   # Repeat the procedure in Exercise 61, but for the titration of 25.0 mL of 0.100 M pyridine with 0.100 M hydrochloric acid ( K b for pyridine is 1.7 × 10 −9 ). Do not calculate the points at 24.9 and 25.1 mL. ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 14, Problem 64E
Textbook Problem
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## Repeat the procedure in Exercise 61, but for the titration of 25.0 mL of 0.100 M pyridine with 0.100 M hydrochloric acid (Kb for pyridine is 1.7 × 10−9). Do not calculate the points at 24.9 and 25.1 mL.

Interpretation Introduction

Interpretation: The pH of the solution when 25mL of 0.100M pyridine is titrated with various given volumes of 0.100MHCl and graph between calculated pH and milliliters of NaOH added is to be stated.

Concept introduction: Pyridine is a weak base and HCl is a strong acid. When mixed together in a solution, neutralization reaction takes place with the formation of a salt and water molecule.

Concept introduction: Pyridine is a weak base and HCl is a strong acid. When mixed together in a solution, neutralization reaction takes place with the formation of a salt and water molecule.

### Explanation of Solution

Explanation

Given

The value of Kb for pyridine is 1.7×109.

At 0.0mL, addition of HCl to the 25mLof0.100Mpyridine, the concentration of pyridinium ion at equilibrium is calculated by using ICE (Initial Change Equilibrium) table.

C5H5N+H2OC5H5NH++OHInitial(M):0.10000Change(M):xxxEquilibrium(M):0.100xxx

The value of base dissociation constant at equilibrium is calculated by the formula.

Kb=[Concentrationofproducts][Concentrationofreactants]

Therefore, for the above reaction the value of base dissociation constant at equilibrium is,

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

Substitute the values of concentration of reactants, products and Kb in the above expression.

1.7×109×109=x×x0.100xx2+1.7×1091.7×1010=0x=0.000013

Hence, the value of [OH] is 0.000013M

The ionic-product of water is,

[H+][OH]=1.0×1014

Substitute the value of [OH] in the above expression.

[H+]×0.000013=1.0×1014[H+]=1.0×10140.000013[H+]=7.69×1010M

The value of pH is calculated by the formula.

pH=log[H+]

Substitute the value of [H+] in the above expression.

pH=log[7.69×1010]pH=9.11_

At 4.0mL, addition of HCl to the 25mLof0.100Mpyridine, the concentration of pyridinium ion at equilibrium is calculated by using the stoichiometry of the reaction.

C5H5N+HClC5H5NHClNumberofMolesbeforethereaction0.00250.00040NumberofMolesafterthereaction0.002100.0004

The value of pH is calculated by the Henderson-Hasselbalch equation.

pH=pKa+log[Base][Acid]

Where,

• pKa is negative logarithm of Ka.
• pH is power of [H+].
• [Base] is concentration of base after the reaction.
• [Acid] is concentration of acid before the reaction.

The standard value of pKa for pyridine is 5.2.

Substitute the values of concentration of base, acid and pKa in the above expression.

pH=5.2+log[0.00210.0004]pH=5.92

At 8.0mL, addition of HCl to the 25mLof0.100Mpyridine,the concentration of pyridinium ion at equilibrium is calculated by using the stoichiometry of the reaction.

C5H5N+HClC5H5NHClNumberofMolesbeforethereaction0.00250.00080NumberofMolesafterthereaction0.001700.0008

The value of pH is calculated by the Henderson-Hasselbalch equation.

pH=pKa+log[Base][Acid]

Where,

• pKa is negative logarithm of Ka.
• pH is power of [H+].
• [Base] is concentration of base after the reaction.
• [Acid] is concentration of acid before the reaction.

The standard value of pKa for pyridine is 5.2.

Substitute the values of concentration of base, acid and pKa in the above expression.

pH=5.2+log[0.00170.0008]pH=5.53

At 12.5mL, addition of HCl to the 25mLof0.100Mpyridine,the concentration of pyridinium ion at equilibrium is calculated by using the stoichiometry of the reaction.

C5H5N+HClC5H5NHClNumberofMolesbeforethereaction0.00250.001250NumberofMolesafterthereaction0.0012500.00125

The value of pH is calculated by the Henderson-Hasselbalch equation.

pH=pKa+log[Base][Acid]

Where,

• pKa is negative logarithm of Ka.
• pH is power of [H+].
• [Base] is concentration of base after the reaction.
• [Acid] is concentration of acid before the reaction.

The standard value of pKa for pyridine is 5.2.

Substitute the values of concentration of base, acid and pKa in the above expression.

pH=5.2+log[0.001250.00125]pH=5.2

At 20.0mL addition of HCl to the 25mLof0.100Mpyridine,the concentration of pyridinium ion at equilibrium is calculated by using the stoichiometry of the reaction.

C5H5N+HClC5H5NHClNumberofMolesbeforethereaction0.00250.00200NumberofMolesafterthereaction0.00500.0020

The value of pH is calculated by the Henderson-Hasselbalch equation.

pH=pKa+log[Base][Acid]

Where,

• pKa is negative logarithm of Ka.
• pH is power of [H+].
• [Base] is concentration of base after the reaction.
• [Acid] is concentration of acid before the reaction.

The standard value of pKa for pyridine is 5.2.

Substitute the values of concentration of base, acid and pKa in the above expression.

pH=5.2+log[0.00050.002]pH=4.6

At 24.5mL addition of HCl to the 25mLof0.100Mpyridine,the concentration of pyridinium ion at equilibrium is calculated by using the stoichiometry of the reaction.

C5H5N+HClC5H5NHClNumberofMolesbeforethereaction0

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