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A catalyst makes a chemical reaction proceed faster by participating in an alternative reaction mechanism and being regenerated. This makes repeated conversion possible: the same catalyst can help many successive reactants become reaction products. An enzyme is a biological catalyst. The overall standard change in Gibbs free energy is unchanged; see the IUPAC Gold Book: catalyst.

Regeneration does not mean that the catalyst sits unchanged throughout every step. For example, a schematic reaction mechanism can temporarily bind A to a catalyst C in a reaction intermediate AC:Adding the steps cancels both the reaction intermediate AC and the catalyst C from the net chemical equation. AC is produced and later consumed; C is used and later recovered. These letters describe possible roles, not a measured reaction mechanism. The alternative route is useful only if its chemical kinetics permits faster conversion. See Flowers et al., Chemistry 2e, §12.7: catalysts and alternative mechanisms.

For a worked barrier example, consider an idealized reversible A-to-B conversion at fixed temperature $T=300\mathrm{K}$. Stipulate standard molar Gibbs free energy levels $G_{\mathrm{A}}^{\circ }=0$ and $G_{\mathrm{B}}^{\circ }=-20\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$, with effective transition state levels $80$ without a catalyst and $50\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ with a catalyst. Use the same standard states and the same zero for both routes; the catalytic route includes the regenerated C at both endpoints. Unlike the separate potential-energy profile example, this illustration uses Gibbs free energy at one specified temperature.

The Gibbs energy of activation is measured from the starting endpoint in the chosen direction. Thus the four barriers areBoth directional barriers fall by $30\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. The endpoint difference stays ${\Delta }_r{G}^{\circ }=-20\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$.

To turn those barriers into rate constants, use the Eyring form of transition state theory for this effective first-order reaction pair, with a common prefactor and unit probability of successful passage:Here $k_{\mathrm{B}}$ is the Boltzmann constant, $h$ is the Planck constant, and $R$ is the molar gas constant. The common prefactor cancels from the comparisons. These Gibbs energies of activation are not the empirical activation energy $E_a$ of the Arrhenius equation. With the stipulated value $R=8.314462618\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$, the speedup isThe exponent is a dimensionless quantity: convert $30\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ to $30000\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$ before dividing by $RT$. The exponential function explains why a modest barrier reduction can have a dramatic kinetic effect. This number follows from the chosen free-energy barriers and common prefactor; it is not a universal catalytic multiplier or a prediction across temperatures.

Assume ideal thermodynamic activities with a common concentration standard state, and directional rate laws $J_f=k_f[\mathrm{A}]$ and $J_r=k_r[\mathrm{B}]$. At chemical equilibrium, equal reaction rates give $K=[\mathrm{B}]/[\mathrm{A}]=k_f/k_r$ for this model. The barriers giveThe catalyst speeds up both directions, so it changes how quickly chemical equilibrium is approached without changing this equilibrium constant. For a general chemical reaction, the thermodynamic relation is $K=\exp [-{\Delta }_r{G}^{\circ }/(RT)]$ with consistent standard states; an arbitrary multistep reaction mechanism need not permit $K=k_f/k_r$ for two measured rate constants. At the same temperature, pressure, and thermodynamic activities, a catalyst also leaves the actual overall reaction change in Gibbs free energy unchanged. See 13.3 Shifting Equilibria: Le Châtelier’s Principle (§13.3: a catalyst does not shift equilibrium). To drive an unfavourable overall conversion, a process such as energy coupling must change the combined transformation.

Catalyst deactivation is loss of catalytic activity during use. Regeneration in an ideal catalytic cycle does not rule out a separate damaging chemical reaction, blocking of a surface, or physical loss of catalyst from the apparatus. A route can also be slowed by diffusion to the surface or by accumulation of a reaction intermediate. Such effects limit the usefulness of a low schematic reaction energy barrier: measured throughput depends on the whole reaction mechanism and on transport. Demonstrating regeneration requires more evidence than observing a faster first run. For a recovered solid catalyst, a useful further comparison is a second run with fresh reactants, matched recovered amount, and the same mixing and temperature, alongside a fresh-solid run and a blank. Loss during recovery or residual reactants can otherwise imitate catalyst deactivation or apparent reuse. This is a proposed control, not an observation reported for the catalysed peroxide comparison.

