Solubility Equilibrium: Saturated Solutions, Common Ion, and Precipitation

Calculate Ksp and predict precipitation

Solubility Equilibrium: Saturated Solutions, Common Ion, and Precipitation

Solubility equilibrium describes the dynamic balance between a slightly soluble ionic solid and its ions in solution. At equilibrium, the dissolution rate equals the precipitation rate, and the solution is saturated. We quantify this balance with the solubility product constant, Ksp, which depends on temperature and the stoichiometry of the dissolution reaction.

You should be able to write Ksp expressions, calculate molar solubility from Ksp (and vice versa), apply the common ion effect, and decide whether a precipitate forms when two solutions are mixed.

Key definitions

  • Saturated solution: Contains the maximum amount of dissolved solute at given conditions; excess solid present.
  • Unsaturated: Less than the equilibrium amount dissolved; no excess solid; more can dissolve.
  • Supersaturated: More than equilibrium amount temporarily dissolved; unstable; precipitates with disturbance.
  • Solubility (s): Equilibrium amount that dissolves (often in mol·L⁻¹). Related to Ksp via stoichiometry.

Writing Ksp expressions

For a generic salt A_pB_q(s) ⇌ p A^m+(aq) + q B^n−(aq):

  • Ksp = [A^m+]^p [B^n−]^q
  • Pure solids are omitted. Only aqueous ion activities (approximated by concentrations in dilute solutions) appear.

Examples:

  • AgCl(s) ⇌ Ag+(aq) + Cl−(aq) → Ksp = [Ag+][Cl−]
  • CaF2(s) ⇌ Ca2+(aq) + 2 F−(aq) → Ksp = [Ca2+][F−]^2
  • PbI2(s) ⇌ Pb2+(aq) + 2 I−(aq) → Ksp = [Pb2+][I−]^2

From Ksp to molar solubility (s)

  1. 1:1 salts, MX with Ksp = [M+][X−] = s·s = s^2 → s = √Ksp
  2. 1:2 salts, MX2 with Ksp = [M2+][X−]^2 = s·(2s)^2 = 4s^3 → s = (Ksp/4)^(1/3)
  3. 2:3 salts, M2X3 with Ksp = [M3+]^2[X2−]^3 = (2s)^2(3s)^3 = 108 s^5 → s = (Ksp/108)^(1/5)

Always set up an ICE table when in doubt, matching coefficients to ionic concentrations.

Common ion effect

Adding a source of a common ion reduces solubility by shifting equilibrium left (Le Châtelier). For CaF2(s) in a solution containing F−, the term [F−]^2 in Ksp is already large, so less CaF2 dissolves and s decreases.

Quick example: CaF2(s) in 0.10 M NaF. Write Ksp = [Ca2+][F−]^2. Let s' be the solubility of CaF2 in the presence of 0.10 M F−. Then [F−] ≈ 0.10 (if s' is small), so [Ca2+] = Ksp / [F−]^2. This is much smaller than the pure-water solubility.

Precipitation criteria and Q vs Ksp

When two salt solutions are mixed, compute the reaction quotient Qsp using the resulting ion concentrations after mixing. Compare to Ksp:

  • Qsp < Ksp → Unsaturated; no precipitate forms.
  • Qsp = Ksp → At equilibrium; just saturated.
  • Qsp > Ksp → Supersaturated; precipitate forms until equilibrium is restored.

Key steps:

  • Determine moles of each ion contributed.
  • Account for dilution to the final total volume.
  • Compute Qsp with the correct exponents.
  • Compare Qsp to Ksp and conclude.

Selective precipitation

If multiple cations are present, choose an anion that forms a very insoluble salt with one cation (lowest Ksp) so it precipitates first. By adjusting [X−], ions precipitate in order of their Ksp values, enabling separation.

Problem walkthroughs

  1. Will a precipitate form when 25.0 mL of 0.020 M AgNO3 is mixed with 25.0 mL of 0.020 M NaCl? (Ksp AgCl = 1.8×10^−10)
  • Moles Ag+ = 0.0250 L × 0.020 M = 5.00×10^−4 mol; moles Cl− same.
  • Total volume = 0.0500 L; [Ag+] = [Cl−] = (5.00×10^−4 mol)/(0.0500 L) = 0.0100 M.
  • Qsp = [Ag+][Cl−] = (0.0100)(0.0100) = 1.0×10^−4 ≫ 1.8×10^−10 → Qsp > Ksp → AgCl precipitates.
  1. Molar solubility of PbI2 (Ksp = 7.1×10^−9) in pure water:
  • PbI2(s) ⇌ Pb2+ + 2 I−; Ksp = [Pb2+][I−]^2 = s(2s)^2 = 4s^3
  • s = (Ksp/4)^(1/3) ≈ (7.1×10^−9 / 4)^(1/3) ≈ 1.2×10^−3 M (approximate)
  1. Qualitative: How does adding KI affect PbI2 solubility?
  • Adds common ion I−, increases [I−] term in Ksp expression, shifts equilibrium left, reduces s markedly.

