Calculate molar solubility from Ksp and explore common ion effects
Use scientific notation (e.g., 1.77e-10)
For AgCl → Ag⁺ + Cl⁻, enter 1
For PbCl₂ → Pb²⁺ + 2Cl⁻, enter 2
Ksp is the equilibrium constant for the dissolution of a sparingly soluble ionic compound.
For a compound AaBb:
AaBb(s) ⇌ aA+(aq) + bB-(aq)
Ksp = [A+]a × [B-]b
Lower Ksp = less soluble | Higher Ksp = more soluble
Solubility is the maximum amount of a solute that can dissolve in a given amount of solvent at a specific temperature. For ionic compounds, solubility is often expressed as molar solubility (moles per liter).
The solubility product constant (Ksp) is the equilibrium constant for the dissolution of a sparingly soluble ionic compound. It helps predict whether a precipitate will form and calculate the solubility of ionic compounds.
Key Concept:
For a sparingly soluble salt AaBb, the solubility product is: Ksp = [A]a × [B]b, where [A] and [B] are the molar concentrations of the ions at equilibrium.
The Ksp expression is written for the dissolution equilibrium of a sparingly soluble salt:
General Form:
AaBb(s) ⇌ aAn+(aq) + bBm-(aq)
Ksp = [An+]a × [Bm-]b
AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)
Ksp = [Ag⁺][Cl⁻]
Ksp = 1.77 × 10-10
PbCl₂(s) ⇌ Pb²⁺(aq) + 2Cl⁻(aq)
Ksp = [Pb²⁺][Cl⁻]²
Ksp = 1.17 × 10-5
Ag₂CrO₄(s) ⇌ 2Ag⁺(aq) + CrO₄²⁻(aq)
Ksp = [Ag⁺]²[CrO₄²⁻]
Ksp = 1.12 × 10-12
Ca₃(PO₄)₂(s) ⇌ 3Ca²⁺(aq) + 2PO₄³⁻(aq)
Ksp = [Ca²⁺]³[PO₄³⁻]²
Ksp = 2.07 × 10-33
Important Note:
The solid compound does NOT appear in the Ksp expression because its activity is considered to be 1 (pure solid). Only aqueous ions are included.
Calculate the molar solubility of silver chloride (AgCl) in pure water. Ksp = 1.77 × 10-10
Step 1: Write the Dissolution Equation
AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)
Step 2: Write the Ksp Expression
Ksp = [Ag⁺][Cl⁻] = 1.77 × 10⁻¹⁰
Step 3: Set Up ICE Table
Let s = molar solubility of AgCl
Initial: [Ag⁺] = 0, [Cl⁻] = 0
Change: [Ag⁺] = +s, [Cl⁻] = +s
Equilibrium: [Ag⁺] = s, [Cl⁻] = s
Step 4: Substitute into Ksp Expression
Ksp = s × s = s²
1.77 × 10⁻¹⁰ = s²
Step 5: Solve for s
s = √(1.77 × 10⁻¹⁰)
s = 1.33 × 10⁻⁵ M
Answer:
The molar solubility of AgCl in pure water is 1.33 × 10-5 M. This means that at equilibrium, [Ag⁺] = [Cl⁻] = 1.33 × 10-5 M.
The common ion effect is the decrease in solubility of an ionic compound when a common ion (an ion that's already present in the solution) is added.
Scenario:
What is the solubility of AgCl in 0.10 M NaCl solution?
Solution:
AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)
Ksp = [Ag⁺][Cl⁻] = 1.77 × 10⁻¹⁰
Initial [Cl⁻] from NaCl = 0.10 M
Let s = solubility of AgCl
[Ag⁺] = s
[Cl⁻] = 0.10 + s ≈ 0.10 M (s is very small)
Ksp = s × 0.10 = 1.77 × 10⁻¹⁰
s = 1.77 × 10⁻⁹ M
Result:
Solubility in 0.10 M NaCl: 1.77 × 10⁻⁹ M
Solubility in pure water: 1.33 × 10⁻⁵ M
Solubility decreased by a factor of ~7,500!
Le Châtelier's Principle:
Adding a common ion shifts the equilibrium to the left (toward the solid), decreasing solubility. This is an application of Le Châtelier's principle: the system responds to the stress (added ions) by shifting to reduce that stress.
The reaction quotient (Q) helps predict whether a precipitate will form when solutions are mixed.
