A dissolved metal can be present as a free ion or bound inside a complex. A solubility product constrains the relevant free-ion activities; an analytical measurement may report the total metal across several dissolved forms. Confusing those quantities can make a perfectly balanced-looking calculation describe the wrong chemistry.
This guide develops a small coupled-equilibrium model and then removes its most convenient assumption: that free ligand concentration stays fixed. The worked examples use fictional constants chosen for transparent arithmetic. They extend the qualitative common-ion and complex-ion explanation into a quantitative test of free ions, total solubility and finite ligand supply.
eduKateSG · The Gold Standard of Chemistry
Choose your reading route
Build species and mass balances · Chapters 1–2. Build the quantities before using a formula.
Compare maintained and finite ligand · Chapters 3–5. Follow original examples, assumptions and changed conditions.
Audit the inventory and transfer · Chapters 6–8. Test transfer, answer common questions and follow connected reading.
Full chapter index · Practice and evidence checks · Return to Chemistry
Chapter index
Build species and mass balances · Chapters 1–2
Compare maintained and finite ligand · Chapters 3–5
Audit the inventory and transfer · Chapters 6–8
CHAPTER 1 OF 8 · Build species and mass balances
1. Give every dissolved form its own place
Consider a hypothetical solid MX that dissolves to M⁺ and X⁻. A neutral ligand L binds M⁺ to make ML⁺. We will track three equilibrating dissolved species associated with the metal and ligand: free M⁺, free L and the complex ML⁺. X⁻ remains uncomplexed in this teaching model.
The metal does not disappear when it binds L. It changes dissolved form. An instrument that measures total dissolved metal could count both M⁺ and ML⁺, whereas a measurement sensitive primarily to free M⁺ would answer a different question.
Use square brackets for molar concentration in this article. At equilibrium with solid present, the dilute-solution approximation gives Ksp = [M⁺][X⁻]. Complex formation is described by β = [ML⁺]/([M⁺][L]). These concentration forms approximate the activity-based thermodynamic expressions under the stated conditions.
Our model excludes hydrolysis, protonation of L, additional complexes, ion pairing with X⁻ and other precipitates. Those exclusions are part of the model, not claims about every real metal solution.
The Chemistry root connects species, reactions and evidence. The existing Ksp and common-ion guide provides the introductory dissolution equations. Here we follow the mass balances after complex formation becomes quantitatively important.
CHAPTER 2 OF 8 · Build species and mass balances
2. Connect free metal with total dissolved metal
Let s be the total dissolved metal concentration originating from MX. In this model, s = [M⁺] + [ML⁺]. From the formation expression, [ML⁺] = β[M⁺][L]. Substitute this into the metal balance:
s = [M⁺](1 + β[L]).
Define α = 1 + β[L]. Then [M⁺] = s/α and the free-metal fraction is 1/α. If α = 100, one hundredth of the dissolved metal is free M⁺ and the rest is complexed.
If MX is the only source of X⁻ and no X⁻ is removed by another process, [X⁻] = s. The solubility condition becomes Ksp = s²/α, giving s = √(αKsp).
This result has a precise scope. It uses a 1:1 solid, one 1:1 complex, no initial common ion and a known equilibrium free-ligand concentration. Changing any of those features changes the balances or the algebra.
The conditional expression s² = αKsp does not mean the thermodynamic solubility product has changed. It packages complex formation into a relation involving total dissolved metal. Its apparent constant depends on free ligand and on the other conditions that determine that free ligand.
OpenStax’s Multiple Equilibria chapter explains how complex formation couples with dissolution. The equations above apply that general framework to our deliberately simpler fictional system.
CHAPTER 3 OF 8 · Compare maintained and finite ligand
3. Work a case with maintained free ligand
Use Ksp = 1.00 × 10⁻⁸ and β = 1.00 × 10⁶ L mol⁻¹ in the concentration approximation. Suppose the equilibrium free ligand is maintained at 9.90 × 10⁻⁵ mol L⁻¹ by a sufficiently large supply. We are specifying free ligand, not merely the amount originally poured in.
