MCAT content library / Chemistry & physics / Solubility, common ions, and complexes
Compare the ion product with Ksp, then ask whether common ions, protonation, or complex formation remove or add free ions and thereby shift dissolution.
About 17 minutes, including common-ion and complexation examples.
Useful foundations
On this page
- Ksp uses free dissolved-ion concentrations
- Concept: the common-ion effect suppresses dissolution
- Worked application: a hypothetical 1:1 salt
- Pattern: protonation can increase solubility of basic anions
- Rule: complex formation can pull a metal ion out of the free-ion pool
- Check your reasoning
- Continue learning
- Sources
Ksp uses free dissolved-ion concentrations
For the dissolution
MX(s) ⇌ M⁺(aq) + X⁻(aq),
the solubility-product expression is
Ksp = [M⁺][X⁻].
The pure solid is omitted from the equilibrium expression. The ion concentrations refer to free dissolved species in the model, not the total analytical amount if substantial complex formation or protonation changes their chemical form.
For a current mixture, calculate the ion product Qsp. If Qsp < Ksp, additional solid can dissolve. If Qsp > Ksp, precipitation is thermodynamically favored until the free-ion product returns to Ksp.
Concept: the common-ion effect suppresses dissolution
Adding one of the ions already produced by dissolution raises the corresponding term in Qsp. Equilibrium responds by reducing how much additional solid dissolves.
This does not mean Ksp changes. At fixed temperature, the equilibrium constant remains the same; the equilibrium concentrations change to satisfy it.
Worked application: a hypothetical 1:1 salt
An original salt MX has Ksp = 1.0 × 10−8 at the stated temperature.
In pure water: if molar solubility is s, then [M⁺] = [X⁻] = s, so:
s² = 1.0 × 10−8, giving s = 1.0 × 10−4 M.
In 0.010 M NaX: the solution already contains 1.0 × 10−2 M X⁻. If s is much smaller than 0.010 M,
Ksp ≈ s(1.0 × 10−2),
so s ≈ 1.0 × 10−6 M.
| Medium | Approximate molar solubility |
|---|---|
| Pure water | 1.0 × 10−4 M |
| 0.010 M X⁻ common ion | 1.0 × 10−6 M |
The common ion lowers the molar solubility by roughly a factor of 100 under the approximation, while Ksp remains 10−8.
Pattern: protonation can increase solubility of basic anions
If X⁻ is the conjugate base of a weak acid HX, lowering pH can protonate X⁻ to form HX. That removes free X⁻ from the Ksp expression. More MX can then dissolve to restore [M⁺][X⁻] to Ksp.
This is why salts containing basic anions such as carbonate can become more soluble under acidic conditions. The acid does not need to change Ksp; it couples dissolution to a second equilibrium that consumes one dissolved product.
Rule: complex formation can pull a metal ion out of the free-ion pool
A ligand L can bind a dissolved metal ion:
M⁺ + L ⇌ ML⁺.
If complex formation is favorable, it lowers free [M⁺] even though total dissolved metal rises. Lowering free M⁺ pulls the dissolution equilibrium toward more dissolved salt.
This can look paradoxical if total concentration is confused with free-ion concentration. A solution may contain a large amount of dissolved metal while free M⁺ remains small because most metal is bound in complexes.
Selective precipitation uses the same equilibrium logic in reverse: if one ion pair reaches its Ksp threshold before another as a precipitating reagent is added, the less soluble salt can precipitate preferentially within a controlled concentration window.
ACTIVE PRACTICE · CHOOSE, TEST, TRANSFER
Change a Common-Ion Concentration While Keeping Ksp Constant
The lesson’s hypothetical salt MX has Ksp = 1.0×10⁻⁸ at fixed temperature, with 1:1 dissociation. A background X⁻ ion reduces additional MX solubility; to avoid overextending the lesson’s approximation, this model solves s(c+s) = Ksp exactly.
New MX molar solubility: .
Free [M⁺] after equilibration:; free [X⁻]:.
Check [M⁺][X⁻] =; Ksp is 1.0e−8 at every setting.
In pure water, s=√Ksp=1.0×10⁻⁴ M.
Assumes an ideal solution with no acid–base consumption or metal–ligand complexation; free ion concentrations, not total dissolved metal, enter Ksp.
Check your reasoning
Choose an answer before opening its feedback.
Check 1
For MX(s) ⇌ M⁺ + X⁻, what happens to molar solubility when a large amount of soluble NaX is added at constant temperature?
A. It decreases.
Correct. The common X⁻ ion suppresses further dissolution.
B. It increases because Ksp doubles.
Review this choice. Ksp does not change merely because a common ion is added.
C. It becomes independent of ion concentration.
Review this choice. Dissolution must satisfy the ion-product relation.
D. The solid becomes a gas.
Review this choice. The process is dissolution/precipitation.
Check 2
A salt contains a basic anion that is protonated strongly at low pH. What usually happens to the salt’s solubility as pH falls?
A. It can increase.
Correct. Protonation removes free anion and pulls dissolution forward.
B. It must fall to zero.
Review this choice. Consuming a dissolution product tends to favor dissolution.
C. Ksp must become zero.
Review this choice. The coupled acid–base equilibrium changes concentrations, not the fixed-temperature Ksp.
D. Proton concentration can never affect solubility.
Review this choice. Protonation of dissolved ions can couple directly to solubility.
Check 3
A ligand binds M⁺ strongly while MX(s) is present. What can happen?
A. Total dissolved metal can rise while free M⁺ stays relatively low.
Correct. Complexation removes free M⁺ and can pull more solid into solution.
B. Complex formation must precipitate all metal immediately.
Review this choice. Favorable soluble complex formation can increase apparent solubility.
C. Total and free metal concentration are always identical.
Review this choice. Bound metal contributes to total but not free M⁺.
D. The ligand changes the atomic number of M.
Review this choice. Coordination does not change nuclear identity.
Continue learning
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- Weak acids, bases, and salt pH
- Titration curves and equivalence
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