Last updated: 19 August 2026
Calling a salt “insoluble” can be misleading. In most cases, a small amount still dissolves. If undissolved solid remains, the dissolved ions eventually reach concentrations at which dissolution and precipitation occur at equal rates.
This produces a solubility equilibrium. Instead of simply saying that one salt is soluble and another is insoluble, we can use the solubility product, Ksp, to describe the equilibrium quantitatively and calculate how much solid can actually dissolve.
The same equilibrium reasoning also explains two apparently opposite observations. Adding an ion already present in the equilibrium can make a salt less soluble, while removing a dissolved ion by forming a stable complex can make the salt more soluble. The key is to follow what happens to the concentrations of the free ions.
Try these before reading the chapter. They check the equilibrium and stoichiometric ideas that Ksp calculations build on.
- Check your equilibrium foundations.
- Spot common stoichiometric mistakes early.
- Get an explanation for every option.
Solubility equilibria and Ksp
Consider silver chloride, AgCl. When AgCl is placed in water, a small amount dissolves to form Ag⁺ and Cl⁻ ions:
If undissolved AgCl remains, the system eventually reaches dynamic equilibrium. AgCl continues to dissolve, and Ag⁺ and Cl⁻ continue to recombine to form solid AgCl, but the two processes occur at equal rates.
The concentrations of Ag⁺ and Cl⁻ therefore remain constant at equilibrium even though particles continue to move between the solid and aqueous phases.
| Term | Meaning | What it does not mean |
|---|---|---|
| Sparingly soluble | Only a small amount of the solid dissolves | That absolutely none of the solid dissolves |
| Saturated solution | A solution in equilibrium with undissolved solute at that temperature | That dissolution has stopped |
| Dynamic equilibrium | Dissolution and precipitation occur at equal rates | That no microscopic change is occurring |
What is the solubility product?
The solubility product, Ksp, is the equilibrium constant for the dissolution of a sparingly soluble ionic solid in a saturated solution at a specified temperature.
For AgCl:
The square brackets represent equilibrium concentrations in mol dm⁻³.
Adding another ion, removing an ion or forming a complex can change the equilibrium concentrations and the amount of solid that dissolves. These changes do not alter Ksp unless the temperature changes.
Why is the solid omitted?
A pure solid has effectively constant concentration. Its contribution is therefore incorporated into the numerical value of the equilibrium constant.
For the equilibrium
we therefore write:
and not an expression containing [AgCl].
The amount of solid can determine whether solid remains present, but its concentration does not appear explicitly in the Ksp expression.
Writing Ksp expressions
The safest approach is to write the balanced dissolution equation before writing Ksp. The stoichiometric coefficient of each dissolved ion becomes the power of its concentration in the equilibrium expression.
For a general salt MaXb:
the general form is:
| Dissolution equilibrium | Ksp expression |
|---|---|
| AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq) | Ksp = [Ag⁺][Cl⁻] |
| CaF₂(s) ⇌ Ca²⁺(aq) + 2F⁻(aq) | Ksp = [Ca²⁺][F⁻]² |
| Mg(OH)₂(s) ⇌ Mg²⁺(aq) + 2OH⁻(aq) | Ksp = [Mg²⁺][OH⁻]² |
| Al(OH)₃(s) ⇌ Al³⁺(aq) + 3OH⁻(aq) | Ksp = [Al³⁺][OH⁻]³ |
Coefficients affect the calculation twice
Consider Mg(OH)₂:
The coefficient 2 has two related consequences.
- If s mol dm⁻³ of Mg(OH)₂ dissolves, then [OH⁻] = 2s.
- In the Ksp expression, [OH⁻] is squared.
Therefore:
Incorrect: Ksp = s(2s) = 2s².
Correct: [OH⁻] is 2s, and that entire concentration is squared: Ksp = s(2s)² = 4s³.
Can Ksp values always be compared directly?
Care is needed. If two salts dissolve with the same stoichiometry, their Ksp values can be related fairly directly to their molar solubilities.
For two 1:1 salts:
But a 1:2 salt such as MX₂ gives a different relationship:
Therefore, when salts have different dissolution stoichiometries, do not simply assume that the numerically larger Ksp must correspond to the larger molar solubility. Calculate s before comparing them.
Calculating Ksp and molar solubility
The molar solubility, s, is the amount in moles of a substance that dissolves per dm³ of solution to form a saturated solution.
The central calculation skill is not substitution into a formula. It is translating one value of s into the equilibrium concentration of every ion using the balanced dissolution equation.
