Charge neutrality and mass action

So far we have worked in one direction: given the concentrations, we read off the species voltages. Real problems usually run the other way around. Some of the voltages are pinned for us, whether by a contact, a reservoir, or a reaction with a solid phase, and what we are left to determine is where the standard-state ladder $V^\circ_i$ sits. Pinning the $V_i$ does not pin the ladder, and pinning the ladder does not pin the $V_i$; the two are tied together only through a constraint that we have so far set aside: bulk charge neutrality.

Handled algebraically this quickly becomes tedious, since every concentration depends exponentially on the gap between a carrier's voltage and its standard state. On a $V_i$ diagram, though, the whole business reduces to a single motion: sliding the rigid $V^\circ_i$ ladder up or down until the net charge balances.

As a reminder, the ideal-dilute relation from the solutions topic can be rearranged to express a concentration in terms of that gap:

$$c_i = c^\circ \exp\!\left(\frac{z_i F\,(V_i - V^\circ_i)}{RT}\right).$$

A carrier's concentration is fixed entirely by how far its $V_i$ floats away from its own rung on the ladder, or as we will see it here: where the $V^\circ_i$ moves relative to the fixed $V_i$.

Neutrality sets the ladder

Suppose we fix a single carrier's voltage and then slide the ladder. Because only the gap $V_i - V^\circ_i$ enters the formula, moving the ladder down widens the gap and enriches a pinned cation, while moving it up depletes it; for an anion the directions reverse. Every mobile carrier responds to the slide this way at once, exponentially, each with the sign of its own charge.

This is exactly the handle that bulk charge neutrality needs. Away from interfaces, a conductor carries essentially no net charge, so

$$\sum_i z_i F c_i + \rho_{\mathrm{bg}} = 0,$$

where $\rho_{\mathrm{bg}}$ is any fixed, immobile background charge. As the ladder slides down, every cation in this sum enriches and every anion depletes, so the net charge climbs monotonically and passes through zero at exactly one position. Once the $V_i$ are given, the ladder floats to precisely the offset that neutrality requires, and to no other.

Electrostatic mass action: a pair invariant

Neutrality moves the ladder around: dissolve something new, or change the background charge, and the ladder must slide to a new neutral offset, carrying every concentration with it. Certain combinations of concentration, however, come through untouched. Pin two carriers, a species $A$ ($V_{\mathrm{A}}$) and a species $B$ ($V_{\mathrm{B}}$), and let the ladder go wherever neutrality sends it: one particular combination of the two concentrations stays fixed. To find it, expand the pinned difference $V_A - V_B$ with the ideal-dilute relation:

$$V_A - V_B = \left(V^\circ_A - V^\circ_B\right) + \frac{RT}{z_A F}\ln\frac{c_A}{c^\circ} - \frac{RT}{z_B F}\ln\frac{c_B}{c^\circ}.$$

The left side is pinned, and the rung spacing $V^\circ_A - V^\circ_B$ is a rigid property of the medium, blind to the ladder's offset. The two concentration terms are therefore left holding a constant between them, and exponentiating turns that into

$$\left(\frac{c_A}{c^\circ}\right)^{1/z_A}\!\left(\frac{c_B}{c^\circ}\right)^{-1/z_B} = K,$$

with

$$K = \exp\!\left(\frac{F\big[(V_A - V_B) - (V^\circ_A - V^\circ_B)\big]}{RT}\right).$$

$K$ depends only on the pinned difference $V_A - V_B$ and a material constant, and so it holds fixed no matter where the ladder happens to float. (More rigorously, activities can be used^[1].)

This is a mass action law, though with one important difference from its chemical namesake: nothing here is actually reacting.^[2] The two concentrations are tied together purely electrostatically, both responding to the same shift of the ladder. For a $+1$ cation paired with a $-1$ anion both exponents reduce to $+1$, and the law takes the familiar product form,

$$c_A\, c_B = \text{const},$$

so the cation can be enriched only at the anion's expense, and the reverse.

Note the same combination holds just as well for two carriers of the same sign. Pairing $\mathrm{Na}^+$ with $\mathrm{Al}^{3+}$, for instance, gives $c_{\mathrm{Na}^+}/(c_{\mathrm{Al}^{3+}})^{1/3} = \text{const}$. The fractional power^[3] records that a slide of the ladder moves each log-concentration in proportion to $z_i$, three times as far for the $\mathrm{Al}^{3+}$ as for the $\mathrm{Na}^+$: the combination above is locked, yet the plain ratio between the two ions remains free to range widely. This goes for every pair: if all three of $V_{\mathrm{Na}^+}$, $V_{\mathrm{Al}^{3+}}$, $V_{\mathrm{Cl}^-}$ were fixed, then we would have a mass-action law pinning all three pairwise products: $c_{\mathrm{Na}^+}/(c_{\mathrm{Al}^{3+}})^{1/3}$ and $(c_{\mathrm{Al}^{3+}})^{1/3}c_{\mathrm{Cl}^-}$ and $c_{\mathrm{Na}^+}c_{\mathrm{Cl}^-}$.

This pair of tools, neutrality to place the ladder and the invariants that survive its motion, accounts for a whole family of effects that, in their conventional presentations, appear quite unrelated. Each is a disturbance to the neutrality balance, met by a shift of the ladder.

