Worked example 1
The molar solubility of CaF₂ is 2.1 × 10⁻⁴ M. Calculate Ksp.
Try it first: Write out the dissociation and express each ion in terms of s.
0 of 4 steps revealed.
What you'll be able to do: Convert between Ksp and molar solubility and explain quantitatively why a common ion reduces solubility.
Insoluble is a useful lie. Even chalk and limestone dissolve a little, and the equilibrium constant for that tiny amount explains kidney stones, hard water and cave formation.
These are recommended, not required. You can start this lesson at any time.
In a saturated solution, ions leave the crystal surface at the same rate that they rejoin it. Writing that as an equation gives, for example, AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq), and the solid drops out of the expression.
Ksp = [Ag⁺][Cl⁻]
Molar solubility s is the number of moles of the salt that dissolve per litre of saturated solution.
Each formula unit that dissolves releases ions in the ratio of the formula, so for CaF₂ dissolving to a level s the concentrations are [Ca²⁺] = s and [F⁻] = 2s.
Substituting the ion concentrations gives a fixed relationship for each salt type. For AB salts Ksp = s², for AB2 or A2B salts Ksp = 4s³, and for AB3 salts Ksp = 27s⁴. Going backwards means taking the corresponding root.
The pure-water shortcuts assume no initial common ions, no significant hydrolysis or complex formation, and negligible activity-coefficient effects:
| Dissolution | Ion concentrations from salt | Ksp in terms of s |
|---|---|---|
| AB → A + B | s, s | s² |
| AB₂ → A + 2 B | s, 2s | 4s³ |
| A₂B → 2 A + B | 2s, s | 4s³ |
| A₃B₂ → 3 A + 2 B | 3s, 2s | 108s⁵ |
Coefficients create both concentration multipliers and exponents. For CaF₂, Ksp = [Ca²⁺][F⁻]² = s(2s)², not 2s². Solubility in grams per liter additionally requires multiplication by the salt's molar mass; a molar-solubility ranking is not automatically a mass-solubility ranking.
CaF₂: Ksp = (s)(2s)² = 4s³
Do not rank salts of different stoichiometric types by Ksp alone. In ideal pure water at the same temperature, AgCl with Ksp = 1.8 × 10⁻¹⁰ has s ≈ 1.3 × 10⁻⁵ M, whereas Ag₂CrO₄ with Ksp = 1.1 × 10⁻¹² has s ≈ 6.5 × 10⁻⁵ M: the salt with the smaller Ksp is more soluble in this comparison.
For AgCl, Ksp ≈ s²; for Ag₂CrO₄, Ksp ≈ (2s)²s = 4s³. Derive each expression from its dissolution equation before comparing molar solubilities. A common ion or significant activity effects require a different calculation.
Dissolving AgCl in 0.10 M NaCl means starting with chloride already present. That extra product pushes the dissolution equilibrium back toward the solid, so far less AgCl dissolves. Put the supplied common-ion concentration in the initial row. Estimate s from the equilibrium expression, then test whether the added stoichiometric ion concentration is negligible relative to 0.10 M.
For AgCl in 0.100 M NaCl, the exact dilute-model relation is Ksp = s(0.100 + s). Estimate s = 1.80 × 10⁻¹⁰/0.100 = 1.80 × 10⁻⁹ M, then test s/0.100 = 1.80 × 10⁻⁸, far below 5%. The test, not a rule that common-ion approximations always work, justifies neglecting s. A much lower initial common-ion concentration may require the full equation.
To decide whether mixing two solutions produces a solid, calculate Q from the diluted ion concentrations and compare with Ksp. Q greater than Ksp means a precipitate forms until Q falls back to Ksp.
pH and ligands can change free-ion concentrations. Acid consumes CO₃²⁻ from a carbonate dissolution equilibrium, often increasing total dissolved salt; a complexing ligand can similarly lower free Ag⁺ while increasing total dissolved silver. Ksp still relates the free-ion activities at the fixed temperature. Do not insert the total concentration of free plus complexed metal into a free-ion Ksp expression. At high electrolyte concentration, activities can depart from concentrations and simple common-ion predictions need refinement.
