Chemistry
Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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Chapter 15, Problem 67E

Lactic acid is a common by-product of cellular respiration and is often said to cause the “burn” associated with strenuous activity. A 25.0-mL sample of 0.100 M lactic acid (HC3H5O3, pKa = 3.86) is titrated with 0.100 M NaOH solution. Calculate the pH after the addition of 0.0 mL, 4.0 mL, 8.0 mL, 12.5 mL, 20.0 mL, 24.0 mL, 24.5 mL, 24.9 mL, 25.0 mL, 25.1 mL, 26.0 mL, 28.0 mL, and 30.0 mL of the NaOH. Plot the results of your calculations as pH versus milliliters of NaOH added.

Expert Solution & Answer
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Interpretation Introduction

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

Concept introduction: Lactic acid (HC3H5O3) is a weak acid and NaOH is a strong base. When mixed together in a solution, neutralization reaction takes place with the formation of a salt and water molecule.

Answer to Problem 67E

  • The pH when 0.0mLof0.100MNaOH is added to 25mLof0.100MHC3H5O3 is 2.43_.
  • The pH when 4.0mLof0.100MNaOH is added to 25mLof0.100MHC3H5O3 is 3.06_.
  • The pH when 8.0mLof0.100MNaOH is added to 25mLof0.100MHC3H5O3 is 3.53_.
  • The pH when 12.5mLof0.100MNaOH is added to 25mLof0.100MHC3H5O3 is 3.86_.
  • The pH when 20.0mLof0.100MNaOH is added to 25mLof0.100MHC3H5O3 is 4.46_.
  • The pH when 24.5mLof0.100MNaOH is added to 25mLof0.100MHC3H5O3 is 4.24_.
  • The pH when 25.0mLof0.100MNaOH is added to 25mLof0.100MHC3H5O3 is 8.27_.
  • The pH when 25.1mLof0.100MNaOH is added to 25mLof0.100MHC3H5O3 is 10.29_.
  • The pH when 26.0mLof0.100MNaOH is added to 25mLof0.100MHC3H5O3 is 11.29_.
  • The pH when 28.0mLof0.100MNaOH is added to 25mLof0.100MHC3H5O3 is 11.74_.
  • The pH when 30.0mLof0.100MNaOH is added to 25mLof0.100MHC3H5O3 is 11.95_.

Explanation of Solution

Explanation

On 0.0mL addition of NaOH to the 25mLof0.100MHC3H5O3, the concentration of lactate ion at equilibrium is calculated by using ICE (Initial Change Equilibrium) table.

HC5H5O3+H2OC5H5O3+H3O+Initial(M):0.10000Change(M):xxxEquilibrium(M):0.100xxx

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,

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

The standard value of Ka for lactic acid is 1.38×104.

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

1.38×104=x×x0.100xx2+1.38×1041.38×105=0x=3.64×103

Hence, the value of [H+] is 3.64×103M.

The value of pH is calculated by the formula.

pH=log[H+]

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

pH=log[3.64×103]pH=2.43_

on 4.0mL addition of NaOH to the 25mLof0.100MHC3H5O3,the concentration of lactate ion at equilibrium is calculated by using reaction stoichiometric coefficients.

HC5H5O3+NaOHC5H5O3NaNumberofMolesbeforethereaction0.00250.00040NumberofMolesafterthereaction0.002100.0004

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

pH=pKa+log[Salt][Acid]

Where,

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

The standard value of pKa for pyridine is 3.86.

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

pH=3.86+log[0.00040.0025]pH=3.06_

On 8.0mL addition of NaOH to the 25mLof0.100MHC3H5O3, the concentration of lactate ion at equilibrium is calculated by using reaction stoichiometric coefficients.

HC5H5O3+NaOHC5H5O3NaNumberofMolesbeforethereaction0.00250.00080NumberofMolesafterthereaction0.001700.0008

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

pH=pKa+log[Salt][Acid]

Where,

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

The standard value of pKa for pyridine is 3.86.

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

pH=3.86+log[0.00080.0017]pH=3.53_

On 12.5mL addition of NaOH to the 25mLof0.100MHC3H5O3, the concentration of lactate ion at equilibrium is calculated by using reaction stoichiometric coefficients.

