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: pOH = 3.14 is equivalent to:       A) [OH −] = 3.14 × 10 −7 M B) pH = 11. C) [H +] = 1.4 × 10 −10 M D) [OH −] = 7.2 × 10 −4 M E) [H +] = 7.0 × 10 −4 M

Problem

pOH = 3.14 is equivalent to:      

A) [OH ] = 3.14 × 10 −7 M

B) pH = 11.

C) [H +] = 1.4 × 10 −10 M

D) [OH ] = 7.2 × 10 −4 M

E) [H +] = 7.0 × 10 −4 M