Stability constants of complexes — chelate effect: Worked examples

This page is the practice arena for the parent topic. We will not re-derive the theory — instead we hunt down every kind of number this topic can hand you and work each one to the ground. Read the matrix first, then do the examples with your hand covering the steps.

The scenario matrix

Every problem in this topic is one (or a blend) of these cells. The last column names the example that kills that cell.

#	Cell (the "scenario class")	What makes it tricky	Killed by
A	Stepwise $K_{i}$ → overall $β_{n}$ (multiply)	remembering multiply, not add	Ex 1
B	Logs given → $lo g β$ (add)	the classic "add vs multiply" trap	Ex 2
C	$β$ → $Δ G^{\circ}$ (or back)	sign of $ln$ , units of $R$	Ex 3
D	Split $Δ G$ into $Δ H$ and $- T Δ S$	which term wins?	Ex 4
E	Chelate vs monodentate — predict sign of $Δ S$ by particle count	zero net change vs positive	Ex 5
F	Degenerate / limiting: $n = 1$ , or $K_{i} = 1$ , or $Δ S = 0$	"nothing happens" cases	Ex 6
G	Real-world word problem (EDTA titration, water hardness)	translating words → equilibrium	Ex 7
H	Exam twist: ring-size / macrocyclic reasoning, no big numbers	conceptual, no calculator	Ex 8

Constants used everywhere below: $R = 8.314 J K^{- 1} mol^{- 1}, T = 298 K, 2.303 R T = 5708 J/mol \approx 5.71 kJ/mol .$

Figure — Stability constants of complexes — chelate effect

Example 1 — Cell A: stepwise → overall (the multiply case)

Worked example

For $Cd^{2 +}$ binding ammonia, the first two stepwise constants are $K_{1} = 400$ and $K_{2} = 130$ . Find the overall constant $β_{2}$ and its log.

Forecast: Guess — is $β_{2}$ near $530$ or near $52000$ ? Write your guess before reading on.

Write what $β_{2}$ means. $β_{2} = \frac{[ M L _{2} ]}{[ M ] [ L ] ^{2}}$ , the constant for $M + 2 L ⇌ M L_{2}$ formed directly. Why this step? We must know which quantity we're chasing before combining anything — $β$ is the whole-journey constant.
Use the multiply bridge. $β_{2} = K_{1} K_{2} = 400 \times 130 = 52000$ . Why this step? The two steps happen in sequence; equilibrium constants of sequential reactions multiply (the shared $[M L]$ cancels).
Take the log. $lo g β_{2} = lo g 400 + lo g 130 = 2.602 + 2.114 = 4.716$ . Why this step? Logs turn a product into a sum, which is how tables report these numbers.

Verify: $1 0^{4.716} \approx 5.20 \times 1 0^{4} ✓$ . The guess " $530$ " (adding) would be the classic mistake. Units: $K$ has units, but they cancel cleanly in ratios of concentrations here; we quote the number.

Example 2 — Cell B: logs given → add them (the trap case)

Worked example

A ligand forms $M L$ , $M L_{2}$ , $M L_{3}$ with $lo g K_{1} = 4.3$ , $lo g K_{2} = 3.5$ , $lo g K_{3} = 2.6$ . Find $lo g β_{3}$ and $β_{3}$ .

Forecast: Will $lo g β_{3}$ be bigger or smaller than any single $lo g K$ ? By roughly how much?

Add the logs. $lo g β_{3} = 4.3 + 3.5 + 2.6 = 10.4$ . Why this step? Because $β_{3} = K_{1} K_{2} K_{3}$ , taking $lo g$ converts the product to a sum. This is the only place addition is correct.
Un-log to get $β_{3}$ . $β_{3} = 1 0^{10.4} \approx 2.5 \times 1 0^{10}$ . Why this step? $lo g$ is the inverse of $1 0^{(\cdot)}$ ; it undoes the log so we recover the actual constant.
Sanity on decreasing $K_{i}$ . Note $4.3 > 3.5 > 2.6$ : stepwise constants fall as sites fill (statistical + crowding), exactly as theory predicts. Why this step? A physical check — if $K_{i}$ had risen, we'd suspect a spin-state change or a data error.

Verify: $lo g (K_{1} K_{2} K_{3}) = 10.4$ ; and $1 0^{10.4} = 2.51 \times 1 0^{10} ✓$ . Bigger than any single step ✓ (products of numbers $> 1$ grow).

