Module 5 Acids, Bases, and Solubility Equilibrium
Welcome to the practical world of solubility equilibria, where we explore how sparingly soluble compounds dissolve and precipitate under different conditions. You'll learn to predict when precipitation occurs, how pH affects solubility, and how to separate ions—skills essential for analytical chemistry, environmental monitoring, and industrial applications.
In this lesson, you'll master two essential learning objectives that complete your understanding of equilibrium systems:
We begin with solubility product constants (Ksp) and learn to calculate whether precipitation will occur when solutions are mixed. You'll discover how common ions affect solubility through Le Châtelier's principle, then explore how pH changes can dissolve insoluble compounds. Finally, we examine complex ion equilibria, showing how Lewis acid-base chemistry can dramatically increase solubility.
Why This Matters: Solubility equilibria control water quality (hard water, scaling, pollution), geological processes (cave formation, mineral deposits), biological systems (kidney stones, bone formation), pharmaceutical development (drug solubility, bioavailability), and analytical chemistry (qualitative analysis, separations, purifications). Understanding these principles is crucial for environmental chemistry, materials science, and biochemistry.
How to Succeed: Focus on the relationships between Qsp and Ksp to predict precipitation. Practice pH-solubility problems by identifying which species are affected by H⁺ or OH⁻ addition. For complex ions, remember they follow Lewis acid-base theory—metal cations accept electron pairs from ligands. Work through separation problems systematically by comparing Ksp values and calculating the required conditions.
Overby/Chang: Chemistry, 14th Ed. - Chapter 16: Sections 16.6-16.12
Solubility Equilibria Fundamentals
Advanced Applications
Master the foundation concepts of solubility equilibria through systematic study of Ksp relationships and precipitation predictions.
Systematic Approach to Precipitation Problems: Follow this decision tree to solve any Ksp problem systematically.
Step 1: What are you calculating?
Step 2: What's the equilibrium expression?
Step 3: Key Decision Points:
Apply LO5.1.5: Master Ksp calculations and precipitation predictions. Work through problems systematically using the decision tree approach.
PROBLEM 1 - BASIC Calculate the molar solubility of AgCl in pure water. Ksp for AgCl = 1.8 × 10⁻¹⁰
Answer & Step-by-Step Solution
Answer: Molar solubility = 1.3 × 10⁻⁵ M
Step 1: Write the equilibrium and ICE table:
AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)
Let s = molar solubility = [Ag⁺] = [Cl⁻] at equilibrium
Step 2: Write Ksp expression:
Ksp = [Ag⁺][Cl⁻] = (s)(s) = s²
Step 3: Solve for s:
1.8 × 10⁻¹⁰ = s²
s = √(1.8 × 10⁻¹⁰) = 1.3 × 10⁻⁵ M
Interpretation: Only 1.3 × 10⁻⁵ mol of AgCl dissolves per liter - truly "sparingly soluble"!
PROBLEM 2 - INTERMEDIATE Will a precipitate form when 100 mL of 0.010 M Pb(NO₃)₂ is mixed with 200 mL of 0.025 M KI? Ksp for PbI₂ = 7.1 × 10⁻⁹
Answer & Step-by-Step Solution
Answer: YES, precipitation occurs because Qsp > Ksp
Step 1: Calculate concentrations after mixing (before reaction):
Total volume = 100 + 200 = 300 mL = 0.300 L
[Pb²⁺] = (0.010 M)(0.100 L)/0.300 L = 0.00333 M
[I⁻] = (0.025 M)(0.200 L)/0.300 L = 0.0167 M
Step 2: Calculate reaction quotient for PbI₂:
PbI₂(s) ⇌ Pb²⁺ + 2I⁻
Qsp = [Pb²⁺][I⁻]² = (0.00333)(0.0167)² = 9.3 × 10⁻⁷
Step 3: Compare Qsp to Ksp:
Qsp = 9.3 × 10⁻⁷ > Ksp = 7.1 × 10⁻⁹
Conclusion: Since Qsp > Ksp, PbI₂ will precipitate!
Explore how environmental conditions affect solubility through pH changes and common ion effects—crucial for understanding natural systems.
Understanding Nature's pH-Solubility Laboratory: Caves demonstrate pH effects on calcium carbonate solubility in dramatic fashion.
Step 1: Acid Formation
Step 2: Limestone Dissolution
Step 3: Speleothem Formation
Environmental Factors:
Key Insight: The same equilibrium principles that dissolve kidney stones (acidic conditions) also carved the world's most spectacular caves!
