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Take base-ten logarithms and use dimensionless thermodynamic activities. The supplied concentration notation is interpreted in the ideal dilute-solution approximation, with a consistent reference concentration. PutWith pure solids and unit water activity, the reaction quotients are $Q_I=10^{2y+2x}$ and $Q_{II}=10^{2y+6x}$. At chemical equilibrium, $Q=K$, so the two lines on the mineral activity diagram areThey have slopes $-1$ and $-3$. Restricting the plotted silica activity to $x<0$ is compatible with their intersection,Their full straight extensions are equilibrium relations for the respective phase pairs; not every extension is a stable boundary when the third mineral is available.

At the one-bar standard state, pure mineral endmembers and water vapour treated as an ideal gas all have unit thermodynamic activity. The Gibbs energy of reaction is ${\Delta }_r{G}^{\circ }={\Delta }_r{H}^{\circ }-T{\Delta }_r{S}^{\circ }$. Setting it to zero givesThe quantities are taken per mole of the reaction as written. The simplified result uses the stipulated temperature-independent enthalpy and entropy differences.

The compositions alone do not specify a unique thermodynamic activity model. Adopt the usual ideal, independent mixing on crystallographic sites approximation: three Mg–Fe positions per formula unit, with the other sites fixed at their stated endmember compositions. Then the mineral endmember activities areThe exponent counts the three mixing cations; treating each whole mineral formula as a single mixing site would be a different model. In particular, the magnesian biotite and garnet components here are phlogopite and pyrope.

For the written reaction, the reaction quotient iswith the other solid activities equal to one. The Gibbs energy of reaction balance therefore givesThe smaller product-endmember activity lowers the reaction chemical potential and lowers the equilibrium temperature in this model. Nonideal activity coefficients or unequal occupancies of crystallographic sites would require further information rather than a revised universal numerical answer.^[ref]

The pH of a chemical solution is the negative base-ten logarithm of its hydrogen ion thermodynamic activity:The thermodynamic activity is a dimensionless quantity, so the logarithm has a dimensionless argument. The definition uses thermodynamic activity, rather than total chemical acid concentration. Single-ion thermodynamic activities require a convention; practical pH measurements are tied to calibrated standards. See IUPAC Gold Book, pH.

In a dilute ideal chemical solution in water, a useful approximation isSquare brackets denote concentration in moles per litre. In water, $[{\mathrm{H}}^{+}]$ is shorthand for the hydronium concentration, often written $[{\mathrm{H}}_3{\mathrm{O}}^{+}]$; it does not describe isolated bare protons. Departures from ideal behavior need activity coefficients. The concentration formula approximates the thermodynamic activity definition.

For an ideal strong acid supplying one proton per initial unit, stipulate a total concentration of $0.001\mathrm{mol}{\mathrm{L}}^{-1}$ and negligible self-ionization of water. Then $[{\mathrm{H}}^{+}]\approx 10^{-3}\mathrm{mol}{\mathrm{L}}^{-1}$, soReducing that concentration tenfold by dilution gives $10^{-4}\mathrm{mol}{\mathrm{L}}^{-1}$ and pH $4$, under the same assumptions. More generally, dividing the hydrogen ion thermodynamic activity by ten raises pH by one. This is why the scale compresses an enormous range: equal steps of pH represent multiplicative changes, rather than equal differences in concentration. These are model calculations, rather than measured samples.

Neutrality depends on temperature. In the dilute ideal water model, define the dimensionless ion product $K_w=([{\mathrm{H}}^{+}]/c^{\circ })([{\mathrm{OH}}^{-}]/c^{\circ })$ and $\mathrm{p}{K}_w=-{\log }_{10}{K}_w$. Equal hydronium and hydroxide concentrations implyAt $25{}^{\circ }\mathrm{C}$, the textbook approximation $K_w\approx 10^{-14}$ gives neutral pH $7$. Since $K_w$ varies with temperature, neutrality need not mean pH $7$. Nor is pH restricted to $0$–$14$: the activity definition permits negative pH when $a_{{\mathrm{H}}^{+}}>1$, and pH above $14$ when $a_{{\mathrm{H}}^{+}}<10^{-14}$. High concentrations can invalidate the dilute ideal approximation. See Flowers et al., Chemistry 2e, pH and pOH.

S. P. L. Sørensen introduced pH at the Carlsberg Laboratory in 1909, as recorded in the Carlsberg Foundation historical timeline.

The pKa is the negative base-ten logarithm of the dimensionless acid dissociation constant. A lower pKa means a larger acid dissociation constant and a stronger tendency to donate a proton under the same standard state, solvent, and temperature conventions. The Henderson-Hasselbalch equation places equal weak acid and conjugate base thermodynamic activities at pH equal to pKa. See Flowers et al., Chemistry 2e, the Henderson-Hasselbalch equation.

