   Chapter 14, Problem 102E

Chapter
Section
Textbook Problem

# Calculate the percentage of pyridine (C5H5N) that forms pyridinium ion, C5H5NH+, in a 0.10-M aqueous solution of pyridine (Kb = 1.7 × 10−9).

Interpretation Introduction

Interpretation: The percent of pyridine (C5H5N) that leads to the formation of pyridium ion (C5H5NH+) in a 0.10M aqueous solution of pyridine is to be calculated.

Concept introduction: The pH of a solution is defined as a figure that expresses the acidity of the alkalinity of a given solution.

The pH of a solution is calculated by the formula, pH=log[H+]

At equilibrium, the equilibrium constant expression is expressed by the formula,

Kb=ConcentrationofproductsConcentrationofreactants

The percent ionization of an acid is calculated by the formula,

Percentionization=Equilibriumconcentrationof[OH]Initialconcentrationofthebase×100

Explanation

Explanation

To determine: The percent of pyridine that leads to the formation of pyridium ion (C5H5NH+) in the given aqueous solution of pyridine.

The equilibrium constant expression for the stated ionization reaction is,

Kb=[C5H5NH+][OH][C5H5N]

C5H5N is a weaker base.

The dominant equilibrium reaction for the given case is,

C5H5N(aq)+H2O(l)C5H5NH+(aq)+OH(aq)

At equilibrium, the equilibrium constant expression is expressed by the formula,

Kb=ConcentrationofproductsConcentrationofreactants

Where,

• Kb is the base ionization constant.

The equilibrium constant expression for the given reaction is,

Kb=[C5H5NH+][OH][C5H5N] (1)

The [OH] is 1.30×10-5M_ .

The change in concentration of C5H5N is assumed to be x .

The liquid components do not affect the value of the rate constant.

The ICE table for the stated reaction is,

C5H5N(aq)C5H5NH+(aq)+OH(aq)Inititialconcentration0.1000Changex+x+xEquilibriumconcentration0.10xxx

The equilibrium concentration of [C5H5N] is (0.10x)M .

The equilibrium concentration of [C5H5NH+] is xM

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