Apply LO5.1.6: Master how pH changes affect solubility of compounds with basic anions. Focus on identifying which species react with H⁺ or OH⁻.
PROBLEM 3 - INTERMEDIATE Will Mg(OH)₂ be more soluble in pure water or in a solution with pH = 3? Explain using Le Châtelier's principle. Ksp for Mg(OH)₂ = 1.8 × 10⁻¹¹
Answer & Step-by-Step Solution
Answer: MUCH more soluble at pH 3 due to H⁺ consumption of OH⁻ ions
Step 1: Write the dissolution equilibrium:
Mg(OH)₂(s) ⇌ Mg²⁺(aq) + 2OH⁻(aq)
Step 2: Consider what happens at low pH:
At pH 3: [H⁺] = 10⁻³ M (high acid concentration)
H⁺ + OH⁻ → H₂O (neutralization reaction)
Step 3: Apply Le Châtelier's principle:
Removing OH⁻ (through neutralization) shifts equilibrium RIGHT
More Mg(OH)₂ dissolves to replace consumed OH⁻ ions
Step 4: Calculate solubility in pure water vs pH 3:
Pure water: s = ∛(Ksp/4) = ∛(1.8×10⁻¹¹/4) = 1.6×10⁻⁴ M
At pH 3: OH⁻ constantly consumed → much higher solubility
Real-world Application: This is why antacids (Mg(OH)₂) work—stomach acid increases their solubility!
Connect Lewis acid-base theory to complex ion formation and discover how it dramatically enhances solubility in analytical and industrial applications.
Building on M5L1 Lewis Theory: Complex ion formation is Lewis acid-base chemistry in action, creating dramatic solubility changes.
Drag each species to identify its role in complex ion formation. Remember Lewis theory: acids accept electron pairs, bases donate electron pairs.
Apply LO5.1.6: Master complex ion calculations and understand how they enhance solubility. Connect to Lewis theory principles from M5L1.
PROBLEM 4 - ADVANCED Calculate the molar solubility of AgCl in 1.0 M NH₃ solution. Ksp for AgCl = 1.8 × 10⁻¹⁰, Kf for [Ag(NH₃)₂]⁺ = 1.7 × 10⁷
Answer & Step-by-Step Solution
Answer: Molar solubility = 0.055 M (4000× higher than in pure water!)
Step 1: Identify the coupled equilibria:
AgCl(s) ⇌ Ag⁺ + Cl⁻ Ksp = 1.8 × 10⁻¹⁰
Ag⁺ + 2NH₃ ⇌ [Ag(NH₃)₂]⁺ Kf = 1.7 × 10⁷
Step 2: Write the overall reaction:
AgCl(s) + 2NH₃ ⇌ [Ag(NH₃)₂]⁺ + Cl⁻
Koverall = Ksp × Kf = (1.8 × 10⁻¹⁰)(1.7 × 10⁷) = 3.1 × 10⁻³
Step 3: Set up ICE table (let s = molar solubility):
Initial: [NH₃] = 1.0 M, [[Ag(NH₃)₂]⁺] = 0, [Cl⁻] = 0
Change: [NH₃] = -2s, [[Ag(NH₃)₂]⁺] = +s, [Cl⁻] = +s
Equilibrium: [NH₃] = 1.0-2s, [[Ag(NH₃)₂]⁺] = s, [Cl⁻] = s
Step 4: Apply equilibrium expression:
Koverall = [[Ag(NH₃)₂]⁺][Cl⁻]/[NH₃]² = (s)(s)/(1.0-2s)² = 3.1 × 10⁻³
Taking the square root: s/(1.0-2s) = 0.056
Solving: s = 0.055 M
Key Insight: Complex formation increases solubility by removing Ag⁺ from solution!
Connecting All Learning Objectives:
Real-World Problem Solving:
You now have the complete toolkit for understanding aqueous equilibria in all their forms!
Apply solubility principles to real analytical chemistry problems—from separating metal ions to understanding environmental chemistry.
Master Integration: Apply both LO5.1.5 and LO5.1.6 in realistic scenarios that combine multiple equilibrium effects.
PROBLEM 5 - ADVANCED A solution contains 0.10 M Pb²⁺ and 0.10 M Ag⁺. Can these ions be separated by slowly adding KI solution? Ksp values: PbI₂ = 7.1 × 10⁻⁹, AgI = 8.3 × 10⁻¹⁷
Answer & Step-by-Step Solution
Answer: YES - excellent separation! AgI precipitates first, leaving Pb²⁺ in solution.
