized thermal balance produces one fold on the melting branch and, with the freezing branch, an interval of three possible interface speeds.

The solidus bounds the entirely solid equilibrium region of a phase diagram. For a linear liquidus $T_L(C_l)=-m{C}_l$ and a constant segregation coefficient $C_s=k_D{C}_l$, the corresponding solidus is $T_S(C_s)=-m{C}_s/k_D$. The concentration argument is the solid concentration, not the neighboring liquid concentration.

Solubility specifies the amount that can remain dissolved at chemical equilibrium under stated conditions; it is different from how quickly a sample dissolves. Separating ions from an ionic crystal costs energy, but solvation establishes new interactions with the solvent. In water, hydration of ions couples charged ions to polar water molecules. The net enthalpy of dissolution can have either sign; entropy and concentration also enter the Gibbs free energy balance. Consequently, being a salt does not guarantee high solubility in water or any other solvent.^[ref]

Low solubility must also be distinguished from weak electrolyte behavior. A small amount of an ionic compound can dissolve essentially completely as ions, giving few charge carriers because little material dissolved. Calling that a weak electrolyte solely because its chemical solution conducts poorly would conflate two different questions.^[ref]

A solubility product constrains a product of thermodynamic activities; the available inventories determine which combinations of concentrations are possible. Their intersection determines the molar solubility. Consider an original, abstract AB solid, with the preceding one-to-one dissolution chemical equation. Stipulate $c^{\circ }=1\mathrm{M}$, ${\gamma }_{\mathrm{A}}={\gamma }_{\mathrm{B}}=1$ and $K_{\mathrm{sp},c}=10^{-4}{\mathrm{M}}^2$ at fixed temperature and pressure. Thus $K_{\mathrm{sp}}=10^{-4}$ is dimensionless. These are chosen model parameters, not data for an identified material.

Initially, a completely dissociated soluble XB salt supplies B anions at molar concentration $c_0\ge 0$ and X cations at the same molar concentration; A is absent. All three ions have unit electric charge magnitude. X remains a spectator ion, and only AB can form a solid. Assume constant volume, no additional chemical reactions, and negligible contributions from self-ionization of water. Dissolving AB raises both reacting-ion concentrations by the same $s$:Charge neutrality is satisfied: $c_{\mathrm{A}}+c_{\mathrm{X}}-c_{\mathrm{B}}=s+c_0-(s+c_0)=0$. Adding B alone without its accompanying cation would omit the initial electric charge balance.

In the plane of the two reacting-ion concentrations, this inventory constraint is the line $c_{\mathrm{B}}=c_{\mathrm{A}}+c_0$. Saturation supplies the hyperbola $c_{\mathrm{B}}=K_{\mathrm{sp},c}/c_{\mathrm{A}}$ for $c_{\mathrm{A}}>0$. Their positive intersection solves a quadratic equation:The quadratic formula supplies two roots of a polynomial; only the positive one is a physical concentration. Uniqueness also follows because $s(s+c_0)$ strictly increases for $s\ge 0$, starting at zero and growing without bound. The second expression, obtained by multiplying by the conjugate, avoids subtraction of nearly equal numbers at large $c_0$.

For pure water, $c_0=0$ gives $s_{\ast }=\sqrt{K_{\mathrm{sp},c}}=0.010\mathrm{M}$. For $c_0=0.030\mathrm{M}$,Their product is $10^{-4}{\mathrm{M}}^2$, although neither concentration equals the pure-water value. This is the useful geometrical distinction: an inventory line and an equilibrium constant specify different constraints.

Check the solid inventory before imposing saturation. If the initially added AB has amount of substance $n_0$ in volume $V$, the saturated branch requires $n_0>V{s}_{\ast }$ for excess solid to remain. For example, choose $n_0=0.0020\mathrm{mol}$ and $V=0.100\mathrm{L}$. In pure water, $0.0010\mathrm{mol}$ dissolves and the same amount of substance remains as solid. At $c_0=0.030\mathrm{M}$, approximately $0.0003027756\mathrm{mol}$ dissolves, leaving $0.0016972244\mathrm{mol}$ as solid. In this model, an insufficient supply instead givesIf $n_0/V<s_{\ast }$, all AB dissolves and the ion product is below the solubility product. At $n_0/V=s_{\ast }$, the last solid disappears exactly at saturation.