The magnesian aluminium-bearing garnet mineral endmember. In an ideal Mg–Fe site model with other positions fixed, its thermodynamic activity is $X_{\mathrm{Mg}}^3$ because three magnesium positions occur per formula unit.

For a chemical reaction with signed stoichiometric vector entries ${\nu }_i$, positive for products and negative for reactants, the reaction quotient is the product of the species' thermodynamic activities raised to these powers. It equals the equilibrium constant at chemical equilibrium, with all quantities using the same standard state convention.

Here is an explicitly stipulated elementary kinetic model, not experimental data. A well-mixed ideal closed system has constant temperature and volume. Alongside an uncatalysed elementary reaction, allow an illustrative catalytic encounter that returns C immediately:Both channels obey the law of mass action. In the second, the forward reaction rate is ${\alpha }_f[\mathrm{A}][\mathrm{C}]$ and the reverse reaction rate is ${\alpha }_r[\mathrm{B}][\mathrm{C}]$. C cancels in each encounter, so its free concentration $c_C$ stays constant without an external supply. This toy elementary reaction encodes regeneration without binding; it is not a claim that real catalytic cycles universally consist of one encounter. A binding cycle can accumulate a reaction intermediate, in which case constant total C does not imply constant free $c_C$.

Adding the parallel reaction rates gives effective first-order reaction rate constantsBecause $k_f^u/k_r^u={\alpha }_f/{\alpha }_r=3$, each channel balances at the same $[\mathrm{B}]/[\mathrm{A}]=3$, and so does their sum. At $c_C=0$ the pair is $(0.030,0.010){\mathrm{s}}^{-1}$; at $c_C=0.001\mathrm{mol}{\mathrm{L}}^{-1}$ it is $(0.300,0.100){\mathrm{s}}^{-1}$. Both give $K=3$ with ideal thermodynamic activities and a common concentration standard state. At $T=300\mathrm{K}$ their compatible standard reaction change in Gibbs free energy is ${\Delta }_r{G}^{\circ }=-RT\ln 3\approx -2.7403\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. Choosing an unrelated reverse rate constant would break this consistency.

Let $S=[\mathrm{A}]+[\mathrm{B}]$ be the conserved concentration, and define the dimensionless quantity $x=[\mathrm{B}]/S$. ThenSolving this first-order linear differential equation givesThe chemical relaxation time is $25\mathrm{s}$ without C and $2.5\mathrm{s}$ with the chosen C. From either $x_0=0$ or $x_0=1$, the exponential decay of the deviation leads to the same chemical equilibrium $x=3/4$. Reducing that deviation to one percent takes $\ln (100)\tau$: about $115.13\mathrm{s}$ versus $11.51\mathrm{s}$.

At this chemical equilibrium, conversion continues in both directions. For $S=1\mathrm{mol}{\mathrm{L}}^{-1}$, the uncatalysed channel has equal forward and reverse reaction rates $0.0075\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$. Adding C gives total directional reaction rates $0.075\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$ each, yet zero net reaction rate. Faster molecular exchange and unchanged composition coexist. This exact calculation explains how catalysis can matter enormously within an observation window while leaving the final equilibrium constant intact.

A high-temperature potassium feldspar mineral, with idealized composition $\mathrm{KAlS}{\mathrm{i}}_3{\mathrm{O}}_8$. Substitution on its K and Al sites changes the thermodynamic activity of that mineral endmember and therefore the chemical equilibrium of reactions involving it.

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.

The solubility product is the equilibrium constant for dissolution of a specified solid into its dissolved ions. For a pure bulk AB solid at the chosen temperature, pressure and standard states,The pure bulk solid's thermodynamic activity is one in this convention. Coexisting solid and chemical solution at chemical equilibrium give this equality. If all available solid dissolves, the final chemical solution may instead be undersaturated; do not impose equality merely because a salt was initially added.^[ref]

Using the thermodynamic activity definition $a_i={\gamma }_i{c}_i/c^{\circ }$, with activity coefficients ${\gamma }_i$ and reference molar concentration $c^{\circ }$, gives the independent conversionThus $K_{\mathrm{sp}}$ is dimensionless, while the conditional concentration form $K_{\mathrm{sp},c}$ has units ${\mathrm{M}}^2$ for this one-to-one model. Treating $K_{\mathrm{sp},c}$ as constant requires a justified fixed or approximately fixed activity coefficient product. Changing ionic strength can change that product even at fixed temperature; changing temperature can change the equilibrium constant itself. Specify the solid thermodynamic phase too: different crystal polymorphs need not have the same solubility product.

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.

