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: Consider a 0.10 M solution of a weak polyprotic acid (H2 A) with the possible values of Ka1  and Ka2  given below. Calculate the contributions to [H3O+] from each ionization step.Ka1  = 1.0 10 - 4 ; Ka2  = 1.0 10 - 5

Solution: Consider a 0.10 M solution of a weak polyprotic acid (H2 A) with the possible values of Ka1  and Ka2  given below. Calculate the contributions to [H3O+] from each ionization step.Ka1  = 1.0 10 - 4 ; K

Problem

Consider a 0.10 M solution of a weak polyprotic acid (H2 A) with the possible values of Ka1  and Ka2  given below. Calculate the contributions to [H3O+] from each ionization step.

Ka1  = 1.0 10 - 4 ; Ka2  = 1.0 10 - 5

Solution

We’re given a weak diprotic acid and we are asked to calculate the contributions to [H3O+] from each ionization step.


Given:

Ka1 = 1.0 x10-4

1st equilibrium reaction:                H2A(aq) + H2O(l) → HA­-(aq)H3O+(aq

Ka2 = 1.0 x10-5

2nd equilibrium reaction:               HA-(aq) + H2O(l) → A2­-(aq)H3O+(aq)

H3O+(aq) is also produced from the 2nd equilibrium reaction

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