Ch.15 - Acid and Base 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:  A monoprotic acid, HA, is dissolved in water: HA + H2O ⇋ H3O+ + A-The equilibrium concentrations of the reactants and products are [HA] = 0.230 M[H3O+] = 2.00x10-4 M [A-] = 2.00x10-4 MCalculate the K

Solution:  A monoprotic acid, HA, is dissolved in water: HA + H2O ⇋ H3O+ + A-The equilibrium concentrations of the reactants and products are [HA] = 0.230 M[H3O+] = 2.00x10-4 M [A-] = 2.00x10-4 MCalculate the K

Problem

 A monoprotic acid, HA, is dissolved in water: 

HA + H2O ⇋ H3O+ + A-

The equilibrium concentrations of the reactants and products are 

[HA] = 0.230 M

[H3O+] = 2.00x10-4

[A-] = 2.00x10-4 M

Calculate the Ka value for the acid HA.