The shape of the lesson
By the end of the lesson, learners can
- Writing K stab: Watch the Kstab panel on the ligand exchange bench as you add ammonia and chloride.
- Predicting exchange: Compare log K stab values on the figure.
- Kstab = [complex] / ([Mn+][L]x), water left out.
- Units from the powers: 4 ligands give mol−4 dm12.
- Larger Kstab = more stable; exchange goes towards it.
- Chelate effect: more particles, ΔS positive.
One picture every idea returns to
The stability constant, Kstab, is the equilibrium constant for the formation of a complex ion in a solvent from its constituent ions or molecules. For Cu2+(aq) + 4NH3(aq) ⇌ [Cu(NH3)4(H2O)2]2+(aq), Kstab = [complex] / ([Cu2+][NH3]4), units mol−4 dm12. Water is left out because it is the solvent and its concentration is effectively constant. Values are huge, so they are often quoted as log Kstab. A ligand exchange happens if the new complex has a larger Kstab: the larger the value, the more stable the complex. Multidentate ligands such as EDTA4− form very stable complexes (the chelate effect) because one ligand releases several water molecules, so the number of particles and the entropy increase.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, The more stable complex wins.. Ligand exchange is an equilibrium, so it has an equilibrium constant. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 17 min | Teach and check: Writing K stab | Watch the Kstab panel on the ligand exchange bench as you add ammonia and chloride. Project tab 3 of the lab (Transition elements lab, ligand exchange bench) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: Predicting exchange | Compare log K stab values on the figure. Project tab 3 of the lab (Transition elements lab, ligand exchange bench) and let a student drive it while the class predicts. | Round Table |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 15 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Predicting exchange, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Round Table
One sheet travels round the team. Each student adds one line of working or one idea and passes it on, so the solution is built four hands at a time and every contribution is visible.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| Leave water out of K stab expressions. | Units depend on the number of ligands: four monodentate ligands give mol −4 dm 12, but one EDTA 4− gives mol −1 dm 3. A larger K stab means a more stable complex, and ligand exchange goes towards it. The chelate effect is mainly an entropy effect, not a stronger bond. |
Stability constants, Kstab: worksheet
6 questions, 15 marks. Written in the style of the papers; no past paper question is reproduced.
Define the stability constant of a complex, and write the Kstab expression and units for [Co(H2O)6]2+ + 4Cl− ⇌ [CoCl4]2− + 6H2O.
At equilibrium, [Cu(NH3)4(H2O)22+] = 0.0500 mol dm−3, [NH3] = 0.200 mol dm−3 and [Cu2+] = 2.60 × 10−12 mol dm−3. Calculate Kstab and log Kstab.
Kstab for [Ag(NH3)2]+ is 1.7 × 107 mol−2 dm6. In a solution at equilibrium [Ag(NH3)2+] = 0.0100 mol dm−3 and [NH3] = 0.100 mol dm−3. Calculate [Ag+].
Explain, in terms of entropy, why [Cu(EDTA)]2− is much more stable than [Cu(NH3)4(H2O)2]2+.
Using log Kstab values ([CuCl4]2− 5.6, [Cu(NH3)4(H2O)2]2+ 13.1, [Cu(EDTA)]2− 18.8), predict what is seen when EDTA4− is added to the deep blue ammine complex, and explain.
Calculate the equilibrium constant for [CuCl4]2− + 4NH3 ⇌ [Cu(NH3)4(H2O)2]2+ + 4Cl− using the log Kstab values 5.6 and 13.1.
Answer key and marking notes
- The equilibrium constant for the formation of a complex ion in a solvent from its constituent ions or molecules.
- Kstab = [CoCl42−] / ([Co(H2O)62+][Cl−]4) (water omitted).
- Units: mol−4 dm12.
- Kstab = 0.0500 ÷ (2.60 × 10−12 × 0.2004).
- 0.2004 = 1.60 × 10−3, so Kstab = 0.0500 ÷ 4.16 × 10−15 = 1.20 × 1013 mol−4 dm12.