For transfer of the same species $X$ from a gaseous reference state into a solvent $S$, define the molar solvation free energy byHere ${\mu }^{\circ }$ is a standard chemical potential. Negative solvation free energy means that transfer into the stated solvent lowers the Gibbs free energy relative to the stated gaseous reference. Include solute-solvent interactions, making room in the solvent, and its reorganization; counting only a favorable pair interaction omits part of the transfer.^[ref]

State the temperature, gaseous reference pressure, dissolved reference concentration or other standard state convention, and species identity. Reference-state changes alter the quoted solvation free energy. For example, with an ideal dilute thermodynamic activity $a=c/c^{\circ }$, replacing $c^{\circ }$ by $\alpha c^{\circ }$ while preserving the physical chemical potential requires ${\mu }^{\circ }$ to increase by $RT\ln \alpha$. Thus the corresponding solvation free energy increases by $RT\ln \alpha$ if the gaseous reference stays fixed. This follows directly from $\mu ={\mu }^{\circ }+RT\ln a$, with $R$ the molar gas constant.

At fixed temperature and pressure, compare fully specified initial and final states usingThe entropy change is for the whole selected system, including both solute and solvent. Dispersing a solute can increase accessible arrangements while orienting nearby solvent molecules restricts others. One picture of an ordered hydration shell cannot determine the sign of the total entropy change. A negative Gibbs free energy change favors the specified process; it does not determine its reaction rate.^[ref]

An illustrative enthalpy accounting path separates solute constituents, reorganizes the solvent, and establishes their interactions:This is a chosen decomposition between compatible states, not a measured sequence. If a gaseous-ion-to-chemical solution solvation term already includes solvent reorganization, do not add that reorganization a second time. Lattice enthalpy from solution calorimetry gives the alternative Hess's law route that combines lattice separation with the complete hydration of ions term.^[ref]

The invented inputs are $+100$ for separation and $+20$ for solvent reorganization, in $\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. Interaction terms of $-150$ in solvent A and $-90$ in solvent B give $\Delta H_A=-30$ and $\Delta H_B=+30$. Stipulated total entropy changes of $-50$ and $+150\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ give $-T\Delta S_A=+15$ and $-T\Delta S_B=-45\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ at $300\mathrm{K}$. Both selected processes therefore have $\Delta G=-15\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. This deliberately constructed comparison shows why the signs of the enthalpy and Gibbs free energy changes need not agree; it does not predict either solvent's solubility.

For incremental dissolution at a particular composition, the relevant chemical potential difference also depends on thermodynamic activities. At chemical equilibrium that driving difference vanishes. A quoted standard Gibbs free energy change and the finite change between arbitrary mixtures answer different questions; use compatible standard states and the reaction quotient when discussing solubility.

The motivating calorimetry experiment is especially instructive. A weighed potassium chloride sample dissolves in a known amount of water in a platinum-lined, shielded calorimeter; a quartz-oscillator thermometer and measured electrical heating provide the readouts. Electrical calibration before and after the reaction, comparison with a resistance thermometer, and checks on residual water and particle size constrain errors. The inspected authored article reports positive enthalpy of solution at $298.15\mathrm{K}$ for the specified dilution, despite complete dissolution. This is a combined enthalpy balance, not an isolated solvation free energy or direct single-ion hydration of ions measurement. Shield tracking and final concentration matter; the existing lattice enthalpy from solution calorimetry discussion gives the numerical result and the conversion limits.^[ref]

The chemical potential assigned to a specified standard state. With dimensionless thermodynamic activity $a$, the per-particle convention is $\mu ={\mu }^{\circ }+k_{B}T\ln a$; a molar convention uses $RT$. The standard state changes both ${\mu }^{\circ }$ and the activity definition without changing observable equilibrium.

A standard state is a specified reference state used to define a species' standard chemical potential ${\mu }_i^{\circ }$. The reference convention, including temperature, must be consistent when calculating thermodynamic activities, reaction quotients, and equilibrium constants.

Standard-state activity conversion changes the numerical thermodynamic activity assigned to a fixed physical state when its reference chemical potential changes. The observable equilibrium and physical chemical potential stay the same. The activity and its standard chemical potential must therefore be converted together.

At fixed temperature $T$, writeIf ${\delta }_i={\tilde{\mu }}_i^{\circ }-{\mu }_i^{\circ }$, subtraction givesHere $R$ is the molar gas constant. This follows from the IUPAC definition of relative activity and its specified standard state.^[ref] Changing just the activity in the first equation would instead change the calculated physical chemical potential.

For an invented conversion ${\tilde{a}}_i=10a_i$, the compensating reference shift is ${\delta }_i=-RT\ln 10$. An old activity $0.03$ becomes $0.30$; the two expressions for ${\mu }_i$ are identical. An activity of one consequently identifies a reference condition under the chosen convention, not an invariant amount of material.