Step 1: Calculate [I⁻] needed to start precipitation of each ion:
For AgI: Ksp = [Ag⁺][I⁻] → [I⁻] = Ksp/[Ag⁺] = (8.3×10⁻¹⁷)/(0.10) = 8.3×10⁻¹⁶ M
For PbI₂: Ksp = [Pb²⁺][I⁻]² → [I⁻] = √(Ksp/[Pb²⁺]) = √(7.1×10⁻⁹/0.10) = 8.4×10⁻⁴ M
Step 2: Compare precipitation thresholds:
AgI starts precipitating at [I⁻] = 8.3×10⁻¹⁶ M
PbI₂ starts precipitating at [I⁻] = 8.4×10⁻⁴ M
Separation factor = (8.4×10⁻⁴)/(8.3×10⁻¹⁶) = 10¹² - EXCELLENT!
Step 3: Determine separation strategy:
Add KI slowly until [I⁻] ≈ 10⁻⁴ M
AgI precipitates completely, PbI₂ remains dissolved
Filter to separate AgI(s) from solution containing Pb²⁺
Application: This principle is used in qualitative analysis schemes to systematically identify metal ions!
PROBLEM 6 - ADVANCED Environmental Chemistry: Acid rain (pH 4.0) flows over limestone (CaCO₃). Calculate the [Ca²⁺] concentration in the resulting solution. Ka₁ for H₂CO₃ = 4.3 × 10⁻⁷, Ksp for CaCO₃ = 3.4 × 10⁻⁹
Answer & Step-by-Step Solution
Answer: [Ca²⁺] = 1.3 × 10⁻² M (much higher than in neutral water!)
Step 1: Identify the key reaction at low pH:
CaCO₃(s) + H⁺ → Ca²⁺ + HCO₃⁻ (acid dissolves limestone)
Step 2: Write equilibrium expressions:
CaCO₃(s) ⇌ Ca²⁺ + CO₃²⁻ Ksp = 3.4 × 10⁻⁹
CO₃²⁻ + H⁺ ⇌ HCO₃⁻ K = 1/Ka₂ ≈ 2 × 10⁶
Step 3: Calculate [H⁺] and approach:
pH = 4.0 → [H⁺] = 1.0 × 10⁻⁴ M
High [H⁺] converts CO₃²⁻ to HCO₃⁻, increasing solubility
Step 4: Apply combined equilibrium:
K_combined = Ksp × (1/Ka₂) = (3.4×10⁻⁹) × (2×10⁶) = 6.8 × 10⁻³
At equilibrium: [Ca²⁺] = [HCO₃⁻], and [HCO₃⁻] ≈ total dissolved CO₂
K_combined = [Ca²⁺][HCO₃⁻]/[H⁺] = [Ca²⁺]²/(1.0×10⁻⁴) = 6.8 × 10⁻³
Solving: [Ca²⁺] = √(6.8×10⁻³ × 1.0×10⁻⁴) = 1.3 × 10⁻² M
Environmental Impact: This explains cave formation and limestone erosion in acid rain areas!
Historical Analytical Chemistry: Before modern instruments, chemists identified ions using systematic precipitation schemes based on Ksp differences.
Group I - HCl Addition
Precipitates: AgCl, PbCl₂, Hg₂Cl₂
All other ions remain in solution
Group II - H₂S Addition
Precipitates: CuS, PbS, CdS (low pH)
Relies on pH control of [S²⁻]
Group III - NH₃/NH₄⁺ Buffer
Precipitates: Fe(OH)₃, Al(OH)₃
Buffer controls pH for selective precipitation
Group IV & V - Carbonate & Soluble
Remaining cations identified by specific tests
Na⁺, K⁺, NH₄⁺ detected by flame tests, etc.
Modern Relevance: While instruments now do this work, understanding these principles is essential for water treatment, environmental analysis, and materials purification!
Match each real-world scenario with the underlying solubility principle. Consider which equilibrium effects (Ksp, pH, complex formation) are most important.
Complete your mastery of aqueous equilibria by connecting all the concepts across acid-base, buffer, and solubility systems.
Problem-Solving Approach:
Integration Skills:
You now have a complete understanding of aqueous equilibria systems and can predict the behavior of acids, bases, buffers, and sparingly soluble compounds in real-world scenarios!