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: The Ka of a monoprotic weak acid is  2.08 × 10−3. What is the percent ionization of a .194 M solution of this acid?

Problem

The Ka of a monoprotic weak acid is  2.08 × 10−3. What is the percent ionization of a .194 M solution of this acid?

Solution

We’re being asked to calculate the percent ionization of a 0.194 M aqueous solution of a monoprotic acid HA.


Since HA has a low Ka value, it’s a weak acid. Remember that weak acids partially dissociate in water and that acids donate H+ to the base (water in this case). The dissociation of HA is as follows:


HA(aq) + H2O(l)  H3O+(aq) + A(aq); Ka = 2.08 × 10−3


From this, we can construct an ICE table. Remember that liquids are ignored in the ICE table.

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