Conserve a ligand before claiming quantitative complexation
A large formation constant cannot supply ligand that is not present.
Learning outcome: Vary total ligand through the metal amount and vary beta. Recover both free inventories, compare the excess-ligand approximation and reject any complex amount above either supply.
Starting recap
Refresh normalized free-ion activities, charge-balanced stoichiometry, finite analytical inventories, coupled constants and independent versus shared assay errors.
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Try
For Mtotal=Ltotal=.001 M and concentration beta10⁴ M⁻¹, solve x=beta(.001−x)² to obtain ML≈.000729844 M, not all .001 M.
Open the experimentPredict
Can treating total ligand as free ligand overstate the bound-metal fraction near equal supplies?
Experiment
Vary total ligand through the metal amount and vary beta. Recover both free inventories, compare the excess-ligand approximation and reject any complex amount above either supply.
Conserve both metal and a finite ligand supply
Worked teaching model · synthetic values · Fictional M++L⇌ML+, neutral monodentate ligand, total M=.001 M and total L selected. Concentration-form beta=[ML]/([M][L]) has units M⁻¹; thermodynamic normalized formation constant is beta*c°. Ideal solution, fixed temperature, no other complexes, protonation or solids.
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- Exact bound fraction versus total ligand: solid
Axes rescale. A solubility comparison, depletion constraint and conditional reservoir scan have different horizontal meanings. Read the axes; these are not measured time traces.
All calculated plot coordinates
| total ligand (mmol/L) | metal bound fraction |
|---|---|
| 0.5 | 0.4258343 |
| 0.5416667 | 0.4573767 |
| 0.5833333 | 0.4880151 |
| 0.625 | 0.5176721 |
| 0.6666667 | 0.5462709 |
| 0.7083333 | 0.5737366 |
| 0.75 | 0.6 |
| 0.7916667 | 0.625 |
| 0.8333333 | 0.6486869 |
| 0.875 | 0.6710253 |
| 0.9166667 | 0.6919958 |
| 0.9583333 | 0.7115967 |
| 1 | 0.7298438 |
| 1.041667 | 0.7467695 |
| 1.083333 | 0.7624209 |
| 1.125 | 0.776857 |
| 1.166667 | 0.7901457 |
| 1.208333 | 0.8023608 |
| 1.25 | 0.8135792 |
| 1.291667 | 0.8238781 |
| 1.333333 | 0.8333333 |
| 1.375 | 0.8420177 |
| 1.416667 | 0.85 |
| 1.458333 | 0.8573444 |
| 1.5 | 0.8641101 |
| 1.541667 | 0.8703513 |
| 1.583333 | 0.8761174 |
| 1.625 | 0.881453 |
| 1.666667 | 0.8863983 |
| 1.708333 | 0.8909896 |
| 1.75 | 0.8952595 |
| 1.791667 | 0.8992372 |
| 1.833333 | 0.9029488 |
| 1.875 | 0.906418 |
| 1.916667 | 0.9096658 |
| 1.958333 | 0.9127112 |
| 2 | 0.9155711 |
| 2.041667 | 0.918261 |
| 2.083333 | 0.9207945 |
| 2.125 | 0.9231842 |
| 2.166667 | 0.9254413 |
| 2.208333 | 0.927576 |
| 2.25 | 0.9295974 |
| 2.291667 | 0.931514 |
| 2.333333 | 0.9333333 |
| 2.375 | 0.9350623 |
| 2.416667 | 0.9367072 |
| 2.458333 | 0.9382739 |
| 2.5 | 0.9397675 |
| 2.541667 | 0.9411929 |
| 2.583333 | 0.9425545 |
| 2.625 | 0.9438563 |
| 2.666667 | 0.9451021 |
| 2.708333 | 0.9462954 |
| 2.75 | 0.9474393 |
| 2.791667 | 0.9485366 |
| 2.833333 | 0.9495902 |
| 2.875 | 0.9506026 |
| 2.916667 | 0.951576 |
| 2.958333 | 0.9525126 |
| 3 | 0.9534144 |
| Quantity | Value | Unit |
|---|---|---|
| Complex ML concentration | 0.0007298438 | mol/L |
| Unbound metal concentration | 0.0002701562 | mol/L |
| Unbound ligand concentration | 0.0002701562 | mol/L |
| Metal-bound fraction | 0.7298438 | fraction |
| Recovered concentration formation constant | 10000 | L/mol |
| Excess-ligand approximation to bound metal | 0.0009090909 | mol/L |
A strong formation constant does not eliminate a finite ligand shortage.
Solve x=beta(Mtotal−x)(Ltotal−x) within both inventories. Replacing free ligand by total ligand assumes ligand depletion is negligible; it may overpredict bound metal near equal supplies. The ligand donates an electron pair and the metal accepts it; a coordinate bond origin is not a separate weaker bond class once formed. Other stoichiometries and competing ligands require extra equations.
These calculated packets are not performed chemical measurements or separation procedures. Keep feasible inventories, activity conventions, fitting observations and independent checks distinct.
A strong formation constant does not eliminate a finite ligand shortage.
Solve x=beta(Mtotal−x)(Ltotal−x) within both inventories. Replacing free ligand by total ligand assumes ligand depletion is negligible; it may overpredict bound metal near equal supplies. The ligand donates an electron pair and the metal accepts it; a coordinate bond origin is not a separate weaker bond class once formed. Other stoichiometries and competing ligands require extra equations.
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Which additional equations would a second ligand or a protonated ligand require?
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