Then α = 1 + (1.00 × 10⁶)(9.90 × 10⁻⁵) = 100. The total solubility is √(100 × 1.00 × 10⁻⁸) = 1.00 × 10⁻³ mol L⁻¹.
| Quantity | Without ligand | With maintained free ligand |
|---|---|---|
| Total dissolved metal, s | 1.00 × 10⁻⁴ M | 1.00 × 10⁻³ M |
| Free metal, [M⁺] | 1.00 × 10⁻⁴ M | 1.00 × 10⁻⁵ M |
| Complex, [ML⁺] | 0 | 9.90 × 10⁻⁴ M |
| Free X⁻ | 1.00 × 10⁻⁴ M | 1.00 × 10⁻³ M |
Check the solubility product: (1.00 × 10⁻⁵)(1.00 × 10⁻³) = 1.00 × 10⁻⁸. Check formation: (9.90 × 10⁻⁴)/[(1.00 × 10⁻⁵)(9.90 × 10⁻⁵)] = 1.00 × 10⁶ L mol⁻¹.
The ligand case contains ten times more dissolved metal in total but one tenth as much free M⁺. Those statements are consistent because the complex stores most of the dissolved metal.
A final balance check exposes the cost of maintaining that free ligand. The dissolved phase needs 9.90 × 10⁻⁴ M bound ligand in addition to 9.90 × 10⁻⁵ M free ligand. An initial supply of only 9.90 × 10⁻⁵ M total ligand could not produce this state. A convenient free-ligand assumption must be supported by an adequate inventory.
CHAPTER 4 OF 8 · Compare maintained and finite ligand
4. Remove the fixed-ligand assumption
Now place the same hypothetical solid in a solution containing a finite total ligand concentration Cᴸ = 1.53 × 10⁻⁴ M. There is no external ligand reservoir and the only ligand-containing species are L and ML⁺.
The ligand balance is Cᴸ = [L] + [ML⁺]. Because binding consumes free ligand, substituting Cᴸ directly for [L] in α is generally unjustified.
One equilibrium state is s = 2.00 × 10⁻⁴ M. Verify it rather than accepting it on trust. The solubility product requires [M⁺] = Ksp/s = 5.00 × 10⁻⁵ M. The metal balance gives [ML⁺] = s − [M⁺] = 1.50 × 10⁻⁴ M. The ligand balance then gives [L] = 3.00 × 10⁻⁶ M.
Insert those values into the formation expression: β[M⁺][L] = (1.00 × 10⁶)(5.00 × 10⁻⁵)(3.00 × 10⁻⁶) = 1.50 × 10⁻⁴ M, exactly the calculated complex concentration.
If we had treated all added ligand as free, α would have been 154 and s would have been approximately 1.24 × 10⁻³ M. That shortcut predicts a complex concentration far larger than the available ligand. It fails the inventory before it deserves a detailed equilibrium interpretation.
A useful solution strategy is to express everything in terms of s: [M⁺] = Ksp/s, [ML⁺] = s − Ksp/s and [L] = Cᴸ − s + Ksp/s. Substitute these into β = [ML⁺]/([M⁺][L]) and solve for a physically admissible positive value. Require every species concentration to be non-negative and check all balances afterwards.
CHAPTER 5 OF 8 · Compare maintained and finite ligand
5. Add a common ion and keep the species separate
If a compatible soluble salt supplies X⁻ at an initial concentration c, the X balance becomes [X⁻] = c + s. With a known maintained free ligand, the equation is s(c + s) = αKsp.
Use our earlier maintained-ligand α = 100 and let c = 0.0100 M. The equation is s(0.0100 + s) = 1.00 × 10⁻⁶. Solving the quadratic gives s ≈ 9.90 × 10⁻⁵ M.
| Conditions | Total solubility |
|---|---|
| No ligand, no added X⁻ | 1.00 × 10⁻⁴ M |
| Maintained ligand, no added X⁻ | 1.00 × 10⁻³ M |
| Maintained ligand and 0.0100 M initial X⁻ | About 9.90 × 10⁻⁵ M |
Here the ligand still binds metal, but the added X⁻ suppresses the amount that can dissolve. Comparing only the phrase “ligand added” misses the second constraint.
