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

A diprotic acid is an acid that can donate two hydronium ions (H+). 

Diprotic Acids

Since a diprotic acid has two acidic hydrogens it would have 2 equilibrium equations. 

Concept #1: Diprotic Acids

The diprotic bases would also have 2 equilibrium equations. 

Concept #2: Diprotic Acids

Concept #3: As a result of these equations for diprotic acids and bases the relationship between Ka and Kb can be established. 

Example #1: Sulfurous acid, H2SO3, represents a diprotic acid with a Ka1 = 1.6 x 10-2 and Ka2 = 4.6 x 10-5. Calculate the pH and concentrations of H2SO3, HSO3 and SO32– when given 0.250 M H2SO3

Example #2: Determine the pH of 0.115 M Na2S. Hydrosulfuric acid, H2S, contains Ka1 = 1.0 x 10-7 and Ka2 = 9.1 x 10-8.