Consider an original geometric model of $N$ separate nonporous solid spheres, each with radius $r$. Hold their total volume $V_s$ fixed. The sphere formulas giveAt fixed mass density, fixed $V_s$ also means fixed total mass. Halving $r$ requires eight times as many particles and doubles their total external surface area $S$; this follows by eliminating $N$, without any assumption about chemical kinetics.

For example, take one sphere of radius $1.00\mathrm{mm}$, so $V_s=4\pi /3\mathrm{m}{\mathrm{m}}^3$. Eight spheres of radius $0.50\mathrm{mm}$ have the same $V_s$. Their total surface areas are respectively $4\pi \mathrm{m}{\mathrm{m}}^2$ and $8\pi \mathrm{m}{\mathrm{m}}^2$, approximately $12.57$ and $25.13\mathrm{m}{\mathrm{m}}^2$. Sixty-four particles of radius $0.25\mathrm{mm}$ would instead give $16\pi \mathrm{m}{\mathrm{m}}^2$, four times the original area.

To obtain a kinetic prediction, add assumptions explicitly. Let $f$ be the accessible fraction of external surface area, $\sigma$ the amount of substance of catalytic surface sites per unit accessible area, and $\nu$ the mean number of completed conversions per site per unit time under the chosen conditions. Then the model's reaction product formation rate isHere $\nu$ has units ${\mathrm{s}}^{-1}$ and $\sigma$ has units $\mathrm{mol}{\mathrm{m}}^{-2}$, giving $\mathrm{mol}{\mathrm{s}}^{-1}$ overall. Halving $r$ doubles this predicted rate only if $f$, $\sigma$ and $\nu$ remain unchanged. Particle contact can hide area; pores can add area; diffusion, altered surface structure and catalyst deactivation can change the other factors. Consequently, twice the calculated external area is not a universal promise of twice the measured reaction rate. The geometric argument changes contact, not the net Gibbs free energy or the equilibrium constant of the same overall chemical reaction.

Catalytic selectivity concerns which reaction products form when competing reaction mechanisms are available. Faster conversion is valuable when a process would otherwise be impractically slow; a suitable catalyst can also favour a useful product within the available time. In an illustrative pair of irreversible first-order reactions A to B and A to D, the initial fraction of converted A going to B is $k_B/(k_B+k_D)$. Multiplying both rate constants by the same factor changes speed but leaves this fraction unchanged; changing their relative values changes the kinetic product split. This is a finite-time kinetic illustration, not a new equilibrium constant.

A catalyst need not accelerate every competing chemical reaction equally. Enzymes, for example, can distinguish substrates through their active sites.^[ref] If every relevant reversible channel eventually equilibrates in a closed system, its final chemical equilibrium still obeys the endpoint thermodynamics. Selective acceleration, product removal, or external driving can make an observed product distribution depend on time and operating conditions. A claim of improved yield must specify those conditions.

The cathode is the electrode at which the net electrode chemical reaction is reduction: the reacting chemical species gains electrons. This definition is independent of terminal sign. During Daniell cell discharge, ${\mathrm{Cu}}^{2+}$ cations accept electrons arriving at the copper cathode and become neutral copper. The cathode is then the positive terminal. In an externally driven electrolytic cell, the source's negative terminal supplies electrons to the negative cathode. See Pingarrón et al., §3.4: cathode and Daniell cell.

An ion carrying electric current need not undergo reduction at the cathode. For example, in the ideal Daniell cell charge-balance picture, potassium cations arriving from a salt bridge remain dissolved while ${\mathrm{Cu}}^{2+}$ cations are consumed. Follow the half-reaction to identify the reacting species; follow electric charge conservation to account for the other moving ions.

In chain-growth polymerization, each growth step joins a monomer to a reactive site on a polymer chain and regenerates that reactive site. This is the precise mechanistic restriction in IUPAC's definition: growth occurs exclusively through these monomer–chain-site reactions. The word chain refers to the succession of site-regenerating chemical reactions. It does not merely mean that the product looks like a long chain. Chain-growth initiation starts this succession; chain-growth propagation continues it; chain-growth termination and chain transfer in chain polymerization can change its fate.^[ref]

The contrast with step-growth polymerization concerns the permitted joining partners. There, suitable reactive groups can join two existing chains, for example a two-unit chain and a three-unit chain. In chain-growth propagation, a free monomer adds to a polymerization-active site. Release of a small byproduct is a separate classification question: IUPAC explicitly allows condensative chain polymerization.^[ref] The useful consequence is that a few growing chains can incorporate many monomer molecules while much of the feed remains unreacted. Their degree of polymerization depends on how those units are shared among chains, as finite chain-growth bookkeeping makes explicit.

