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.
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Chemical Equilibrium & SolubilitySolubility EquilibriaContent level: Core 26 min
Solubility Product Ksp, Molar Solubility and the Common Ion Effect
What you'll be able to do: Convert between Ksp and molar solubility and explain quantitatively why a common ion reduces solubility.
Introduction
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.
Learning objectives
- Write Ksp expressions for sparingly soluble salts
- Convert molar solubility to Ksp and back
- Explain when Ksp values can be compared directly
- Calculate solubility in the presence of a common ion
Lesson
Dissolution as an equilibrium
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
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.
s is the amount of salt dissolved, not the concentration of any one ion.
From s to Ksp
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.
CaF₂: Ksp = (s)(2s)² = 4s³
Comparing solubilities
A smaller Ksp only means a less soluble salt when the salts have the same ion ratio. AgCl with Ksp = 1.8 × 10⁻¹⁰ has s = 1.3 × 10⁻⁵ M, whereas Ag₂CrO₄ with a larger Ksp of 1.1 × 10⁻¹² still works out at s = 6.5 × 10⁻⁵ M, five times more soluble. Convert to molar solubility before ranking salts of different types.
Different ion ratios? Convert both to s before comparing.
The common ion effect
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. In the ICE table the common ion goes in the initial row, and since Ksp is small the approximation 0.10 + s is about 0.10 is almost always safe.
Precipitation
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.
Key ideas
Definition
Solubility product Ksp
The equilibrium constant for a solid dissolving into its ions, with the solid omitted.
Definition
Molar solubility s
Moles of salt that dissolve per litre of saturated solution.
Rule
Salt type sets the formula
AB gives Ksp = s², AB₂ gives 4s³, AB₃ gives 27s⁴.
Rule
Common ion effect
A shared ion suppresses dissolution and lowers solubility.
Key concept
Ksp is temperature dependent
Like any K, Ksp changes only with temperature, not with a common ion.
Equation
Ksp for an AB salt
Ksp = s²
- s = molar solubility in mol/L
Equation
Ksp for an AB₂ or A₂B salt
Ksp = 4s³
- s = molar solubility in mol/L
Equation
Precipitation test
Q > Ksp means a precipitate forms
- Q = ion product using current concentrations
Worked examples
Worked example 2
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.
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Worked example 3
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.
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Common mistakes
Including the solid salt in the Ksp expression.
Why it's wrong: Pure solids are omitted from every equilibrium expression.
Check instead: Write only the aqueous ion terms.
Forgetting to raise the 2s term to a power as well as multiplying by 2.
Why it's wrong: The coefficient acts twice, once as a multiplier of s and once as an exponent.
Check instead: Write (2s)², not 2s².
Ranking salts of different types by Ksp alone.
Why it's wrong: The relationship between Ksp and s depends on the ion ratio.
Check instead: Convert each Ksp to a molar solubility first.
Saying a common ion lowers Ksp.
Why it's wrong: Ksp changes only with temperature; it is the solubility that falls.
Check instead: State that the equilibrium shifts toward the solid.
Reporting an ion concentration when the question asked for molar solubility.
Why it's wrong: For AB₂ salts the anion concentration is 2s, not s.
Check instead: Identify s explicitly at the end.
Practice this skill
No practice questions are available for this topic yet. You can still practice the whole unit.
What you should now know
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.
- Ksp is the equilibrium constant for a salt dissolving, solid omitted
- Molar solubility s is moles of salt dissolved per litre
- AB gives Ksp = s² and AB₂ gives Ksp = 4s³
- Compare solubilities by converting to s, not by comparing Ksp
- A common ion lowers solubility but never changes Ksp
Sources and further reading
This lesson is original Chem Help content. No external sources were adapted.