Example 3 — Cell C: $β$ ↔ $Δ G^{\circ}$

Example 4 — Cell D: split $Δ G$ into enthalpy and entropy

Worked example

A chelate reaction has $Δ H^{\circ} = - 12 kJ/mol$ and $Δ S^{\circ} = + 100 J K^{- 1} mol^{- 1}$ . Find $Δ G^{\circ}$ at $298$ K, then $lo g β$ . Which term dominates?

Forecast: Guess whether enthalpy or entropy contributes more to $- Δ G$ .

Compute the entropy term. $- T Δ S^{\circ} = - (298) (100) = - 29800 J/mol = - 29.8 kJ/mol$ . Why this step? $Δ S$ is in J, $T$ in K, so $T Δ S$ comes out in J — watch the unit match before adding.
Add the two contributions. $Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ} = - 12 + (- 29.8) = - 41.8 kJ/mol$ . Why this step? This is the master split $Δ G = Δ H - T Δ S$ ; both terms are negative here so both help.
Convert to $lo g β$ . $lo g β = \frac{- Δ G ^{\circ}}{2.303 R T} = \frac{41800}{5708} = 7.32$ . Why this step? Rearranged Cell-C bridge; gives the observable ranking number.

Verify: Entropy piece $- 29.8$ vs enthalpy piece $- 12.0$ ⇒ entropy dominates ( $71%$ of $- Δ G$ ), exactly the chelate story ✓. Check: $- 41800/ - 5708 = 7.32 ✓$ .

Example 5 — Cell E: predict the sign of $Δ S$ by counting particles

Worked example

Two reactions replace six coordinated waters on $M^{2 +}$ : $(a) [M(H_{2} O)_{6}] + 6 NH_{3} ⇌ [M(NH_{3})_{6}] + 6 H_{2} O$ $(b) [M(H_{2} O)_{6}] + 3 en ⇌ [M(en)_{3}] + 6 H_{2} O$ Predict the sign of $Δ S^{\circ}$ for each, and roughly estimate the extra $lo g β$ that (b) gains over (a).

Forecast: Which reaction increases the number of free-floating particles?

Count free particles, left vs right, reaction (a). Left: $1 + 6 = 7$ free species. Right: $1 + 6 = 7$ . Net change $= 0$ . Why this step? Entropy of solution tracks how many independent particles wander freely; equal counts ⇒ $Δ S^{\circ} \approx 0$ .
Count free particles, reaction (b). Left: $1 + 3 = 4$ . Right: $1 + 6 = 7$ . Net change $= + 3$ . Why this step? Three linked en molecules release six waters — you free more than you consume ⇒ $Δ S^{\circ} > 0$ .
Estimate the entropy bonus. Take $\sim + 84 J K^{- 1} mol^{- 1}$ for those 3 extra particles. Extra $- T Δ S = - (298) (84) = - 25000 J/mol$ . So $Δ (lo g β) = \frac{25000}{5708} = 4.4$ . Why this step? Convert the entropy edge into observable log-units via the same bridge — this is the size of the chelate effect.

Verify: Reaction (a) $Δ S \approx 0$ ; (b) $Δ S ≫ 0$ ✓. The $+ 4.4$ jump matches the observed gap ( $lo g β_{6} \approx 8.6$ for NH₃ vs $lo g β_{3} \approx 18.3$ for en — a difference dominated by entropy) $✓$ .

Example 6 — Cell F: degenerate & limiting cases

Worked example

Three quick edge cases. In each, decide what happens before computing. (i) A metal binds only one ligand: $n = 1$ , $K_{1} = 1 0^{5}$ . What is $β$ ? (ii) A step has $K_{i} = 1$ . How does it change $lo g β$ ? (iii) A reaction has $Δ S^{\circ} = 0$ and $Δ H^{\circ} = 0$ . What is $β$ ?

Forecast: Which of these leaves $lo g β$ unchanged, and which gives $β = 1$ ?

(i) Single ligand. With $n = 1$ the product has one factor: $β_{1} = K_{1} = 1 0^{5}$ , so $lo g β_{1} = 5$ . Why this step? $β$ is literally defined as $K_{1}$ when $n = 1$ ; there is nothing to multiply. This is the smallest non-trivial case.
(ii) A step of $K_{i} = 1$ . $lo g K_{i} = lo g 1 = 0$ , so adding this step changes $lo g β$ by $0$ — that ligand is bound as often as it is released; it neither helps nor hurts. Why this step? $K_{i} = 1$ means the $i$ -th step's forward and reverse are equally likely — a "do-nothing" step, useful as a reference line.
(iii) Both thermodynamic terms zero. $Δ G^{\circ} = 0 - 0 = 0$ , so $ln β = 0$ and $β = e^{0} = 1$ ; $lo g β = 0$ . Why this step? $Δ G^{\circ} = 0$ is the balance point — complex and free forms coexist $50/50$ at unit activities.