Precipitation occurs
Solution is supersaturated. The system will shift left, forming a precipitate until Q = Ksp.
At equilibrium
Solution is saturated. No net change occurs. System is at equilibrium.
No precipitation
Solution is unsaturated. More solid can dissolve if present. No precipitate forms.
Will a precipitate form if 50.0 mL of 0.0020 M AgNO₃ is mixed with 50.0 mL of 0.0015 M NaCl? (Ksp of AgCl = 1.77 × 10-10)
Step 1: Calculate diluted concentrations
[Ag⁺] = (0.0020 M × 50.0 mL) / 100.0 mL = 0.0010 M
[Cl⁻] = (0.0015 M × 50.0 mL) / 100.0 mL = 0.00075 M
Step 2: Calculate Q
Q = [Ag⁺][Cl⁻] = (0.0010)(0.00075) = 7.5 × 10-7
Step 3: Compare Q to Ksp
Q = 7.5 × 10-7
Ksp = 1.77 × 10-10
Q > Ksp
Conclusion:
Since Q > Ksp, precipitation of AgCl will occur.
Water softening uses precipitation to remove Ca²⁺ and Mg²⁺ ions by adding carbonate or phosphate. Understanding Ksp helps optimize the amount of reagent needed to precipitate hardness ions.
In analytical chemistry, selective precipitation is used to separate and identify metal ions. Different Ksp values allow chemists to precipitate one ion while keeping others in solution.
Kidney stones are often calcium oxalate (CaC₂O₄) or calcium phosphate. Understanding solubility helps develop treatments to prevent stone formation and dissolve existing stones.
Tooth enamel is hydroxyapatite [Ca₅(PO₄)₃OH]. Fluoride treatment works by forming fluorapatite, which has lower solubility, making teeth more resistant to acid dissolution (cavities).
Silver halides (AgBr, AgCl) are light-sensitive and used in photographic film. Their low solubility keeps them stable until light exposure triggers chemical changes.
Heavy metal contamination can be treated by precipitation. Adding sulfide ions can precipitate toxic metals like Pb²⁺, Cd²⁺, and Hg²⁺ as insoluble sulfides for removal.
| Compound | Formula | Ksp | Dissolution Equation |
|---|---|---|---|
| Silver chloride | AgCl | 1.77 × 10⁻¹⁰ | Ag⁺ + Cl⁻ |
| Silver bromide | AgBr | 5.38 × 10⁻¹³ | Ag⁺ + Br⁻ |
| Barium sulfate | BaSO₄ | 1.08 × 10⁻¹⁰ | Ba²⁺ + SO₄²⁻ |
| Calcium carbonate | CaCO₃ | 3.36 × 10⁻⁹ | Ca²⁺ + CO₃²⁻ |
| Calcium fluoride | CaF₂ | 3.45 × 10⁻¹¹ | Ca²⁺ + 2F⁻ |
| Lead(II) chloride | PbCl₂ | 1.17 × 10⁻⁵ | Pb²⁺ + 2Cl⁻ |
| Magnesium hydroxide | Mg(OH)₂ | 5.61 × 10⁻¹² | Mg²⁺ + 2OH⁻ |
| Iron(III) hydroxide | Fe(OH)₃ | 2.79 × 10⁻³⁹ | Fe³⁺ + 3OH⁻ |
The solid compound does not appear in the Ksp expression.
Correct: For AgCl, Ksp = [Ag⁺][Cl⁻], NOT Ksp = [AgCl][Ag⁺][Cl⁻]
Stoichiometric coefficients become exponents in the Ksp expression.
Correct: For PbCl₂, Ksp = [Pb²⁺][Cl⁻]², NOT [Pb²⁺][Cl⁻]
Common ions dramatically reduce solubility. Don't assume pure water conditions.
Correct: Account for initial ion concentrations from other sources
Can't directly compare Ksp values for compounds with different formulas to determine relative solubility.
Correct: Calculate and compare molar solubilities, not Ksp values
For AB: s = √Ksp
For AB₂: s = ∛(Ksp/4)
For A₂B: s = ∛(Ksp/4)
For AB₃: s = ⁴√(Ksp/27)
Q > Ksp: Precipitation occurs
Q = Ksp: At equilibrium
Q < Ksp: No precipitation
Adding common ion → decreases solubility
Equilibrium shifts toward solid
Le Châtelier's principle applies
Lower Ksp = less soluble
Higher Ksp = more soluble
(Only for same stoichiometry!)