Ch.14 - Chemical EquilibriumWorksheetSee 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 following equilibrium constants have been determined for oxalic acid at 25 °C.H2C2O4 (aq) ⟺ H+ (aq) + HC2O4 - (aq)    K1 = 6.5 x 10 -2HC2O4- (aq) ⟺ H+ (aq) + C2O4 2- (aq)        K 2 = 6.1 x 10 -5C

Problem

The following equilibrium constants have been determined for oxalic acid at 25 °C.

H2C2O4 (aq) ⟺ H+ (aq) + HC2O- (aq)    K= 6.5 x 10 -2

HC2O4- (aq) ⟺ H+ (aq) + C2O2- (aq)        K = 6.1 x 10 -5

Calculate the equilibrium constant for the following reaction at the same temperature: 

                    C  2O2− + 2H + ⟺ H2C2O4 (aq)