Stability & Thermodynamics
Every complex in this book exists only to the extent that an equilibrium constant allows it to. This part is about those constants: how they are defined, why they almost always fall as more ligands are added, how they are measured, and what controls their size. It treats the chelate and macrocyclic effects quantitatively rather than as slogans — separating the entropy term from the enthalpy term and showing where the usual textbook explanation is incomplete — and it ends in the analytical laboratory, with the conditional constants that make EDTA titrations work. Two layers on every section: a slow, hand-held beginner path and a research-grade advanced/reference path.
The 9 sections in Part 7
- 1Stepwise constants Kn, overall constants βn, and moving between them Free below
- 2Why K₁ > K₂ > K₃ … — and the instructive cases where it is not
- 3Measuring stability constants — pH titration, spectrophotometry and Job’s method
- 4Metal-ion factors — charge, size and the Irving–Williams series
- 5Ligand factors — basicity, ring size, sterics and π-bonding
- 6Hard and soft acids and bases, applied properly
- 7The chelate effect quantified
- 8The macrocyclic and cryptate effects, and preorganisation
- 9Conditional constants, masking and the EDTA titration
Stepwise constants Kn, overall constants βn, and moving between them
Free extractSection J.1 of Part 7, reproduced in full from the book — figures and all. No sign-in, no paywall on this section.
What the two kinds of constant mean, how they are related, and the arithmetic of converting one into the other — done slowly, because everything later in this part depends on getting it automatic.
Put a metal ion in water and add a ligand. The ligands do not all arrive at once. They arrive one at a time, each in its own equilibrium, and each of those equilibria has its own constant. Consider a metal that ends up with four ligands:
ML + L ⇌ ML₂ K₂ = [ML₂] / ([ML][L])
ML₂ + L ⇌ ML₃ K₃ = [ML₃] / ([ML₂][L])
ML₃ + L ⇌ ML₄ K₄ = [ML₄] / ([ML₃][L])
Now write the same chemistry a second way — as if each complex were assembled from the bare metal ion and free ligands in a single act:
M + 2L ⇌ ML₂ β₂ = [ML₂] / ([M][L]²)
M + 3L ⇌ ML₃ β₃ = [ML₃] / ([M][L]³)
M + 4L ⇌ ML₄ β₄ = [ML₄] / ([M][L]⁴)
The two sets are not independent. Add the first two stepwise equilibria together — and remember that when you add equilibria you multiply their constants:
ML + L ⇌ ML₂ (K₂)
―――――――――――――――――――
M + 2L ⇌ ML₂ (K₁K₂ = β₂)
You can see it in the algebra directly: multiply the two expressions and the [ML] cancels.
And going the other way, each stepwise constant is the ratio of two consecutive overall constants:
Two conventions you must know before you read any table
(a) The coordinated water is not written. The honest equation for the first step of ammine formation on nickel is
Every table in every book compresses this to Ni²⁺ + NH₃ ⇌ [Ni(NH₃)]²⁺. The reason is that water is the solvent: it is present at about 55.5 mol L⁻¹ and its activity is taken as 1, so it drops out of the equilibrium expression. This is a convention, it is universal, and it is harmless — until J.8, where the whole argument about the chelate effect turns on exactly this choice. Remember that it was made.
(b) The constants are concentration constants at a stated ionic strength. Strictly, equilibrium constants are written in activities. In practice, stability constants are measured in a swamping inert electrolyte — typically 0.1 M or 1.0 M KNO₃, NaClO₄ or KCl — so that activity coefficients stay constant while the reagent concentrations are varied. The number that comes out is a concentration constant valid at that ionic strength, and it will differ, usually by a few tenths of a log unit, from the thermodynamic constant at I = 0.