Example 1: finding Ksp from molar solubility
The molar solubility of Mg(OH)₂ is 2.50 × 10⁻⁴ mol dm⁻³ at a particular temperature. Calculate Ksp.
First write the dissolution equation:
If s = 2.50 × 10⁻⁴ mol dm⁻³:
Now write the equilibrium expression:
Substituting:
Answer: Ksp = 6.25 × 10⁻¹¹ mol³ dm⁻⁹.
Example 2: finding molar solubility from Ksp
The Ksp of AgCl is 1.8 × 10⁻¹⁰ mol² dm⁻⁶ at a stated temperature. Calculate its molar solubility in pure water.
If s mol dm⁻³ of AgCl dissolves:
Hence:
Answer: molar solubility = 1.34 × 10⁻⁵ mol dm⁻³.
Useful stoichiometric patterns
| Salt type | Ion concentrations | Ksp in terms of s |
|---|---|---|
| MX | [M] = s, [X] = s | Ksp = s² |
| MX₂ | [M] = s, [X] = 2s | Ksp = 4s³ |
| M₂X | [M] = 2s, [X] = s | Ksp = 4s³ |
| M₂X₃ | [M] = 2s, [X] = 3s | Ksp = 108s⁵ |
That shortcut works only for a 1:1 salt dissolving in pure water. For every new formula, derive the relationship between Ksp and s from the balanced dissolution equation.
The common ion effect
The common ion effect is the decrease in the solubility of a sparingly soluble ionic compound when another substance supplies an ion already present in the dissolution equilibrium.
Consider AgCl:
Suppose NaCl is added. Since NaCl is soluble, it dissociates and increases [Cl⁻]. The chloride ion is a common ion because Cl⁻ is already present in the AgCl equilibrium.
According to Le Chatelier’s Principle, increasing the concentration of a product favours the reverse reaction:
More AgCl therefore forms and less AgCl can remain dissolved.
- NaCl is added.
- [Cl⁻] increases.
- The AgCl equilibrium shifts to the left.
- More AgCl precipitates.
- The molar solubility decreases.
- Ksp remains unchanged at constant temperature.
After the disturbance, the concentrations adjust until the product [Ag⁺][Cl⁻] once again equals the same Ksp.
Calculating solubility in the presence of a common ion
Suppose Ksp for AgCl is 1.8 × 10⁻¹⁰ mol² dm⁻⁶. Calculate its approximate molar solubility in 0.100 mol dm⁻³ NaCl.
Let the molar solubility of AgCl be s:
The NaCl already supplies approximately 0.100 mol dm⁻³ Cl⁻. Dissolving AgCl adds another s mol dm⁻³:
If s is much smaller than 0.100, then:
Therefore:
Answer: molar solubility ≈ 1.8 × 10⁻⁹ mol dm⁻³.
Compare this with the value in pure water:
The chloride common ion has therefore reduced the molar solubility of AgCl by several orders of magnitude.
Why can we sometimes ignore the extra s?
Strictly,
rather than exactly 0.100 mol dm⁻³. However, our calculated value of s is only 1.8 × 10⁻⁹ mol dm⁻³, which is tiny compared with 0.100 mol dm⁻³.
Adding 1.8 × 10⁻⁹ to 0.100 makes no meaningful difference at the precision of the calculation, so the approximation is justified.
Do not assume the approximation automatically.
Treat the common-ion concentration as approximately constant only when the amount contributed by dissolution is negligible in comparison.
Complex ion formation and solubility
The common ion effect makes a salt less soluble by increasing the concentration of one of its dissolved ions. Complex ion formation can have the opposite effect by removing a free ion from the dissolution equilibrium.
Silver chloride provides an important example.
When aqueous ammonia is added, NH₃ acts as a ligand and forms a complex ion with Ag⁺:
As Ag⁺ is converted into [Ag(NH₃)₂]⁺, the concentration of free Ag⁺ decreases.
The AgCl equilibrium responds by shifting to the right:
More solid AgCl therefore dissolves.
In Ksp = [Ag⁺][Cl⁻], [Ag⁺] means the concentration of free Ag⁺ ions. Silver contained in [Ag(NH₃)₂]⁺ is a different species and is not counted as free Ag⁺.
Following the two linked equilibria
It helps to separate the process into two equilibria.
First, AgCl dissolves:
Then NH₃ removes some of the free Ag⁺:
Because free [Ag⁺] falls, the first equilibrium moves to the right to replace it. As long as the complex is sufficiently stable and enough ligand is present, the overall result is that more silver salt dissolves.