The common-ion approximation c + s ≈ c gives s ≈ αKsp/c = 1.00 × 10⁻⁴ M. Since the exact s is about one percent of c, that approximation is reasonably close for this teaching calculation. If s were comparable with c, keep the full equation.
In real solutions, pH may protonate the ligand, ionic strength may change activities, or additional complexes may form. A larger total ligand concentration is not necessarily a proportionally larger free ligand concentration. Record temperature and chemical conditions before transporting a conditional expression to another solution.
CHAPTER 6 OF 8 · Audit the inventory and transfer
6. Practise with a species and inventory audit
Case A: a total-metal result
An analysis reports 0.80 millimolar dissolved metal. A separate speciation model estimates a free-metal fraction of 0.025. What free-metal concentration is implied?
Multiply: 0.80 × 0.025 = 0.020 millimolar, or 2.0 × 10⁻⁵ M. This is a model-dependent conversion. It does not prove the fraction if that fraction was itself assumed without validation.
Case B: an impossible ligand balance
A proposed solution contains 0.30 millimolar ML⁺, but the initial total ligand was only 0.20 millimolar and no other ligand source exists. Is the result plausible?
No. A 1:1 complex needs one ligand per complex. The proposed bound amount exceeds the complete ligand inventory. Recheck the stoichiometry, dilution, units and solution of the equations before interpreting the chemical mechanism.
Case C: all solid has dissolved
A calculation predicts a saturated solubility of 1.00 millimolar, but only enough MX was added to produce 0.10 millimolar if it dissolves completely. Must the solution reach Ksp?
No. The finite solid inventory may be exhausted before saturation. Then use the amount added as a mass-balance constraint and test whether the resulting free-ion product is below Ksp. The equality Ksp = [M⁺][X⁻] assumes equilibrium with the appropriate solid present.
For an evidence check, write the solid formula, all relevant dissolved species, mass balances, charge balance where needed, equilibrium expressions, activity approximation and finite inventories. State which concentration an instrument actually reports. These steps make a result reproducible without turning the exercise into a recipe for unsupervised chemical handling.
CHAPTER 7 OF 8 · Audit the inventory and transfer
7. Questions about free ions and dissolved totals
Does a larger total-metal concentration mean more free metal?
Not necessarily. Complex formation can increase dissolved total while decreasing the free fraction. The worked case shows both quantities moving in opposite directions.
Is Ksp the same as molar solubility?
No. Their relationship depends on the dissolution stoichiometry, other sources of ions, complexation and activity assumptions. Derive it from the balances for the actual system.
Can I put total ligand into the formation expression?
Only when it is an adequately justified approximation to free ligand or the expression has explicitly been redefined for the intended conditions. Check ligand depletion and other binding or protonation reactions.
Do these fictional constants describe silver and ammonia?
No. They are chosen for an original teaching model. Real systems need verified constants, temperature, activities and the appropriate set of complexes. Familiar silver-ammonia chemistry illustrates the principle but has different stoichiometry from our 1:1 complex.
Why does a calculator sometimes return several roots?
Algebra can produce roots outside the physical domain. Keep only solutions compatible with non-negative concentrations, the inventories and all equilibrium conditions; then verify by substitution.
CHAPTER 8 OF 8 · Audit the inventory and transfer
8. Follow the chemical species into wider reading
Connect dissolution and coordination
- The Gold Standard of Chemistry supplies the wider particle, reaction and measurement framework.
- Ksp and the Common-Ion Effect develops introductory dissolution and precipitation calculations.
- Transition Metals and Complex Ions connects coordination chemistry with the learner’s next subject questions.
Check what an analysis establishes
- Spike Timing, Extraction Recovery and Matrix Response distinguishes chemical recovery from instrumental response.
- Recovery Corrections and Uncertainty in a Reported Concentration helps interpret a measured total without treating a correction as certainty.
Consult the teaching sources
- OpenStax: Lewis Acids and Bases explains complex formation and formation constants.
- OpenStax: Multiple Equilibria connects complex formation, dissolution and coupled equilibrium.
The transferable habit is to ask what form a quantity is in before combining or comparing it. Continue through Learning That Transfers into Life and How X Works for other systems where a total conceals several physically different parts.