Lesson 8 · 30 min

Stability and the chelate effect

Formation constants measure how stable a complex is. Why does one ethylenediamine hold on far better than two ammonias? The answer lies in entropy — and EDTA takes it to the limit.

In this lesson

Write stepwise constants Kₙ and overall constants βₙ, and convert between them
Compute the fraction of each complex from the free-ligand concentration and read a distribution diagram
Explain the chelate effect with ΔH° and ΔS° data
Use conditional formation constants to explain why calcium and magnesium are titrated with EDTA at pH 10

Lead poisoning is treated with an intravenous drip of calcium disodium edetate, $\ce{Na2[Ca(edta)]}$ . In the blood, lead ions push the calcium out of the EDTA, and the $\ce{[Pb(edta)]^2-}$ formed leaves in the urine. Why does EDTA prefer lead to calcium? And why is the drug loaded with calcium first instead of being given as a plain sodium salt? Both answers lie in formation constants: $\ce{[Pb(edta)]^2-}$ is about twenty million times more stable than $\ce{[Ca(edta)]^2-}$ , while calcium-free EDTA would also strip the blood of the calcium it needs.

Predict first

$\ce{Cd^2+}$ can bind four methylamines, $\ce{CH3NH2}$ , or two ethylenediamines — four Cd–N bonds either way, releasing almost the same heat (ΔH° about −57 kJ/mol). Which complex has the larger formation constant?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

\ce{[Cd(CH3NH2)4]^2+}

, by far

\ce{[Cd(en)2]^2+}

, by about 10⁴ About the same, since the bonds are equally strong It depends only on the ligand concentration

One ligand at a time: stepwise constants

Complexes form step by step, and every step has its own equilibrium constant. In water, for example, $\ce{NH3}$ replaces the water molecules on $\ce{Cu^2+}$ one at a time (water is left out of the expressions): log K₁ = 4.04, log K₂ = 3.43, log K₃ = 2.80 and log K₄ = 1.48. Multiplying the four steps gives the overall formation constant $\beta_4 = K_1K_2K_3K_4 = 10^{11.75}$ .

Kₙ usually shrinks from step to step: fewer water molecules remain to be replaced while more ligands could leave (a statistical effect), and the ligands crowd each other more and more. A fifth and sixth $\ce{NH3}$ hardly bind to copper at all, because the Jahn–Teller distortion holds the two axial sites far out.

K_n = \frac{[\mathrm{ML}_n]}{[\mathrm{ML}_{n-1}][\mathrm{L}]},\qquad \beta_n = K_1K_2\cdots K_n = \frac{[\mathrm{ML}_n]}{[\mathrm{M}][\mathrm{L}]^n}

A stepwise constant describes adding one more ligand; an overall constant describes going from the aqua ion all the way; in logarithms, log βₙ is the running sum of the log K values.

$\mathrm{M},\ \mathrm{L}$: aqua metal ion and free ligand
$\mathrm{ML}_n$: complex carrying n ligands
$K_n,\ \beta_n$: stepwise and overall constants of step n

\alpha_n = \frac{\beta_n[\mathrm{L}]^n}{\sum_{i=0}^{N}\beta_i[\mathrm{L}]^i}\qquad (\beta_0 = 1)

The fraction of the metal present as MLₙ depends only on the free-ligand concentration [L]; neighbouring curves cross where [L] = 1/Kₙ.

$\alpha_n$: fraction of the metal present as MLₙ
$[\mathrm{L}]$: concentration of free (unbound) ligand

Stepwise constants · the chelate effect · EDTA

How stable is a complex?

