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: In a 0.050 M solution of a weak monoprotic acid, [H  +] = 1.8 x 10 –3 . What is the Ka? (A) 3.6 x 10 –2 (B) 9.0 x 10 –5 (C) 6.7 x 10 –5 (D) 1.6 x 10 –7

Problem

In a 0.050 M solution of a weak monoprotic acid, [H  +] = 1.8 x 10 –3 . What is the Ka?

(A) 3.6 x 10 –2

(B) 9.0 x 10 –5

(C) 6.7 x 10 –5

(D) 1.6 x 10 –7