- log Kstab = 13.1.
- [Ag+] = [complex] ÷ (Kstab × [NH3]2) = 0.0100 ÷ (1.7 × 107 × 0.0100).
- [Ag+] = 5.88 × 10−8 mol dm−3.
- [Cu(H2O)6]2+ + EDTA4− → [Cu(EDTA)]2− + 6H2O: 2 particles become 7.
- With NH3, 5 particles become 5, so there is little entropy change.
- ΔH is similar (similar bonds broken and made), so the large positive ΔS makes ΔG more negative and Kstab larger for EDTA.
- The deep blue colour fades to the paler blue of [Cu(EDTA)]2−.
- EDTA4− replaces the NH3 and H2O ligands because [Cu(EDTA)]2− has the larger Kstab (18.8 > 13.1).
- K = Kstab(ammine) ÷ Kstab(chloro), so log K = 13.1 − 5.6 = 7.5.
- K = 107.5 = 3.16 × 107 (no units, since the powers cancel); the exchange goes almost to completion.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
Which is the correct expression for the stability constant of [Cu(NH3)4(H2O)2]2+? [Cu(H2O)6]2+ + 4NH3 ⇌ [Cu(NH3)4(H2O)2]2+ + 4H2O
- AKstab = [Cu(NH3)4(H2O)22+] / ([Cu(H2O)62+][NH3]4)
- BKstab = [Cu(NH3)4(H2O)22+][H2O]4 / ([Cu(H2O)62+][NH3]4)
- CKstab = [Cu(NH3)4(H2O)22+] / ([Cu(H2O)62+][NH3])
- DKstab = [Cu(H2O)62+][NH3]4 / [Cu(NH3)4(H2O)22+]
What are the units of Kstab for the formation of [Cu(NH3)4(H2O)2]2+ from [Cu(H2O)6]2+ and four NH3 ligands?
- Amol−4 dm12
- Bmol4 dm−12
- Cmol−1 dm3
- Dno units
The chart shows log Kstab for three copper(II) complexes. Which ligand-exchange reaction goes almost to completion?
- A[Cu(NH3)4(H2O)2]2+ + EDTA4− → [Cu(EDTA)]2− + 4NH3 + 2H2O
- B[Cu(EDTA)]2− + 4NH3 + 2H2O → [Cu(NH3)4(H2O)2]2+ + EDTA4−
- C[Cu(NH3)4(H2O)2]2+ + 4Cl− → [CuCl4]2− + 4NH3 + 2H2O
- D[Cu(EDTA)]2− + 4Cl− → [CuCl4]2− + EDTA4−
What does a large value of Kstab tell you about a complex ion?
- Athe equilibrium for its formation lies far to the right, so the complex is stable and does not easily dissociate
- Bthe complex forms very quickly
- Cthe complex is strongly coloured
- Dthe metal ion has a high oxidation state
Why is the concentration of water left out of a Kstab expression for a complex formed in aqueous solution?
- Awater is the solvent and is in such large excess that its concentration is effectively constant
- Bwater is a weak ligand, so it does not take part in the reaction
- Cwater is a liquid, and liquids never appear in any equilibrium expression
- Dthe water molecules released cancel out with those in the aqua ion
At equilibrium, a solution contains [Cu(NH3)4(H2O)2]2+ = 0.100 mol dm−3, [NH3] = 0.200 mol dm−3 and [Cu(H2O)6]2+ = 5.00 × 10−12 mol dm−3. What is Kstab?
- A1.25 × 1013 mol−4 dm12
- B1.00 × 1011 mol−4 dm12
- C8.00 × 10−14 mol−4 dm12
- D2.50 × 1010 mol−4 dm12
Why is [Cu(H2O)6]2+ + EDTA4− → [Cu(EDTA)]2− + 6H2O so favourable? (This is the chelate effect.)