Common pitfalls

  • Forgetting to recompute concentrations after mixing (dilution!).
  • Omitting exponents from Ksp expressions.
  • Comparing Qsp and Ksp incorrectly or using initial (not mixed) concentrations.
  • Assuming all salts have the same relationship between Ksp and s; always use stoichiometry.

Quick reference

  • Ksp uses ions only; solids don’t appear
  • Convert Ksp ↔ molar solubility with stoichiometry
  • Common ion lowers solubility
  • Compare Qsp to Ksp to decide precipitation
  • Use selective precipitation to separate ions

📚 Practice Problems

1Problem 1easy

❓ Question:

Write the K_sp expression for the dissolution of silver chloride, AgCl(s).

💡 Show Solution

Step 1: Write the dissolution equation AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)

Step 2: Write K_sp expression K_sp = [Ag⁺][Cl⁻]

Note: The solid AgCl is not included in the equilibrium expression (only aqueous species).

The K_sp value for AgCl at 25°C is approximately 1.8 × 10⁻¹⁰.

2Problem 2medium

❓ Question:

The K_sp of PbCl₂ is 1.7 × 10⁻⁵. Calculate the molar solubility of PbCl₂ in pure water.

💡 Show Solution

Step 1: Write dissolution equation PbCl₂(s) ⇌ Pb²⁺(aq) + 2Cl⁻(aq)

Step 2: Set up ICE table (let s = molar solubility) Pb²⁺ 2Cl⁻ I: 0 0 C: +s +2s E: s 2s

Step 3: Write K_sp expression K_sp = [Pb²⁺][Cl⁻]² 1.7 × 10⁻⁵ = (s)(2s)² 1.7 × 10⁻⁵ = 4s³

Step 4: Solve for s s³ = 4.25 × 10⁻⁶ s = 1.6 × 10⁻² M

The molar solubility of PbCl₂ is 0.016 M or 1.6 × 10⁻² M.

3Problem 3medium

❓ Question:

Explain the common ion effect. How does adding NaCl affect the solubility of AgCl?

💡 Show Solution

Common Ion Effect: The solubility of a salt decreases when a common ion is added to the solution. This is an application of Le Chatelier's principle.

For AgCl: AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)

Adding NaCl increases [Cl⁻]: • NaCl fully dissociates: NaCl → Na⁺ + Cl⁻ • Increased [Cl⁻] shifts equilibrium LEFT • More AgCl precipitates out of solution • Solubility of AgCl DECREASES

Quantitative effect: K_sp = [Ag⁺][Cl⁻] remains constant If [Cl⁻] increases, [Ag⁺] must decrease proportionally

Example: If [Cl⁻] doubles, [Ag⁺] is cut in half to maintain K_sp.

4Problem 4hard

❓ Question:

Will a precipitate form if 100 mL of 0.002 M AgNO₃ is mixed with 100 mL of 0.002 M NaCl? K_sp(AgCl) = 1.8 × 10⁻¹⁰.

💡 Show Solution

Step 1: Calculate ion concentrations after mixing Total volume = 200 mL (dilution by factor of 2) [Ag⁺] = 0.002 M / 2 = 0.001 M = 1.0 × 10⁻³ M [Cl⁻] = 0.002 M / 2 = 0.001 M = 1.0 × 10⁻³ M

Step 2: Calculate Q (reaction quotient) Q = [Ag⁺][Cl⁻] Q = (1.0 × 10⁻³)(1.0 × 10⁻³) Q = 1.0 × 10⁻⁶

Step 3: Compare Q to K_sp Q = 1.0 × 10⁻⁶ K_sp = 1.8 × 10⁻¹⁰

Q >> K_sp (by factor of ~5,500)

Conclusion: YES, a precipitate WILL form.

When Q > K_sp, the solution is supersaturated and precipitation occurs until Q = K_sp.

5Problem 5hard

❓ Question:

Calculate the molar solubility of Mg(OH)₂ in a solution buffered at pH 10. K_sp = 1.8 × 10⁻¹¹.

💡 Show Solution

Step 1: Find [OH⁻] from pH pH = 10, so pOH = 14 - 10 = 4 [OH⁻] = 10⁻⁴ M

Step 2: Write dissolution equation Mg(OH)₂(s) ⇌ Mg²⁺(aq) + 2OH⁻(aq)

Step 3: Write K_sp expression K_sp = [Mg²⁺][OH⁻]²

Step 4: Account for buffered [OH⁻] The buffer maintains [OH⁻] = 10⁻⁴ M Let s = solubility of Mg(OH)₂ [Mg²⁺] = s [OH⁻] = 10⁻⁴ M (from buffer, dominates over OH⁻ from Mg(OH)₂)

Step 5: Solve for s 1.8 × 10⁻¹¹ = s(10⁻⁴)² 1.8 × 10⁻¹¹ = s(10⁻⁸) s = 1.8 × 10⁻³ M

In buffered solution: 1.8 × 10⁻³ M

For comparison in pure water: s would be much higher (~1.7 × 10⁻⁴ M) The common ion effect from the high [OH⁻] reduces solubility significantly.

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