Cu²⁺
[Cu(NH₃)]²⁺
[Cu(NH₃)₂]²⁺
[Cu(NH₃)₃]²⁺
[Cu(NH₃)₄]²⁺

Average ligands n̄: 2.11
Main species: [Cu(NH₃)₂]²⁺ (49 %)
Computed colour of 0.010 M Cu²⁺: 2% · 18% · 49% · 31% · 1%

Drag the free-ammonia concentration: each [Cu(NH₃)ₙ]²⁺ takes over in turn, and the computed colour deepens from pale blue to royal blue. Neighbouring curves cross at lg[NH₃] = −log Kₙ — read log K₃ off the graph.

The chelate effect

The cadmium data in the opening question say it all: four Cd–N bonds release almost the same heat, yet the stabilities differ ten-thousand-fold. The difference is entropy. Each ethylenediamine that binds drives out two water molecules at once:

$\ce{[Cd(H2O)6]^2+} + \ce{4CH3NH2}$ → complex + $\ce{4H2O}$ : five particles become five;
$\ce{[Cd(H2O)6]^2+} + \ce{2en}$ → complex + $\ce{4H2O}$ : three particles become five.

More particles moving freely in solution means higher entropy and a more negative ΔG°. There is a probability picture too: once one end of an en is attached, the other end is tethered close to the metal and closes the ring easily; even if one end lets go, it usually rebinds before the ligand can drift away. Five-membered chelate rings (en, oxalate, EDTA) have the least strain and are the most stable.

\Delta G^\circ = -RT\ln\beta = -2.303\,RT\,\lg\beta = \Delta H^\circ - T\Delta S^\circ

At 25 °C, 2.303RT = 5.708 kJ/mol: every extra unit of log β makes ΔG° 5.7 kJ/mol more negative.

$\beta$: overall formation constant
$\Delta H^\circ,\ \Delta S^\circ$: standard enthalpy and entropy changes

Worked example

The chelate effect in numbers

At 25 °C, $\ce{Cd^2+} + \ce{4CH3NH2}$ has log β₄ = 6.55 and ΔH° = −57.3 kJ/mol; $\ce{Cd^2+} + \ce{2en}$ has log β₂ = 10.62 and ΔH° = −56.5 kJ/mol. Find ΔG° and ΔS° for both, and the equilibrium constant of $\ce{[Cd(CH3NH2)4]^2+} + \ce{2en} \mathrel{\ce{<=>}} \ce{[Cd(en)2]^2+} + \ce{4CH3NH2}$ .

EDTA and conditional formation constants

EDTA⁴⁻ (written $\ce{Y^4-}$ ) wraps almost any metal ion 1 : 1 with 2 N and 4 O, closing five five-membered chelate rings. log $K_\text{f}$ runs from 7.3 for $\ce{Ag+}$ through 10.7 for $\ce{Ca^2+}$ and 18.0 for $\ce{Pb^2+}$ up to 25.1 for $\ce{Fe^3+}$ .

But EDTA is also a hexaprotic weak acid, and only the fully deprotonated $\ce{Y^4-}$ binds well. The more acidic the solution, the harder $\ce{H+}$ competes with the metal for the EDTA. Folding this competition in gives the conditional formation constant K′. In an ammonia buffer at pH 10, K′ for both $\ce{Ca^2+}$ and $\ce{Mg^2+}$ exceeds the roughly 10⁸ needed for a titration, which is why water hardness is measured at pH 10; at pH 2 calcium does not react at all, yet $\ce{Fe^3+}$ can still be titrated.

K' = \alpha_{\mathrm{Y}^{4-}}\,K_\mathrm{f},\qquad \lg K' = \lg K_\mathrm{f} + \lg\alpha_{\mathrm{Y}^{4-}}

α(Y⁴⁻) is the fraction of the free EDTA present as Y⁴⁻, computed from EDTA’s six pKa values: 0.30 at pH 10 but only about 3 × 10⁻⁷ at pH 5.

$K_\mathrm{f}$: formation constant of M with Y⁴⁻
$\alpha_{\mathrm{Y}^{4-}}$: fraction of all free EDTA present as Y⁴⁻
$K'$: conditional formation constant at the given pH

Stepwise constants · the chelate effect · EDTA

How stable is a complex?