A worked table: copper(II) and ammonia
| Step | Equilibrium | log Kn | Cumulative | log βn |
|---|---|---|---|---|
| 1 | Cu²⁺ + NH₃ ⇌ [Cu(NH₃)]²⁺ | 4.15 | β₁ = K₁ | 4.15 |
| 2 | [Cu(NH₃)]²⁺ + NH₃ ⇌ [Cu(NH₃)₂]²⁺ | 3.50 | β₂ = K₁K₂ | 7.65 |
| 3 | [Cu(NH₃)₂]²⁺ + NH₃ ⇌ [Cu(NH₃)₃]²⁺ | 2.89 | β₃ = K₁K₂K₃ | 10.54 |
| 4 | [Cu(NH₃)₃]²⁺ + NH₃ ⇌ [Cu(NH₃)₄]²⁺ | 2.13 | β₄ = K₁K₂K₃K₄ | 12.67 |
Given log K₁ = 4.15, log K₂ = 3.50, log K₃ = 2.89 and log K₄ = 2.13 for Cu(II)/NH₃, find β₂ and β₄ as numbers, and state their units. Easy
For Ag(I)/NH₃ at 20 °C and I = 0.1 M, log β₂ = 7.22 and log K₁ = 3.31. Find log K₂, and comment. Easy
⚠ Common mistakes & exam traps
- β₂ is not K₂. β₂ = K₁K₂. If a question gives you “β₂ = 10⁷·⁶₅” and asks for the constant of the second step, you must subtract log K₁ first. This single confusion accounts for more lost marks in this topic than anything else.
- Do not compare βm with βn for m ≠ n. They have different units, so “β₆ = 10⁸·₇ is smaller than β₃ = 10¹⁸·³” is not a meaningful sentence. To compare two ligand systems, write the exchange reaction between them; its constant is dimensionless.
- “Stability constant” unqualified usually means the formation constant. But the older literature also uses instability constant or dissociation constant, which is its reciprocal. If a value looks absurdly small (10⁻¹² for a complex that obviously forms), you are looking at a dissociation constant.
- Water does not appear in the expression, but it did not go away. Every formation constant in water is really a substitution constant, ligand for water. That is not pedantry — it is the reason the chelate effect has a standard-state problem (J.8).
- A large K does not mean fast. Thermodynamic stability and kinetic inertness are independent properties. Part 8 has the vocabulary; here, simply never use “stable” to mean “slow to react”.
Show solution
The ratio asked for is β₄[CN⁻]⁴ because β₄ = [Cd(CN)₄²⁻]/([Cd²⁺][CN⁻]⁴). So log(ratio) = log β₄ + 4 log[CN⁻] = 18.78 + 4(−3.00) = 18.78 − 12.00 = 6.78, i.e. a ratio of 6.0 × 10⁶. Even at millimolar free cyanide the tetracyano complex outnumbers free Cd²⁺ by six million to one — which is why cyanide is the standard masking agent for cadmium (J.9).
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M + 6 L ⇌ ML₆ with β₆(L), and M + 3 L′′ ⇌ M(L′′)₃ with β₃(L′′).
Subtract the first from the second (i.e. reverse the first and add): ML₆ + 3 L′′ ⇌ M(L′′)₃ + 6 L, Kex = β₃(L′′) / β₆(L).
Units: β₃ carries M⁻³, β₆ carries M⁻⁶; the quotient carries M⁻³⁺⁶ = M³. That is not dimensionless — and this is exactly the point. Check the reaction itself: 4 particles on the left, 7 on the right, so Δn = +3 and the constant must carry M³. The claim in the question is false as stated. It becomes true only for a particle-conserving exchange. This is the seed of the standard-state argument in J.8, and being able to spot a units-inconsistent comparison is worth more than memorising any table.
Activity constants, mixed constants, the species-distribution picture, and the free-energy bookkeeping that connects all of it to thermodynamics.
From constant to free energy
Every stability constant is a disguised free energy, and switching between the two languages is what allows the enthalpy/entropy dissection of J.7 to happen at all.
At 298.15 K, 2.303 RT = 2.303 × 8.314 × 298.15 = 5708 J mol⁻¹, so:
Because free energies add when reactions add, and log K adds when reactions add, the two bookkeeping systems are the same system. The relation log βn = ∑ log Ki is nothing more than ΔG°overall = ∑ ΔG°step, which is Hess’s law.