Verify: (i) $1 0^{5}$ ✓; (ii) $lo g 1 = 0$ ✓ (no shift); (iii) $β = 1$ ✓ (perfectly poised equilibrium). These are the "signposts" — memorise that $K = 1 \Rightarrow Δ G = 0 \Rightarrow lo g = 0$ .

Example 7 — Cell G: real-world word problem (EDTA / water hardness)

Worked example

In a water-hardness titration you complex $Ca^{2 +}$ with $EDTA^{4 -}$ (hexadentate). For $Ca-EDTA$ , $lo g K = 10.7$ . (a) Write the reaction and count free particles. (b) Compute $Δ G^{\circ}$ . (c) In one line, say why one EDTA beats six separate donors.

Forecast: Left side vs right side — does the particle count jump a lot or a little?

(a) Reaction + particle count. $[Ca(H_{2} O)_{6}]^{2 +} + EDTA^{4 -} ⇌ [Ca(EDTA)]^{2 -} + 6 H_{2} O$ . Left: $1 + 1 = 2$ free particles; right: $1 + 6 = 7$ . Net $+ 5$ . Why this step? One ligand releasing six waters is the maximum entropy payoff — that's why EDTA is a "super-ligand".
(b) Energy. $Δ G^{\circ} = - 2.303 R T lo g K = - (5708) (10.7) = - 6.11 \times 1 0^{4} J/mol \approx - 61 kJ/mol$ . Why this step? Uses the Cell-C bridge; the strongly negative value means the titration goes essentially to completion — good, we want a sharp endpoint.
(c) One-line reason. One EDTA molecule forms five 5-membered chelate rings and liberates six waters while consuming just one ligand ⇒ huge $+ Δ S$ ⇒ huge $- Δ G$ ⇒ huge $K$ . Why this step? Ties the number back to the physical entropy engine that makes complexometric titration reliable.

Verify: $- 61000/ - 5708 = 10.7 ✓$ . Net particle jump $+ 5$ (biggest of all our examples) matches EDTA being the strongest chelator among them ✓.

Example 8 — Cell H: exam twist (ring size, no big numbers)

Worked example

A student proposes two ligands: (P) a diamine forming a 5-membered chelate ring, and (Q) a diamine forming a 7-membered ring with the same metal and same two N-donors. Both have essentially identical $Δ H^{\circ}$ and release the same number of waters. Which gives the larger $β$ , and why? No numbers given — reason it out.

Forecast: If particle counts are equal, what else can decide?

Note the entropy tie. Same number of waters released, same particles ⇒ the translational entropy advantage is the same for both. So particle-count entropy cannot break the tie. Why this step? Rule out the usual winner first, so we know we need a geometric argument.
Bring in ring strain / bite probability. A 5-membered ring closes with near-ideal bond angles (unstrained), so the second donor reaches the metal easily. A 7-membered ring must span farther; the second tooth has a lower probability of finding the metal (lower effective local concentration) and the ring can flex away. Why this step? When particle-entropy is equal, the ease of ring closure — a geometric/effective-concentration effect — decides.
Conclude. (P), the 5-membered ring, gives the larger $β$ . Chelate stability peaks at 5-membered rings. Why this step? This is the exam's trap: "bigger ring = tighter grip" is wrong; 5 is the sweet spot.

Verify: Consistent with the parent note's rule (5-membered most stable; 4 too strained, large rings can't close) ✓. No arithmetic needed — the reasoning is the answer.

Recall check

Recall

When you're given stepwise K values, do you add or multiply for β? ::: Multiply the K's (or add their logs). Given log β, how do you get ΔG°? ::: ΔG° = −2.303 R T log β = −5708·log β J/mol at 298 K. A step with K_i = 1 changes log β by how much? ::: Zero — log 1 = 0, a do-nothing step. If ΔH° = 0 and ΔS° = 0, what is β? ::: β = 1 (ΔG° = 0, the balance point). Two ligands release equal waters but form 5- vs 7-membered rings — which wins? ::: The 5-membered ring (least strain, best bite probability). Why does EDTA give the biggest entropy jump in these examples? ::: One ligand releases six waters (2 free particles → 7), the largest net +ΔS.