Ch.15 - Acid and Base EquilibriumWorksheetSee all chapters
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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: What is the hydronium ion concentration of a 0.980 M acetic acid solution with Ka = 1.8 x 10-5? The equation for the dissociation of acetic acid is: CH3CO2H (aq) + H2O (l) ⇌ H3O+ (aq) + CH3CO2- (aq). 

Problem

What is the hydronium ion concentration of a 0.980 M acetic acid solution with Ka = 1.8 x 10-5? The equation for the dissociation of acetic acid is: 

CH3CO2H (aq) + H2O (l) ⇌ H3O+ (aq) + CH3CO2- (aq). 

a. 3.0 times 10-2

b. 4.2 times 10-2

c. 3.0 times 10-3

d. 42 times 10-3

e. There is not enough information to solve.