Chemistry
Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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Textbook Question
Chapter 15, Problem 70E

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.

Expert Solution & Answer
Check Mark
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.

Answer to Problem 70E

Answer

  • The pH when 0.0mLof0.100MHCl is added to 25mLof0.100Mpyridine is 9.11_.
  • The pH when 4.0mLof0.100MHCl is added to 25mLof0.100Mpyridine is 5.92_.
  • The pH when 8.0mLof0.100MHCl is added to 25mLof0.100Mpyridine is 5.53_.
  • The pH when 12.5mLof0.100MHCl is added to 25mLof0.100Mpyridine is 5.2_.
  • The pH when 20.0mLof0.100MHCl is added to 25mLof0.100Mpyridine is 4.6_.
  • The pH when 24.5mLof0.100MHCl is added to 25mLof0.100Mpyridine is 3.82_.
  • The pH when 25.0mLof0.100MHCl is added to 25mLof0.100Mpyridine is 0.05_.
  • The pH when 26.0mLof0.100MHCl is added to 25mLof0.100Mpyridine is 2.71_.
  • The pH when 28.0mLof0.100MHCl is added to 25mLof0.100Mpyridine is 2.24_.
  • The pH when 30.0mLof0.100MHCl is added to 25mLof0.100Mpyridine is 2.04_.

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+HClC5H5NHClNumberofMolesbeforethereaction0.00250.00240NumberofMolesafterthereaction0.000100.0024

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 og base, acid and pKa in the above expression.

pH=5.2+log[0.0010.0025]pH=3.82

At 25.0mL addition of HCl to the 25mLof0.100Mpyridine,the concentration of pyridinium ion at equilibrium is calculated by using the stoichiometry of the reaction.

C5H5N+HClC5H5NHClNumberofMolesbeforethereaction0.00250.00250NumberofMolesafterthereaction000.0025

At this stage complete consumption of pyridine and HCl takes place with the formation of salt. The strength of the salt is calculated by considering the dissociation reaction of salt formed as.

C5H5NHClC5H5NH++Cl

The concentration of  pyridium ion (C5Hϕ5NH+)=MolesofpyridiumionVolume

Substitute the value of moles and volume in the above expression.

Concentration=0.00250.05Concentration=0.05M

The value of [H+] is calculated by using ICE table at dissociation reaction of pyridium ion.

C5H5NH++H2OH3O++C5H5NInitial(M):0.0500Change(M):yyyEquilibrium(M):0.05yyy

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

Ka=[Concentrationofproducts][Concentrationofreactants]

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

Ka=[C5H5NH+][OH][C5H5N]

The relation,

Ka=KwKb

The value of Kb is 1.7×109.

Substitute the value of Kw and Kb in the above expression.

Ka=1.0×10141.7×109Ka=5.88×106

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

5.88×106=x×x0.05xx2+5.88×1062.94×107=0x=5.42×104

Hence, the value of [H+] is 0.000013M

The value of pH is calculated by the formula.

pH=log[H+]

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

pH=log[5.42×104]pH=3.3_

At 26.0mL addition of HCl to the 25mLof0.100Mpyridine,the concentration of pyridinium ion at equilibrium is calculated by using the stoichiometry of the reaction.

C5H5N+HClC5H5NHClNumberofMolesbeforethereaction0.00250.00260NumberofMolesafterthereaction00.00010.0025

The [H+] value =0.00010.025+0.026=0.00196M

The value of pH is calculated by the formula.

pH=log[H+]

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

pH=log[0.00196]pH=2.71_

At 28.0mL addition of HCl to the 25mLof0.100Mpyridine,the concentration of pyridinium ion at equilibrium is calculated by using reaction stoichiometric coefficients.

C5H5N+HClC5H5NHClNumberofMolesbeforethereaction0.00250.00280NumberofMolesafterthereaction00.00030.0025

The [H+] value =0.00030.025+0.026=0.00566M

The value of pH is calculated by the formula.

pH=log[H+]

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

pH=log[0.00566]pH=2.24_

At 30.0mL addition of HCl to the 25mLof0.100Mpyridine,the concentration of pyridinium ion at equilibrium is calculated by using reaction stoichiometric coefficients.

C5H5N+HClC5H5NHClNumberofMolesbeforethereaction0.00250.0300NumberofMolesafterthereaction00.00050.0025

The [H+] value =0.00050.025+0.030=0.009M

The value of pH is calculated by the formula.

pH=log[H+]

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

pH=log[0.009]pH=2.04_

The graph between pH and the volumeofHClused is,

Chemistry, Chapter 15, Problem 70E

Conclusion

Conclusion

  • The pH when 0.0mLof0.100MHCl is added to 25mLof0.100Mpyridine is 9.11_.
  • The pH when 4.0mLof0.100MHCl is added to 25mLof0.100Mpyridine is 5.92_.
  • The pH when 8.0mLof0.100MHCl is added to 25mLof0.100Mpyridine is 5.53_.
  • The pH when 12.5mLof0.100MHCl is added to 25mLof0.100Mpyridine is 5.2_.
  • The pH when 20.0mLof0.100MHCl is added to 25mLof0.100Mpyridine is 4.6_.
  • The pH when 24.5mLof0.100MHCl is added to 25mLof0.100Mpyridine is 3.82_.
  • The pH when 25.0mLof0.100MHCl is added to 25mLof0.100Mpyridine is 0.05_.
  • The pH when 26.0mLof0.100MHCl is added to 25mLof0.100Mpyridine is 2.71_.
  • The pH when 28.0mLof0.100MHCl is added to 25mLof0.100Mpyridine is 2.24_.
  • The pH when 30.0mLof0.100MHCl is added to 25mLof0.100Mpyridine is 2.04_.

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Chapter 15 Solutions

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