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

Solution: The Ksp of Al(OH)3 at 25 °C is 1 x 10-33. Consider a solution that is 1.0 x 10-10 M Al(NO3)3 and 2.0  10-8 M NaOH.

Problem

The Ksp of Al(OH)3 at 25 °C is 1 x 10-33. Consider a solution that is 1.0 x 10-10 M Al(NO3)3 and 2.0  10-8 M NaOH.

Solution

We’re being asked to determine if a precipitate will form when combining solutions  1.0 x 10-10 M Al(NO3)3 and 2.0  10-8 M NaOH with  Al(OH)as the expected precipitate.


We will use the reaction quotient, Q, to determine if the chemical reaction would be at equilibrium or not. The formula for Q is:

Q = productsreactants


For this, we need to compare the reaction quotient (Q) vs. the solubility product constant (Ksp)

Recall that when:

• Q > Ksp: the solution is supersaturated and a precipitate will form. Reactants are favored.

• Q = Ksp: the solution is at equilibrium and no precipitate will form.

• Q < Ksp: the solution is unsaturated and no precipitate will form. Products are favored.


Solution BlurView Complete Written Solution