An activity conversion of an equilibrium constant follows the same logic. For the reaction A $\rightleftharpoons$ B, take $K=a_B/a_A$ at equilibrium and convert by ${\tilde{a}}_A=10a_A$, ${\tilde{a}}_B=2a_B$. Then $\tilde{K}=K/5$. If $K=5$, the converted value is $1$, whilebecause originally ${\Delta }_r{G}^{\circ }=-RT\ln 5$. For any physical composition, the corresponding reaction quotient also changes by $1/5$, so ${\Delta }_r{G}^{\circ }+RT\ln Q$ remains unchanged. These are reference changes, not altered reaction chemistry.

The same requirement governs activity equality for coexisting liquids: compare activities only after identifying their standard states, composition scales and reference conventions. Equal numerical values expressed under incompatible conventions do not, by themselves, establish liquid-liquid equilibrium.

Consider an original ideal model: $V_A=10.00\mathrm{mL}$ of hydrochloric acid at $c_A=0.100\mathrm{mol}{\mathrm{L}}^{-1}$, titrated with sodium hydroxide at $c_B=0.100\mathrm{mol}{\mathrm{L}}^{-1}$. Treat these numbers as exact model inputs. Assume complete dissociation of an ionic compound, complete strong acid ionization, additive volumes, uniform mixing, and activity coefficients equal to one. Use stipulated $K_w=10^{-14}$, approximately appropriate near $25{}^{\circ }\mathrm{C}$. These are calculated pH values, not measurements of these materials.

The chemical equation for neutralization isInitial acid inventory is $n_A=1.000\mathrm{mmol}$, so the equivalence point is $v=10.00\mathrm{mL}$. Added water does not change this inventory. But the final volume $V_A+v$, including the titrant, determines all concentrations.

Keep self-ionization of water even at the equivalence point. With $c^{\circ }=1\mathrm{mol}{\mathrm{L}}^{-1}$, define the dimensionless acid excess $d$, hydronium level $h$, and hydroxide level $o$ byThe sodium cation and chloride anion inventories fix the electric charge balance $h-o=d$, while chemical equilibrium gives $ho=K_w$. Eliminating $o$ yields a quadratic equation $h^2-dh-K_w=0$ with exactly one positive root. For stable calculation on either side, writeThe second expression follows by multiplying by $s-d$; it avoids subtracting two nearly equal large quantities on the alkaline side. A logarithm acts on the dimensionless $h$, as required by the pH definition.

At $v=5.00\mathrm{mL}$, the excess is $0.500\mathrm{mmol}$ in $15.00\mathrm{mL}$, giving $d=1/30$ and pH $1.477121\dots$. At $v=15.00\mathrm{mL}$, excess hydroxide is $0.500\mathrm{mmol}$ in $25.00\mathrm{mL}$, giving $d=-1/50$ and pH $12.301030\dots$. Neglecting dilution would give the wrong values even far from the equivalence point.

Exactly at $v=10.00\mathrm{mL}$, $d=0$ and $h=o=\sqrt{K_w}=10^{-7}$, giving pH $7$ in this model. At $9.999\mathrm{mL}$ and $10.001\mathrm{mL}$, the same smooth expression gives pH $5.300835\dots$ and $8.699121\dots$. The familiar excess-only formulas would predict zero hydronium or zero hydroxide at the equivalence point; retaining self-ionization of water removes that failure. A steep curve is still continuous.

The sensitivity is especially revealing. Differentiating $h-K_w/h=d$ giveswhen $c_A=c_B$. At this model's equivalence point, the derivative is about $1.086\times 10^4$ pH units per millilitre. This is a local derivative, not a linear prediction over a whole drop. Real endpoint accuracy also depends on dosing, mixing, measurement uncertainty, temperature, extra reacting chemical species, and nonideal thermodynamic activities.

At fixed temperature and pressure, an interior reacting equilibrium minimizes Gibbs free energy under the constraints imposed by stoichiometry. Each independent reversible reaction then has zero actual Gibbs energy of reaction,This is not a requirement that the standard Gibbs energy of reaction vanish.

Using dimensionless thermodynamic activities $a_i$, define the reaction quotient $Q={\prod }_i{a}_i^{{\nu }_i}$. The relationtherefore givesThe equilibrium constant $K$ is dimensionless because each activity is measured relative to its declared standard state. In general $K\ne 1$ even though the actual equilibrium Gibbs energy of reaction is zero. ^[ref]

These equations determine an equilibrium composition within the model. They do not supply the reaction rate or guarantee rapid equilibration; that requires the chemical kinetics.

The thermodynamic activity is the dimensionless quantity defined by ${\mu }_i={\mu }_i^{\circ }+RT\ln a_i$, where $R$ is the molar gas constant. For an ideal dilute solution with a specified concentration reference, $a_i=c_i/c^{\circ }$; departures from ideality are represented by an activity coefficient multiplying this ratio.