Silver halides and ammonia
The behaviour of the silver halide precipitates with ammonia provides a familiar application of this principle.
| Silver halide | Behaviour with NH₃(aq) | Interpretation |
|---|---|---|
| AgCl | Dissolves in dilute NH₃ | Complex formation removes enough Ag⁺ to drive AgCl dissolution |
| AgBr | Dissolves in concentrated NH₃ | A greater NH₃ concentration is needed to drive sufficient dissolution |
| AgI | Does not dissolve appreciably even in concentrated NH₃ | The equilibrium remains strongly in favour of AgI(s) under these conditions |
Why does concentrated NH₃ dissolve AgBr but not AgI?
AgBr is less soluble than AgCl, so a greater reduction in free [Ag⁺] is required before substantial AgBr dissolves. Increasing the NH₃ concentration favours formation of [Ag(NH₃)₂]⁺ and removes more free Ag⁺, which is enough to dissolve AgBr.
AgI is much more sparingly soluble. Under the usual qualitative-analysis conditions, even concentrated NH₃ does not remove free Ag⁺ sufficiently strongly to cause appreciable AgI dissolution.
Connecting the common ion effect and complex formation
These two effects may initially seem unrelated, but they are mirror images of the same equilibrium idea.
| Change | Effect on a dissolved ion | Direction of dissolution equilibrium | Effect on solubility |
|---|---|---|---|
| Add a common ion | Free-ion concentration increases | Towards the solid | Usually decreases |
| Form a stable complex | Free-ion concentration decreases | Away from the solid | Can increase |
Common ion: increase a product concentration → favour the reverse dissolution reaction → less solid dissolves.
Complex formation: remove a product from its free form → favour further dissolution → more solid can dissolve.
In both cases, Ksp at that temperature remains unchanged. What changes is the set of equilibrium concentrations required to satisfy it.
Worked examples
Example 1: Ksp of CaF₂ from its molar solubility
A saturated solution of CaF₂ has molar solubility 1.50 × 10⁻⁴ mol dm⁻³. Calculate Ksp.
Therefore:
Now:
Answer: Ksp = 1.35 × 10⁻¹¹ mol³ dm⁻⁹.
Example 2: molar solubility of Mg(OH)₂
Mg(OH)₂ has Ksp = 3.20 × 10⁻¹¹ mol³ dm⁻⁹. Calculate its molar solubility in pure water.
Let the molar solubility be s.
Answer: molar solubility = 2.00 × 10⁻⁴ mol dm⁻³.
Example 3: explaining dissolution by complex formation
A student adds aqueous ammonia to a suspension of AgCl and observes that the precipitate dissolves. Explain the observation in terms of equilibria.
AgCl first establishes:
Ammonia then forms a complex with Ag⁺:
Formation of [Ag(NH₃)₂]⁺ lowers free [Ag⁺]. The AgCl dissolution equilibrium shifts to the right to replace the Ag⁺ that has been removed, causing more AgCl to dissolve.
The value of Ksp itself is unchanged because the temperature has not changed.
Common mistakes with solubility equilibria
| Common mistake | Why it is wrong | Correct approach |
|---|---|---|
| Including the pure solid in Ksp | The concentration of a pure solid is constant | Include only the dissolved ions |
| Ignoring stoichiometric powers | The balanced coefficients determine the powers in Ksp | Write the dissolution equation first |
| Using [OH⁻] = s for Mg(OH)₂ | Each mole of Mg(OH)₂ produces two moles of OH⁻ | Use [Mg²⁺] = s and [OH⁻] = 2s |
| Using s = √Ksp for every salt | That relationship applies only to a 1:1 salt in pure water | Derive the relationship from the balanced equation |
| Saying a common ion lowers Ksp | Ksp remains constant at fixed temperature | State that the common ion lowers the molar solubility |
| Saying NaCl causes AgCl equilibrium to shift right | NaCl adds Cl⁻, which is a product of AgCl dissolution | The equilibrium shifts left towards AgCl(s) |
| Using total dissolved silver as [Ag⁺] | Silver in [Ag(NH₃)₂]⁺ is not free Ag⁺ | Use the concentration of the free species appearing in Ksp |
| Saying complex formation increases Ksp | The complex changes free-ion concentrations rather than Ksp | Explain the increased dissolution using the fall in free-ion concentration |
| Comparing Ksp values without considering stoichiometry | Different salt formulae give different relationships between Ksp and s | Calculate molar solubility before comparing different stoichiometries |
Solubility-equilibrium checklist
Write Ksp
- Write the balanced dissolution equation.