A chemical bond associates chemical atoms or groups of chemical atoms into a sufficiently stable structure to treat it as a distinct entity. The concept covers connections within molecules and in extended solids; see the IUPAC Gold Book, chemical bond. It is useful because composition alone does not specify how the chemical atoms are connected, and those connections help determine a chemical substance's properties and possible chemical reactions.

In a covalent bond, electrons are shared between chemical atoms. Two hydrogen chemical atoms can form a molecular hydrogen molecule, with its electrons interacting with both atomic nuclei. In the usual description of this diatomic molecule, the potential energy as a function of nuclear separation has a minimum: moving the nuclei too close or too far raises the energy. See 7.2 Covalent Bonding.

Separating bonded chemical atoms requires energy; forming the corresponding chemical bonds releases energy. Whether an overall chemical reaction releases energy depends on both the chemical bonds broken and those formed. The lines in the stoichiometry diagram indicate connections schematically, rather than literal rods or a sequence of microscopic events.

A chemical element is an identity shared by all chemical atoms whose atomic nuclei contain the same number of protons. That number is its atomic number, $Z$: six means carbon, eleven means sodium. The identity survives differences in neutron count, which distinguish isotopes, and differences in electron count, which distinguish neutral chemical atoms from ions. See the IUPAC Gold Book, chemical element.

The same name can also denote a pure chemical substance containing only that chemical element. These two uses answer different questions. Molecular oxygen, ${\mathrm{O}}_2$, is a chemical substance made entirely of oxygen; water, ${\mathrm{H}}_2\mathrm{O}$, is a chemical compound containing both hydrogen and oxygen. A water molecule contains three chemical atoms but only two chemical elements. Joining chemical atoms through chemical bonds changes their arrangement without changing their atomic numbers.

This distinction makes stoichiometry possible: ordinary chemical reactions rearrange chemical atoms, so one can count each chemical element before and after a transformation. Changing the proton count instead changes the chemical element and requires a nuclear reaction. The periodic table organizes these identities into a useful map, while atomic number bookkeeping keeps identity, isotope, and electric charge separate.

A chemical equation represents a chemical reaction using chemical formulas, with reactants on the left and reaction products on the right. A balanced chemical equation has the same number of chemical atoms of each chemical element on both sides. For molecular hydrogen and molecular oxygen forming water,The coefficients $2,1,2$ give the relative numbers of molecules; the coefficient $1$ is normally omitted. Count hydrogen as $2\times 2=4$ on the left and $2\times 2=4$ on the right, and oxygen as $1\times 2=2$ on the left and $2\times 1=2$ on the right. See Flowers et al., Chemistry 2e, §4.1: writing and balancing chemical equations.

To balance the initial expression ${\mathrm{H}}_2+{\mathrm{O}}_2⟶{\mathrm{H}}_2\mathrm{O}$, first put $2$ before water to match the two oxygen chemical atoms. This creates four hydrogen chemical atoms on the right, so put $2$ before molecular hydrogen. Changing a coefficient multiplies the entire chemical formula. Changing a subscript changes composition and usually the chemical substance: $2{\mathrm{H}}_2\mathrm{O}$ is two water molecules, whereas ${\mathrm{H}}_2{\mathrm{O}}_2$ is one hydrogen peroxide molecule. Never balance a chemical equation by changing its subscripts.

The bookkeeping follows from conservation of the atomic nuclei in ordinary, non-nuclear chemical reactions; chemical bonds and the distribution of electrons can change. For a chemical equation involving ions, total electric charge must also match. For example, the existing self-ionization of water equation has zero total electric charge on each side, since the $+1$ of hydronium cancels the $-1$ of hydroxide. See 2.6 Ionic and Molecular Compounds (§2.6: nuclei and ions).

The same $2:1:2$ ratio relates changes in amount of substance: two moles of molecular hydrogen consumed per mole of molecular oxygen correspond to two moles of water formed. A mole is a defined unit, not one molecule; the little whole-number picture is an abstract counting model, not a claim that a macroscopic sample contains just five molecules. See molecular count and amount of substance for the exact conversion and SI base unit: mole by Bureau International des Poids et Mesures for the definition.

This is why a balanced chemical equation is useful: stoichiometry turns the same atom accounting into relations between measurable quantities. It does not determine reaction rate, the final conversion set by chemical equilibrium, or the actual yield of recovered reaction products. The limiting reagent gives an upper bound under the specified chemical reaction.

