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: a. Using the Ksp value for Cu(OH)2 (1.6 x 10-19) and the overall formation constant for Cu(NH3)42+ (1.0 x 1013), calculate the value for the equilibrium constant for the following reaction:Cu(OH)2 (s) + 4NH3 (aq) ⇌ Cu(NH3)42+(aq) + 2OH -(aq)b. Use the value of the equilibrium constant you calculated in part a to calculate the solubility (in mol/L) of Cu(OH)2 in 5.0 M NH3. In 5.0 M NH3 the concentration of OH - is 0.0095 M.

Problem

a. Using the Ksp value for Cu(OH)2 (1.6 x 10-19) and the overall formation constant for Cu(NH3)42+ (1.0 x 1013), calculate the value for the equilibrium constant for the following reaction:

Cu(OH)2 (s) + 4NH3 (aq) ⇌ Cu(NH3)42+(aq) + 2OH -(aq)

b. Use the value of the equilibrium constant you calculated in part a to calculate the solubility (in mol/L) of Cu(OH)2 in 5.0 M NH3. In 5.0 M NH3 the concentration of OH - is 0.0095 M.