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Ch.16 - Aqueous Equilibrium WorksheetSee all chapters
All Chapters
Ch.1 - Intro to General Chemistry
Ch.2 - Atoms & Elements
Ch.3 - Chemical Reactions
BONUS: Lab Techniques and Procedures
BONUS: Mathematical Operations and Functions
Ch.4 - Chemical Quantities & Aqueous Reactions
Ch.5 - Gases
Ch.6 - Thermochemistry
Ch.7 - Quantum Mechanics
Ch.8 - Periodic Properties of the Elements
Ch.9 - Bonding & Molecular Structure
Ch.10 - Molecular Shapes & Valence Bond Theory
Ch.11 - Liquids, Solids & Intermolecular Forces
Ch.12 - Solutions
Ch.13 - Chemical Kinetics
Ch.14 - Chemical Equilibrium
Ch.15 - Acid and Base Equilibrium
Ch.16 - Aqueous Equilibrium
Ch.17 - Chemical Thermodynamics
Ch.18 - Electrochemistry
Ch.19 - Nuclear Chemistry
Ch.20 - Organic Chemistry
Ch.22 - Chemistry of the Nonmetals
Ch.23 - Transition Metals and Coordination Compounds
Intro to Buffers
Henderson-Hasselbalch Equation
Intro to Acid-Base Titration Curves
Strong Titrate-Strong Titrant Curves
Weak Titrate-Strong Titrant Curves
Acid-Base Indicators
Titrations: Weak Acid-Strong Base
Titrations: Weak Base-Strong Acid
Titrations: Strong Acid-Strong Base
Titrations: Diprotic & Polyprotic Buffers
Solubility Product Constant: Ksp
Ksp: Common Ion Effect
Precipitation: Ksp vs Q
Selective Precipitation
Complex Ions: Formation Constant

Process of separating specific ions out of a solution by using reagents to form a precipitate.

Concept #1: Selective Precipitation

Successful precipitation of a selected ion depends on the solubility (Ksp) of its salt.

Example #1: Sample of a solution contains 0.405 M CrO42- and 0.628 M S2- ions. These two ions can be precipitated with the use of PbF2. Which ion will precipitate out first and at which concentration? 

(PbCrO4 Ksp = 2.0 x 10-16, PbS Ksp = 7.0 x 10-29.)

Practice: Solution contains [Cu2+] = 0.035 M, [Sr2+] = 0.054 M, [Al3+] = 0.23 M. Cu2+ can be separated by selective precipitation using NaOH. What is the minimum concentration of NaOH needed to start precipitation of Cu2+? (Ksp = 2.2 × 1020 of Cu(OH)2).