Dopants

The semiconductor case is the same pair of statements in the physicist's notation, and a simple one. The ionized donors ($N_{\mathrm{D}}^+$) and acceptors ($N_{\mathrm{A}}^-$) cannot move, so they have no $V_i$ of their own; they enter the balance only as background charge:

$$\rho_{\mathrm{bg}} = +eN_{\mathrm{D}}^+ -eN_{\mathrm{A}}^- ,$$

which the mobile charges must compensate for:

$$-e\, n_{\mathrm{e}^-} + e\, n_{\mathrm{h}^+} + \rho_{\mathrm{bg}} = 0.$$

Equilibrium pins the electron and hole rails together, $V_{\mathrm{e}^-} = V_{\mathrm{h}^+}$ (the Fermi level), leaving the band-edge ladder free to seek neutrality. For the two-carrier case, the electrostatic mass-action law above gives us a shortcut to help find neutrality:

$$n_{\mathrm{e}^-}\, n_{\mathrm{h}^+} = K,$$

which one usually sees as the semiconductor mass action law, $np = {n_i}^2$, with essentially the same meaning.^[4]

The two equations with two unknowns close the system: together they combine into a quadratic equation that places $n_{\mathrm{e}^-}$ and $n_{\mathrm{h}^+}$.

(More detailed treatments model the dopants as incompletely ionized sites which, while immobile, are thermodynamically active, their occupation equilibrating with $V_{\mathrm{e}^-}$. Finding neutrality in this case no longer reduces to a simple quadratic equation, yet the ladder slide remains the correct way to find it.)

The common-ion effect

The same reasoning recovers a classic result of solution chemistry. Consider a solution saturated with $\mathrm{AgCl}$, so that the solid pins the difference $V_{\mathrm{Ag}^+} - V_{\mathrm{Cl}^-}$ at its saturation value: a two-carrier mass-action constraint, with the product $c_{\mathrm{Ag}^+} c_{\mathrm{Cl}^-}$ locked at the solubility product. Now dissolve a little $\mathrm{KCl}$. The $\mathrm{K}^+$ has no salt of its own to pin it, and so it behaves as a pure positive disturbance to neutrality, a mobile dopant. The ladder shifts to restore the balance, and because the product is locked, that shift enriches $\mathrm{Cl}^-$ while depleting $\mathrm{Ag}^+$, which precipitates back onto the solid. (The solid serves as the reservoir that holds the $V_{\mathrm{Ag}^+} - V_{\mathrm{Cl}^-}$ gap rigid throughout.) The result is the textbook common-ion effect, in which added chloride suppresses the solubility of silver, and in the $V_i$ picture it appears directly as a shift of the ladder.

Trapped charge and the Donnan potential

If the disturbance is confined to one side of a membrane, the effect becomes spatial. Let a membrane pass the small mobile ions ($\mathrm{Na}^+$ and $\mathrm{Cl}^-$) while blocking a large polyion, say a protein $\mathrm{Pr}^-$, on the right. The mobile ions equilibrate across the membrane, so $V_{\mathrm{Na}^+}$ and $V_{\mathrm{Cl}^-}$ run perfectly flat through it. The trapped $\mathrm{Pr}^-$, however, is a background charge present on the right alone, and so the ladder must sit at a different offset on each side in order to keep both sides neutral. That step in the $V^\circ_i$ ladder across the membrane is the Donnan potential. With the mobile $V_i$ flat and the ladder stepped, the concentrations step as well: the protein side is left richer in $\mathrm{Na}^+$ and poorer in $\mathrm{Cl}^-$, which is precisely Donnan exclusion.

A different solvent on each side

Let the two sides instead be genuinely different solvents, water against an immiscible organic phase such as nitrobenzene. The flat-$V_i$ equilibrium remains the same, but now each side has a distinct internal $V^\circ_i$ ladder structure. This liquid–liquid interface is known by the acronym ITIES, interface between two immiscible electrolyte solutions. The step in each ion's rung, $\Delta V^\circ_i$, folds together an electrostatic and a solvation contribution; a uniform Donnan shift $\Delta \phi$, which would parse the two apart, is neither well defined here nor pertinent to the thermodynamics.^[5] The total steps do carry meaning, measurable in charge-neutral combinations, subject beyond that to the same single-ion ambiguity as the activities of the nonideality topic.

A semiconductor reader will recognize this deforming ladder: a junction between two different media is a heterojunction. There too each species' edge moves by a different amount at the contact (the conduction and valence band offsets differ), and no single potential step can describe what is happening.

Takeaways

Everything in this topic comes down to a single motion: sliding the rigid $V^\circ_i$ ladder to the one offset at which the bulk is neutral. The pinned $V_i$ values and any fixed background charge together decide where that offset falls. Mass-action products emerge independent of the ladder, since the offset cancels, and dopants, the common-ion effect, and the Donnan potential turn out to be the same neutrality-restoring shift in three different guises. A boundary between two solvents generalizes the picture one step further, to a ladder that deforms as well as shifts.

The Donnan case, in which a fixed background charge on one side of a boundary forces a step in the ladder, has an exact counterpart in the solid state. The next topic sets the two side by side: the electrochemical bipolar membrane and its semiconductor twin, the pn junction.

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