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: Write the equation for the reaction associated with:The Ka2 of sulfuric acid (H2SO4).The Kb2 of carbonate in water (CO32-)Explain in your own words why you do not need a K-value to calculate the disso

Solution: Write the equation for the reaction associated with:The Ka2 of sulfuric acid (H2SO4).The Kb2 of carbonate in water (CO32-)Explain in your own words why you do not need a K-value to calculate the disso

Problem

Write the equation for the reaction associated with:

The Ka2 of sulfuric acid (H2SO4).

The Kb2 of carbonate in water (CO32-)

Explain in your own words why you do not need a K-value to calculate the dissociation of a strong acid?

Solution

Recall: Acid dissociation constant (Ka) - the equilibrium constant corresponding to loss of proton of a weak acid. It's counterpart, the base dissociation constant (Kb), is the equilibrium constant corresponding to when a weak base acquires a proton. Di- and polyprotic acids have multiple Ka or Kb values due to them being capable of losing or acquiring multiple protons.

View the complete written solution...