Solubility EquilibriaCourse level: AP Chemistry

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

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.

The pure-water shortcuts assume no initial common ions, no significant hydrolysis or complex formation, and negligible activity-coefficient effects:

DissolutionIon concentrations from saltKsp in terms of s
AB → A + Bs, ss²
AB₂ → A + 2 Bs, 2s4s³
A₂B → 2 A + B2s, s4s³
A₃B₂ → 3 A + 2 B3s, 2s108s⁵

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

Calcium fluoride dissolution and free-ion solubility product. Read this visual alongside the section; the text alternative describes its scientific content and assumptions.

Comparing solubilities

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.

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. 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.

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.

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.

Compare solubilities through stoichiometry

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.

Use the dissolution equation, not Ksp alone, across different formulas.

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 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.

    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.

      0 of 3 steps revealed.

      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.

        0 of 5 steps revealed.

        Worked example 4

        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.

          Common mistakes

          Common mistakes and better checks
          MistakeWhy it's wrongCheck 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.

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          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

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