Ch.15 - Acid and Base EquilibriumWorksheetSee all chapters
All Chapters
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
Problem

What is Kb for pyridine (C5H5N) if Ka for its conjugate acid is 5.9 x 10-6 at 25°C?

A) 1.7 x 10-9 

B) 5.9 x 108

C) 1 x 1014

D) 1.7 x 1019 

Solution

We are asked to find Kb for C5H5N given that the Ka of its conjugate acid (C5H5NH+) is 5.9x10-6

The dissociation of C5H5NH+ with water will appear as:

C5H5NH(aq) + H2O (l)  ⇌ C5H5NH (aq) + OH- (aq)                       Ka = 5.9x10-6

Recall that Ka and Kb are connected by the autoionization constant of water (Kwby the following equation:

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