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: Complete the Ka2 expression for H2CO3 in an aqueous solution. Ka2 = 4.69 x 10-11


Complete the Ka2 expression for H2CO3 in an aqueous solution. Ka2 = 4.69 x 10-11


For this problem, we are asked for the Ka2 expression so we don't actually need to compute anything. We just have to set it up.

Know that H2CO3 (Carbonic acid) is a polyprotic acid which means we have more than 1 acidic hydrogen that can dissociate. In this specific acid, we have 2 acidic hydrogens. Shown below is the form of carbonic acid for every step that it losses each hydrogen. We also have the proper Ka's that we will use for each removal of hydrogen. Since we are asked for Ka2 we focus on the highlighted step and we start with the intermediate form in the center. 

Solution BlurView Complete Written Solution