Fe³⁺
Cu²⁺
Zn²⁺
Ca²⁺
Mg²⁺

pH · α(Y⁴⁻): 10.0 · 0.3
lg K′ · Fe³⁺: 24.58
lg K′ · Cu²⁺: 18.28
lg K′ · Zn²⁺: 15.98
lg K′ · Ca²⁺: 10.17
lg K′ · Mg²⁺: 8.27

K′ ≈ 10⁸: rule-of-thumb minimum for a sharp endpoint

Drag the pH: the conditional constants of all five metals fall together in acid. The dashed line at K′ = 10⁸ is the rule-of-thumb minimum for a sharp end point: find the lowest pH at which Ca²⁺ and Mg²⁺ can be titrated, then the pH range where only Fe³⁺ can.

Concept check

For $\ce{Cu^2+}$ and $\ce{NH3}$ , log K₁ … log K₄ = 4.04, 3.43, 2.80, 1.48. What is the main reason each successive constant is smaller?

Copper is reduced a little by each

\ce{NH3}

bound The bigger the complex, the more easily

\ce{NH3}

is protonated Fewer waters remain to be replaced and more ligands could leave, plus growing crowding The solution darkens and absorbs more light

Try the calculation

For $\ce{[CaY]^2-}$ , log $K_\text{f}$ = 10.69, and α(Y⁴⁻) = 0.30 at pH 10. Find log K′ at pH 10.

Model limits

Formation constants say where equilibrium lies, not how fast it is reached: $\ce{[Ni(CN)4]^2-}$ is extremely stable yet swaps its ligands for isotope-labelled $\ce{CN-}$ within about half a minute (labile), while $\ce{[Co(NH3)6]^3+}$ ought to decompose in acid yet survives for days (inert). The constants here use concentrations instead of activities and hold only at the stated temperature and ionic strength (20 °C and I = 0.1 M for EDTA); the conditional constant allows only for protonation of EDTA — not for side reactions of the metal with $\ce{OH-}$ or with ammonia from the buffer, nor for hydroxide precipitation at high pH. Above pH 12, $\ce{Mg(OH)2}$ precipitates, which is exactly how calcium is titrated on its own.

Key takeaways

Stepwise constants Kₙ usually decrease step by step; βₙ = K₁K₂…Kₙ, so log βₙ is the running sum of log K.
The fractions αₙ = βₙ[L]ⁿ/Σβᵢ[L]ⁱ depend only on the free-ligand concentration; neighbouring curves cross at [L] = 1/Kₙ.
The chelate effect is mainly entropic: [Cd(en)₂]²⁺ and [Cd(CH₃NH₂)₄]²⁺ have almost the same ΔH°, but K differs by 10⁴.
For EDTA, log K′ = log $K_\text{f}$ + log α(Y⁴⁻): 10.17 for CaY²⁻ at pH 10, which is why calcium and magnesium are titrated at pH 10.

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Data and references

A. E. Martell, R. M. Smith, Critical Stability Constants, vols. 1–6, Plenum (1974–1989), as tabulated in D. Harvey, Analytical Chemistry 2.1, Appendix 12.
N. N. Greenwood, A. Earnshaw, Chemistry of the Elements, 2nd ed., Butterworth-Heinemann (1997), p. 910 (the chelate effect), after C. G. Spike, R. W. Parry, J. Am. Chem. Soc. 75, 2726 (1953).
D. C. Harris, Quantitative Chemical Analysis, 9th ed., W. H. Freeman (2016), ch. 11 (EDTA titrations, conditional formation constants).
H. Taube, “Rates and mechanisms of substitution in inorganic complexes in solution”, Chem. Rev. 50, 69–126 (1952).
G. Schwarzenbach, “Der Chelateffekt”, Helv. Chim. Acta 35, 2344–2359 (1952).