At chemical equilibrium, a reversible chemical reaction has equal forward and reverse reaction rates, so its macroscopic composition stays constant. Individual molecules can keep changing identity: the reaction has not stopped. This dynamic balance is the definition described by the ^[ref]. Equal reaction rates also need not mean equal concentrations. A small population converting rapidly can balance a large population converting slowly.

This distinction gives two useful questions. Which composition is favored at the specified temperature and other constraints? Thermodynamics answers through chemical potentials and an equilibrium constant. How quickly does a prepared mixture approach that composition? Chemical kinetics answers through reaction rates. Knowing the destination does not determine the journey's duration. A composition that barely changes during observation may reflect extremely slow chemical kinetics, so constancy alone does not establish chemical equilibrium.

Consider the idealized elementary chemical reactionAssume a well-mixed closed system at fixed temperature and volume, with first-order law of mass action rates in both directions and no competing chemical reactions. Write $a=[\mathrm{A}]$ and $b=[\mathrm{B}]$ for concentrations in $\mathrm{M}=\mathrm{mol}/\mathrm{L}$. The total $a+b=C=1\mathrm{M}$ is conserved because each conversion replaces one molecule by one molecule. This conservation is special to the chosen stoichiometry.

The forward and reverse reaction rates, measured as conversions per unit volume per unit time, are $J_f=k_{f}a$ and $J_r=k_{r}b$. HenceSetting this derivative to zero givesWith $a(0)=1\mathrm{M}$ and $b(0)=0$, solving the ordinary differential equation yieldsAt $t=5\mathrm{s}$, these are approximately $0.3515\mathrm{M}$ and $0.6485\mathrm{M}$. Initially $J_f=0.3M/s$ and $J_r=0$. At chemical equilibrium, both are $0.075M/s$, while the net reaction rate $J_f-J_r$ is zero. The persistent exchange is visible in the lower panel even after the upper panel has nearly flattened.

Why does chemical equilibrium also involve a free energy minimum? For a closed system at fixed temperature and pressure, without imposed non-expansion work, stable equilibrium minimizes Gibbs free energy over compositions allowed by conserved elemental inventories. See W. C. Carter, MIT thermodynamics notes, sections 2-1.14 and 2-1.19. This is constrained Gibbs minimization for chemical equilibrium: freely minimizing over unrelated species amounts would change the inventory. At fixed temperature and volume, the appropriate potential is Helmholtz free energy. In either setting, varying an allowed extent of reaction gives the corresponding free energy slope ${\sum }_i{\nu }_i{\mu }_i$, with molar chemical potentials and signed stoichiometric vector entries. An interior equilibrium has zero slope; stability requires a minimum. For A converting to B, this says ${\mu }_{\mathrm{B}}={\mu }_{\mathrm{A}}$.

The beauty of this connection is that molecular exchange, a macroscopic composition, and a geometric minimum describe the same balance. In the ideal concentration model, let $x=b/C$ and choose the standard molar chemical potential difference consistent with $K=3$. Up to a composition-independent constant, the fixed-volume Helmholtz free energy then obeyswhere $n$ is the conserved total amount of substance and $R$ the molar gas constant. Its derivative is $\ln [x/(1-x)]-\ln 3$, vanishing at $x=3/4$. Its second derivative is $1/x+1/(1-x)>0$ inside $0<x<1$, so this balance is the unique minimum. The mixing entropy terms explain why the favored composition still contains A.

A chemical substance is chemically inert relative to specified conditions when it does not readily undergo the relevant chemical reactions. The conditions and possible reaction partners matter; the term does not assert the impossibility of every chemical reaction.

Chemical kinetics studies how quickly chemical reactions proceed and how their reaction rates depend on composition and conditions. It connects a measured change in concentration to a rate law, then uses a rate equation to predict future concentrations. This makes it useful for choosing how long to wait for conversion and for testing a proposed reaction mechanism. Changing temperature, reactant concentrations, or the presence of a catalyst can change reaction rates; see Flowers et al., Chemistry 2e, §12.2: factors affecting reaction rates.

Chemical equilibrium specifies the eventual balance under given constraints; chemical kinetics determines how rapidly it is approached. At fixed temperature and pressure, a decrease in Gibbs free energy favours conversion but does not guarantee a fast chemical reaction: reaching a transition state can require crossing a substantial barrier. Thus a mixture can change imperceptibly over an observation period without being at chemical equilibrium. The distinction separates which direction is favoured from how long conversion takes. See 12.5 Collision Theory (§12.5: collision theory and reaction barriers) and Chemistry 2e, §16.1: spontaneity and speed.

