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: Calculate the pH of a 0.080 M carbonic acid solution, H 2CO3(aq), that has the stepwise dissociation constants Ka1 = 4.3 × 10 -7 and Ka2 = 5.6 × 10 -11.A) 1.10B) 3.73C) 6.37D) 10.25

Solution: Calculate the pH of a 0.080 M carbonic acid solution, H 2CO3(aq), that has the stepwise dissociation constants Ka1 = 4.3 × 10 -7 and Ka2 = 5.6 × 10 -11.A) 1.10B) 3.73C) 6.37D) 10.25

Problem

Calculate the pH of a 0.080 M carbonic acid solution, H 2CO3(aq), that has the stepwise dissociation constants Ka1 = 4.3 × 10 -7 and Ka2 = 5.6 × 10 -11.

A) 1.10

B) 3.73

C) 6.37

D) 10.25

Solution

We’re being asked to calculate the pH of a 0.080 M carbonic acid (H2CO3). Carbonic acid is a diprotic acid, it can donate two protons (H+and it will have two equilibrium reactions

To calculate for pH, we use the following equation:


Given:

Ka1 = 4.3x10-7

1st equilibrium reaction:                H2CO3(aq) + H2O(l) → HCO3­-(aq)H3O+(aq

Ka2 = 5.6x10-11

2nd equilibrium reaction:               HCO3-(aq) + H2O(l) → CO32­-(aq)H3O+(aq)

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

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