A smaller Ksp does not always mean a smaller molar solubility when salt formulas differ. For a 1:1 salt AB, Ksp=s²; for AB2, Ksp=s(2s)²=4s³ in pure water. Convert each Ksp to s under the same conditions before ranking how many moles of formula units dissolve.
The equilibrium constant for a solid dissolving into its ions, with the solid omitted.
Moles of salt that dissolve per litre of saturated solution.
AB gives Ksp = s², AB2 gives 4s³, AB3 gives 27s⁴.
A shared ion suppresses dissolution and lowers solubility.
Like any K, Ksp changes only with temperature, not with a common ion.
Ksp = s²
Ksp = 4s³
Q > Ksp means a precipitate forms
The molar solubility of CaF₂ is 2.1 × 10⁻⁴ M. Calculate Ksp.
Try it first: Write out the dissociation and express each ion in terms of s.
0 of 4 steps revealed.
Ksp for AgCl is 1.8 × 10⁻¹⁰. Find its molar solubility in pure water.
Try it first: Decide which of the s formulas applies to a 1:1 salt.
0 of 3 steps revealed.
Find the molar solubility of AgCl (Ksp = 1.8 × 10⁻¹⁰) in 0.10 M NaCl.
Try it first: Put the chloride already present into the initial row of the ICE table.
0 of 5 steps revealed.
Compare molar solubilities of AB with Ksp=1.0×10⁻¹⁰ and XY2 with Ksp=4.0×10⁻¹² in pure water.
Try it first: Write Ksp in terms of s for each formula.
0 of 3 steps revealed.
| Mistake | Why it's wrong | Check instead |
|---|---|---|
| Including the solid salt in the Ksp expression. | Pure solids are omitted from every equilibrium expression. | Write only the aqueous ion terms. |
| Forgetting to raise the 2s term to a power as well as multiplying by 2. | The coefficient acts twice, once as a multiplier of s and once as an exponent. | Write (2s)², not 2s². |
| Ranking salts of different types by Ksp alone. | The relationship between Ksp and s depends on the ion ratio. | Convert each Ksp to a molar solubility first. |
| Saying a common ion lowers Ksp. | Ksp changes only with temperature; it is the solubility that falls. | State that the equilibrium shifts toward the solid. |
| Reporting an ion concentration when the question asked for molar solubility. | For AB2 salts the anion concentration is 2s, not s. | Identify s explicitly at the end. |
Scroll horizontally if needed to read all columns.
After the lesson · Question bank
Work through new problems on your own. The lesson's quick checks and worked examples stay above if you need to review a step.
30 topic questions across 3 difficulty levels.
Loading practice questions…
A saturated solution of a sparingly soluble salt is an equilibrium between the undissolved solid and its ions, described by the solubility product Ksp. Because the solid is a pure phase it does not appear in the expression, so Ksp is simply the product of ion concentrations raised to their coefficients. Molar solubility s is the moles of salt that dissolve per litre, and the relationship between s and Ksp depends on the formula: AB gives Ksp = s² while AB2 gives Ksp = 4s³. Comparing Ksp values directly only ranks solubility fairly for salts with the same ion ratio. Adding an ion the salt already contains shifts the dissolution equilibrium back toward the solid, so solubility drops sharply, which is the common ion effect.
Lesson: Not marked completed
Marking a lesson completed records that you finished reading it. Practice establishes mastery separately. Saved on this device.
Try a short set of different questions to check what you can do independently.
An optional video to support this lesson.
Loading video resources…
External resources belong to their respective creators. Chem Help is not affiliated with or endorsed by these providers. Links open the original provider; their terms apply.
Source attribution for this lesson has not been verified. Missing source records do not establish originality or reuse permission.