Consider two internally equilibrated reservoirs A and B at the same imposed temperature and pressure. They may have different compositions or represent different phases. Transfer a small positive amount of substance $\delta n$ of the same chemical species from A to B, holding the other inventories fixed. To first order, their total Gibbs free energy changes byHence exchange toward lower chemical potential lowers Gibbs free energy. If transfer is allowed in both directions and the two reservoirs contain the exchangeable component, thermodynamic equilibrium requires equal chemical potentials for that component. See MIT 5.60, Lecture 14, pages 1–2.

An impermeable partition can maintain unequal chemical potentials; equality is a condition on an allowed exchange. Two pure samples of the same ideal gas already have equal chemical potentials at the same temperature and pressure. Different compositions at a common pressure can instead give different chemical potentials for one component through its thermodynamic activity.

Equality concerns the same exchangeable component in different places, not all different chemical species. For a permitted chemical reaction, the corresponding interior chemical equilibrium criterion is ${\sum }_i{\nu }_i{\mu }_{\mathrm{m},i}=0$, with signed stoichiometric vector entries ${\nu }_i$. The existing chemical equilibrium and constrained Gibbs minimization for chemical equilibrium sections develop those composition constraints.

A chemical reaction interconverts chemical species, as defined by the ^[ref]. The consumed species are reactants; the formed species are reaction products. In ordinary chemical reactions, chemical atoms are rearranged while each chemical element's total atom count is conserved. For a complete chemical equation, net electric charge is conserved too.

For example, the balanced chemical equationdescribes conversion of molecular hydrogen and molecular oxygen into water. Both sides contain four hydrogen chemical atoms and two oxygen chemical atoms. Three reactant molecules become two reaction product molecules: atom counts are conserved, but molecule counts need not be. See Flowers et al., Chemistry 2e, balancing chemical equations.

The coefficients describe net consumption and formation, without asserting that three molecules collide in one step. A balanced chemical equation supplies quantitative constraints; it does not determine how quickly the chemical reaction occurs, or whether chemical equilibrium prevents complete conversion. Those questions need further information. Keeping these roles distinct makes stoichiometry useful even when the microscopic details are complicated.

Chemistry connects the composition and structure of chemical substances to their properties and transformations. Its central surprise is that rearranging the same chemical atoms can produce a different substance: molecular hydrogen and molecular oxygen can become water without changing which chemical elements are present. Chemical bonds explain the connections; chemical reactions describe the transformations; stoichiometry makes quantitative predictions from the bookkeeping.

This gives a useful route from an invisible world to measurable quantities. A molecule has a definite composition, such as two hydrogen chemical atoms and one oxygen chemical atom in water. A sample contains an enormous number of such molecules. Counting them through amount of substance, measured in moles, lets the same small-number ratios describe a macroscopic sample. The beauty is that the bookkeeping survives this huge change of scale.

A chloroplast is a photosynthetic plastid in plant and algae biological cells. The green-plant chloroplast modeled here has a chloroplast envelope surrounding chloroplast stroma and internal thylakoids. Light-dependent reactions occur at the thylakoid membrane; the Calvin cycle operates in the chloroplast stroma.^[ref]

Cyanobacteria perform oxygenic photosynthesis without chloroplasts: the prokaryotic biological cell itself contains the photosynthetic machinery. Thus chloroplast identifies an organelle, while oxygenic photosynthesis identifies a process. Endosymbiotic theory explains their historical connection; organelle protein import explains how many nuclear-encoded proteins reach the modern organelle.

The useful feature is selective separation: enzymes and substrates can share an organelle without sharing a compartment. For a constructed accessibility example, put a soluble enzyme in chloroplast stroma and its substrate in thylakoid lumen. Sharing the same chloroplast does not permit direct contact; contact requires a transport or boundary-disruption step. Placing the substrate between adjacent thylakoids instead puts it on the stromal side. The visual stack alone cannot establish chemical accessibility.

For the simplest boundary-counting model, represent surrounding cytosol, chloroplast intermembrane space, chloroplast stroma and thylakoid lumen by four vertices in a chain. Its three edges represent the outer envelope, inner envelope and thylakoid membrane, respectively. A route from cytosol to chloroplast stroma crosses two barriers; extending it into thylakoid lumen adds a third. This follows by listing the edges along the chain. Counting the two leaflets of each lipid bilayer separately would answer a different question. These counts describe this idealized geometry, not the number of protein targeting signals or the transport itinerary of every protein.

