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: In the presence of CN -, Fe3+ forms the complex ion Fe(CN)63-. The equilibrium concentrations of Fe3+ and Fe(CN)63- are 8.5 X 10 -40 M and 1.5 X 10 -3 M, respectively, in a 0.11-M KCN solution. Calculate the value for the overall formation constant of Fe(CN)63-.Fe3+(aq) + 6CN -(aq) ⇋ Fe(CN)63-(aq)       Koverall = ?

Solution: In the presence of CN -, Fe3+ forms the complex ion Fe(CN)63-. The equilibrium concentrations of Fe3+ and Fe(CN)63- are 8.5 X 10 -40 M and 1.5 X 10 -3 M, respectively, in a 0.11-M KCN solution. Calcul

Problem

In the presence of CN -, Fe3+ forms the complex ion Fe(CN)63-. The equilibrium concentrations of Fe3+ and Fe(CN)63- are 8.5 X 10 -40 M and 1.5 X 10 -3 M, respectively, in a 0.11-M KCN solution. Calculate the value for the overall formation constant of Fe(CN)63-.

Fe3+(aq) + 6CN -(aq) ⇋ Fe(CN)63-(aq)       Koverall = ?