- Atwo particles become seven, so the entropy change of the system is large and positive, making ΔG strongly negative
- Bthe reaction is very exothermic because EDTA forms six ionic bonds to copper
- Cthe number of particles decreases, so entropy decreases and the reaction is driven by enthalpy
- DEDTA oxidises copper(II) to copper(III), which forms a more stable complex
A metal ion M2+(aq) binds four ligands L in four steps, with stepwise values log K1 = 4.20, log K2 = 3.50, log K3 = 2.90 and log K4 = 2.10. What is the overall stability constant Kstab for forming ML42+?
- A5.01 × 1012 mol−4 dm12
- B12.7 mol−4 dm12
- C89.5 mol−4 dm12
- D1.58 × 104 mol−4 dm12
Quiz key
Kstab is the equilibrium constant for forming the complex from the aqueous metal ion and the ligands. Water is the solvent: it is in huge excess and its concentration is effectively constant, so it is left out. Including [H2O]4 is the classic error. Leaving [NH3] without the power 4 ignores the coefficient. The last option is upside down: it is the expression for the dissociation of the complex, 1/Kstab.
Units = (mol dm−3) / (mol dm−3)5 = (mol dm−3)−4 = mol−4 dm12. mol4 dm−12 belongs to the inverted (dissociation) expression. mol−1 dm3 would fit a single ligand binding (stepwise K1), not four. The concentration terms do not cancel, so Kstab here does have units.
Ligand exchange goes towards the complex with the larger stability constant. log Kstab for [Cu(EDTA)]2− (18.8) is much larger than for [Cu(NH3)4(H2O)2]2+ (13.1), so EDTA displaces ammonia almost completely. Ammonia cannot displace EDTA, because that would form a less stable complex. Chloride (log Kstab 5.6) forms the least stable complex, so it cannot displace either ammonia or EDTA at ordinary concentrations.
Kstab is the equilibrium constant for forming the complex from the aqueous metal ion and its ligands. A large value means the position of equilibrium lies well to the right: at equilibrium there is far more complex than free aqua ion. Equilibrium constants say nothing about rate, so speed of formation is not implied. Colour depends on ΔE and the d electrons, not on Kstab. A large Kstab can occur for any oxidation state; [Cu(EDTA)]2− has copper at +2.
Pure water is about 55.5 mol dm−3. The few molecules released by ligand exchange change this by a negligible amount, so it is treated as constant and absorbed into Kstab. Water does take part: it is the ligand being replaced. Liquids can appear in some Kc expressions (for example an esterification where water is a product, not the solvent); the reason here is that water is the solvent. The waters in the aqua ion are part of the complex formula, and nothing 'cancels' between free water and bound water.
Kstab = 0.100 / (5.00 × 10−12 × 0.2004) = 0.100 / (5.00 × 10−12 × 1.60 × 10−3) = 1.25 × 1013 mol−4 dm12, so log Kstab = 13.1. 1.00 × 1011 leaves out the power of 4. 8.00 × 10−14 inverts the expression. 2.50 × 1010 multiplies [NH3] by 4 instead of raising it to the fourth power.
One hexadentate EDTA replaces six water ligands: 2 particles on the left, 7 on the right. That is a big increase in disorder, so ΔS is large and positive and ΔG = ΔH − TΔS is strongly negative even though ΔH is small (similar dative N-Cu and O-Cu bonds are broken and made). This is the chelate effect, and it explains why Kstab for [Cu(EDTA)]2− is so large. The bonds to EDTA are dative covalent, not ionic, and the enthalpy change is not large. The number of particles increases, not decreases. Copper stays at +2: ligand exchange is not redox.
Overall Kstab = K1 × K2 × K3 × K4, so log Kstab = 4.20 + 3.50 + 2.90 + 2.10 = 12.7 and Kstab = 1012.7 = 5.01 × 1012 mol−4 dm12. 12.7 is log Kstab, not Kstab itself. 89.5 multiplies the log values together, which is not how logs combine. 1.58 × 104 is K1 alone. Each later step is smaller because there are fewer water ligands left to replace.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.