The light-dependent reactions article develops ATP, NADPH and electrochemical gradient coupling, and the Calvin cycle article develops carbon fixation bookkeeping. Their different locations make the chloroplast interesting as an organized chemical workspace. Light-independent chemical reactions still depend on supplied carriers and regulation; the name does not prescribe nighttime operation.

For a dilute solute distributed between two liquid regions at local chemical equilibrium, define the chromatographic partition coefficient asHere $c_m,c_s$ are concentrations expressed in the same units, $V_m,V_s$ specify accessible volume in the chromatographic mobile phase and immobilized liquid chromatographic stationary phase, respectively, $n$ is total solute amount of substance, and $f_m$ is the fraction of that amount of substance in the chromatographic mobile phase. Assume a constant chromatographic partition coefficient, rapid exchange, unchanged phases and temperature, and no chemical reaction. A chemical equilibrium population still exchanges molecules; it is not permanently divided into two sets.

The amount of substance balance proves the expression for $f_m$. Increasing either $K$ or $V_s/V_m$ reduces the mobile fraction, illustrating why a concentration ratio alone cannot determine retention. For surface adsorption, specify an adsorption storage law instead of assigning an arbitrary liquid volume to the solid.

An original discrete model makes the separation visible. In each identical cell, first establish local chemical equilibrium, then advance only the mobile fraction by one cell; let the stationary fraction wait. Repeat twelve times, without any additional spreading. With $V_s=V_m$, stipulated chromatographic partition coefficients $K_A=1/3$ and $K_B=3$ give $f_A=3/4$ and $f_B=1/4$. Each cycle contributes mean displacement $f_m$ cells, so the mean positions after twelve cycles are nine cells and three cells. Redistribution itself broadens each band: the cell positions have a binomial distribution with parameters $12,f_m$. These dimensionless chosen values illustrate transport, not measurements of any particular substances.

A small organic helper involved in an enzyme's catalysis. It can be transformed in one chemical reaction and regenerated in another.

An exothermic chemical reaction between a fuel and an oxidizing agent. Oxygen is a common oxidizer, but neither visible flame nor oxygen is required by every broader chemical use of the term.

The common-ion effect is the change in a chemical equilibrium caused by adding an ion already participating in it. In the fixed-activity coefficient AB model, adding the common B anion lowers the molar solubility. The shared ion participates regardless of which salt supplied it.^[ref]

The preceding solubility product and material balance give a short proof without a qualitative slogan. Taking a derivative of $s_{\ast }^2+c_0{s}_{\ast }=K_{\mathrm{sp},c}$ at fixed $K_{\mathrm{sp},c}$ givesThus the molar solubility decreases continuously as $c_0$ increases. At $c_0=0$, the derivative is $-1/2$; for a very large common-ion background, the rationalized expression approaches $K_{\mathrm{sp},c}/c_0$. The solubility product is fixed within this model even while the molar solubility changes.

Neglecting $s_{\ast }$ beside $c_0$ replaces $s_{\ast }(c_0+s_{\ast })=K_{\mathrm{sp},c}$ by $c_0{s}_{\mathrm{app}}=K_{\mathrm{sp},c}$. This approximation always overestimates the exact molar solubility, since it neglects the B anions supplied by AB itself. For $c_0>0$, its relative error has an exact simple form:At $c_0=0.030\mathrm{M}$, $s_{\mathrm{app}}=0.003333333\mathrm{M}$, approximately $10.1\%$ too high. At $c_0=0.100\mathrm{M}$, $s_{\ast }=0.000990195\mathrm{M}$ and $s_{\mathrm{app}}=0.001000000\mathrm{M}$, approximately $0.990\%$ too high. A small solubility product by itself does not justify the approximation: the common-ion background must be large compared with the resulting molar solubility.

For a desired fractional relative error at most $\epsilon >0$, substituting $s_{\ast }/c_0\le \epsilon$ into the exact balance gives the equivalent criterionFor this model, a $5\%$ tolerance requires $c_0\ge 0.0436436\mathrm{M}$. This bound concerns only the algebraic approximation. In a real chemical solution, changes in ionic strength and activity coefficients, or competing chemical reactions, can invalidate a constant $K_{\mathrm{sp},c}$ even when this bound is met.