Thermodynamic, stoichiometric and mixed constants
Three kinds of constant appear in the literature, and a serious candidate should be able to tell which one a table is quoting.
| Kind | Written in | How obtained | Symbol you may see |
|---|---|---|---|
| Thermodynamic (activity) constant | activities; valid at I = 0 | measured at several ionic strengths and extrapolated to I = 0, or corrected with an activity-coefficient model (Davies, SIT, Pitzer) | K°, TK |
| Stoichiometric (concentration) constant | concentrations, at a stated constant ionic strength | measured in a swamping inert electrolyte, e.g. 0.1 M KNO₃ | K, Kc |
| Mixed (Brønsted) constant | concentrations for everything except H⁺, which enters as activity | arises automatically when a glass electrode (which reads aH) is combined with analytical concentrations of everything else | K′, mixedK |
The ionic-strength dependence itself is not mysterious. For a reaction between ions, the Debye–Hückel treatment gives the leading term in the activity-coefficient correction as proportional to the change in the sum of squared charges:
The practical consequence: constants for reactions between highly charged ions are the most ionic-strength sensitive. Ca²⁺ + Y⁴⁻ ⇌ CaY²⁻ has Δ(z²) = 4 − (4 + 16) = −16, so log K for EDTA complexes moves substantially between I = 0 and I = 0.1 M. This is why the analytical literature quotes EDTA constants at a stated ionic strength, usually 0.1 M.
Species distribution — what the constants actually predict
A set of β values is a complete description of the system: given the free-ligand concentration, every species concentration follows. The fraction of total metal present as MLn is
The denominator — 1 + β₁[L] + β₂[L]² + … — recurs constantly and deserves its own name. In the analytical literature it is written αM(L) and it is the side-reaction coefficient that J.9 is built on. Meeting it here, as a normalising denominator, makes it much less mysterious when it reappears as a masking correction.
Polynuclear and protonated species — when MLn is not enough
Real systems are often not a clean ML₁…MLN series. Two extra families appear frequently enough that their notation should be recognised:
- Polynuclear complexes MmLn, whose constants carry two subscripts: βmn = [MmLn]/([M]m[L]n). The commonest examples are hydroxo-bridged species such as [Fe₂(OH)₂]⁴⁺ and [Cu₂(OH)₂]²⁺, which is why iron(III) speciation in water is so much harder than a simple stepwise series suggests.
- Protonated and hydroxo complexes MHjLn and M(OH)kLn, written with a third index, e.g. β1 1 1 for MHL. These matter enormously in EDTA chemistry, where MHY⁻ and MOHY³⁻ species exist at the extremes of pH.
The general convention, which is worth recognising even if you never use it, writes the cumulative constant of the species MpLqHr as βpqr, with negative r denoting hydroxo species (a deprotonated water). So β1 0 −1 is the constant for M + H₂O ⇌ MOH + H⁺, the first hydrolysis constant.
Show solution
β₁[L]: 5.0 + (−4.0) = 1.0 → 101.0 = 10
β₂[L]²: 9.0 + (−8.0) = 1.0 → 101.0 = 10
β₃[L]³: 12.0 + (−12.0) = 0.0 → 100.0 = 1
Denominator = 1 + 10 + 10 + 1 = 22. So α₂ = 10/22 = 0.45, i.e. 45% of the metal is ML₂. (Free metal is 1/22 = 4.5%, ML is 45%, ML₃ is 4.5%.) Note the shortcut: adding log β and n log[L] keeps every quantity in comfortable range and avoids calculator overflow.
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Read the rest of Part 7
The remaining 8 sections of this part — Why K₁ > K₂ > K₃ … — and the instructive cases where it is not, Measuring stability constants — pH titration, spectrophotometry and Job’s method, Metal-ion factors — charge, size and the… — and all nine parts of Coordination Chemistry are part of ChemVidya Full Access, along with the other books, 55 Study Notes and 6,000+ practice questions.
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