- Include aqueous ions only.
- Omit the pure solid.
- Use coefficients as powers.
Calculate with molar solubility
- Let the molar solubility be s.
- Use stoichiometry to determine each ion concentration.
- Substitute the concentrations into Ksp.
- Do not assume s = √Ksp.
Apply the common ion effect
- Identify the common ion.
- Determine which free-ion concentration increases.
- Predict the equilibrium shift.
- State that solubility changes while Ksp stays constant.
Explain complex formation
- Identify the free ion removed by the ligand.
- Write the complex-ion equation where useful.
- Explain why the dissolution equilibrium shifts.
- Distinguish free-ion concentration from total dissolved metal.
These questions move beyond definitions. Use Ksp, stoichiometry, common-ion reasoning and complex formation to solve unfamiliar problems.
- Practise multi-step Ksp calculations.
- Test equilibrium reasoning rather than memorisation.
- Get feedback on every answer option.
MY₂(s) ⇌ M²⁺(aq) + 2Y⁻(aq)
Ag⁺(aq) + 2NH₃(aq) ⇌ [Ag(NH₃)₂]⁺(aq)
Frequently asked questions
Questions students often ask about solubility equilibria
These questions connect the calculations to the equilibrium ideas that often cause confusion in A-Level Chemistry.
Does “insoluble” mean that none of the substance dissolves?
No. A substance described as insoluble or sparingly soluble usually dissolves to a small extent. A saturated solution therefore contains some dissolved ions even while undissolved solid remains.
Is Ksp simply a special type of equilibrium constant?
Yes. Ksp is the equilibrium constant used for the dissolution of a sparingly soluble ionic solid. Like other equilibrium expressions, it is constructed from the equilibrium species, with the pure solid omitted.
Does adding more solid increase the concentration of ions in a saturated solution?
Not if the solution is already saturated and the temperature is unchanged. Additional solid does not change the equilibrium ion concentrations. It simply increases the amount of undissolved solid present.
Why is Ksp = 4s³ for Mg(OH)₂?
If s mol dm⁻³ of Mg(OH)₂ dissolves, then [Mg²⁺] = s and [OH⁻] = 2s. Since Ksp = [Mg²⁺][OH⁻]²:
Ksp = s(2s)² = 4s³.
Why does adding a common ion reduce solubility?
The common ion increases the concentration of one of the dissolved products of the dissolution equilibrium. The equilibrium therefore shifts towards the undissolved solid, so less of the salt remains dissolved.
Does a common ion always change Ksp?
No. At constant temperature, Ksp remains unchanged. The equilibrium concentrations change until the same Ksp value is satisfied again.
Can a salt become more soluble even though Ksp remains unchanged?
Yes. If another equilibrium removes one of the free dissolved ions, more solid can dissolve to replace that ion. Complex ion formation is an important example.
Why is total dissolved silver not the same as [Ag⁺]?
Silver can exist in several chemical forms. Free Ag⁺ is one species, while [Ag(NH₃)₂]⁺ is another. The Ksp expression for AgCl requires the concentration of free Ag⁺ specifically.
Why does AgCl dissolve in dilute NH₃ but AgBr needs concentrated NH₃?
AgBr is less soluble than AgCl, so a greater reduction in free [Ag⁺] is needed to drive substantial AgBr dissolution. Concentrated NH₃ favours greater formation of [Ag(NH₃)₂]⁺ and therefore removes more free Ag⁺.
Why does AgI not dissolve appreciably in concentrated NH₃?
AgI is extremely sparingly soluble. Under the usual qualitative-analysis conditions, formation of [Ag(NH₃)₂]⁺ does not reduce free [Ag⁺] sufficiently to drive appreciable dissolution of AgI.
Can I compare two salts simply by looking at which has the larger Ksp?
Only with care. Direct comparison is most meaningful when the salts have the same dissolution stoichiometry. If the ion ratios differ, calculate the molar solubility of each salt first and compare the values of s.
Solubility-product questions become much easier when every problem begins with the same question: what happens to the concentrations of the free ions? The balanced dissolution equation gives the Ksp expression and connects molar solubility to ion concentrations. A common ion pushes the equilibrium towards the solid, while complex formation can remove a free ion and pull the equilibrium towards further dissolution. Once these ideas are treated as ordinary equilibrium shifts rather than separate rules to memorise, the calculations and explanations follow the same underlying logic.