   Chapter 16, Problem 96GQ

Chapter
Section
Textbook Problem

Pyridine is weak organic base and readily forms a salt with hydrochloric acid. C 5 H 5 N(aq) + HCl(aq)  →  C 5 H 5 NH + (aq) + Cl − (aq)    Pyridine                          Pyridinium ion What is the pH of a 0.025 M solution of pyridinium hydrochloride [C 5 H 5 NH + ]Cl − ?

Interpretation Introduction

Interpretation:

The pH of 0.025M solution of pyridinium hydrochloride has to be determined.

Concept introduction:

Equilibrium constants:

The equilibrium constant is used to quantitative measurement of the strength of the acid and bases in the water.

Ka is an acid constant for equilibrium reactions.

HA + H2OH3O++ A-Ka[H3O+][A-][HA]

Kb is a base constant for equilibrium reaction.

BOH + H2OB++ OH-Ka[B+][OH-][BOH]

Ion product constant for water

Kw= [H3O+][OH-]       =1.00×10-14 pH = -log[H3O+]pOH= -log[OH-]

Relation between pH and pOH

pH + pOH =14

Explanation

Let’s calculate the Ka of C5H5NH+

The Kb of 1.5× 10-9

Ka×Kb = KwKaKwKbKw= 1.0×10-14Kb = 1.5×10-9

Substitute the values,

Ka1.0×10-141.5×10-9     = 6.7×10-6

The equilibrium chemical reaction is as follows.

C5H5NH+(aq) + H2OH3O+(aq) + C5H5N(aq)

The equilibrium expression:

Ka[C5H5N][H3O+][C5H5NH+]

The initial concentration of pyridinium ion is 0.025M.

C5H5NH+(aq) + H2O(l) H3O+(aq) + C5H5N(aq)I         0.025M              --                --                     --C           -x                    --              + x                  + xE        (0

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