HC5H5O3+NaOHC5H5O3NaNumberofMolesbeforethereaction0.00250.001250NumberofMolesafterthereaction0.00012500.000125

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

pH=pKa+log[Salt][Acid]

Where,

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

The standard value of pKa for pyridine is 3.86.

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

pH=3.86+log[0.001250.00125]pH=3.86_

On 20.0mL addition of NaOH to the 25mLof0.100MHC3H5O3, the concentration of lactate ion at equilibrium is calculated by using reaction stoichiometric coefficients.

HC5H5O3+NaOHC5H5O3NaNumberofMolesbeforethereaction0.00250.0020NumberofMolesafterthereaction0.000500.002

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

pH=pKa+log[Salt][Acid]

Where,

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

The standard value of pKa for pyridine is 3.86.

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

pH=3.86+log[0.0020.0025]pH=4.46_

On 24.5mL addition of NaOH to the 25mLof0.100MHC3H5O3, the concentration of lactate ion at equilibrium is calculated by using reaction stoichiometric coefficients.

HC5H5O3+NaOHC5H5O3NaNumberofMolesbeforethereaction0.00250.00240NumberofMolesafterthereaction0.000100.0024

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

pH=pKa+log[Salt][Acid]

Where,

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

The standard value of pKa for pyridine is 3.86.

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

pH=3.86+log[0.00240.0001]pH=4.24_

On 25.0mL addition of NaOH to the 25mLof0.100MHC3H5O3, the concentration of lactate ion at equilibrium is calculated by using reaction stoichiometric coefficients.

HC5H5O3+NaOHC5H5O3NaNumberofMolesbeforethereaction0.00250.00250NumberofMolesafterthereaction000.0025

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

C5H5O3NaC5H5O3+Na+

The concentration of lactate ion (C5H5O3)=MolesoflacateionVolume

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 lactate ion.

C5H5O3+H2OC5H5O3H+OHInitial(M):0.0500Change(M):yyyEquilibrium(M):0.05yyy

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

Kb=[Concentrationofproducts][Concentrationofreactants]

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

Kb=[C5H5OH][OH][C5H5O3]

The relation,

Kb=KwKa

The value of Ka is 1.38×104.

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

Kb=1.0×10141.38×104Kb=7.24×1011

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

7.24×1011=x×x0.05xx2+7.24×10113.62×1012=0x=1.90×106

Hence, the value of [OH] is 0.000013M

The value of [H+] is calculated by the formula.

[H+]=[1.0×1014][OH]

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

[H+]=[1.0×1014][1.90×106][H+]=5.26×109

The value of pH is calculated by the formula.

pH=log[H+]

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

pH=log[5.26×109]pH=8.27_

On 25.1mL addition of NaOH to the 25mLof0.100Mlaticacid, the concentration of lactate ion at equilibrium is calculated by using reaction stoichiometric coefficients.

HC3H5O3+NaOHC3H5O3NaNumberofMolesbeforethereaction0.00250.002510NumberofMolesafterthereaction00.000010.0025

The [OH] value =0.000010.025+0.026=0.000196M

The value of [H+] is calculated by the formula.

[H+]=[1.0×1014][OH]

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

[H+]=[1.0×1014][1.96×104][H+]=5.10×1011

The value of pH is calculated by the formula.

pH=log[H+]

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

pH=log[5.10×1011]pH=10.29_

On 26mL addition of NaOH to the 25mLof0.100Mlacticacid, the concentration of lactate ion at equilibrium is calculated by using reaction stoichiometric coefficients.

HC3H5O3+NaOHC3H5O3NaNumberofMolesbeforethereaction0.00250.00260NumberofMolesafterthereaction00.00010.0025

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

The value of [H+] is calculated by the formula,

[H+]=[1.0×1014][OH]

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

[H+]=[1.0×1014][1.96×103][H+]=5.10×1012

The value of pH is calculated by the formula,

pH=log[H+]

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

pH=log[1.96×104]pH=11.29_

On 28mL addition of NaOH to the 25mLof0.100Mlacticacid,the concentration of lactate ion at equilibrium is calculated by using reaction stoichiometric coefficients.

HC3H5O3+NaOHC3H5O3NaNumberofMolesbeforethereaction0.00250.00280NumberofMolesafterthereaction00.00030.0025

The [OH] value =0.00030.025+0.028=5.6×103M

The value of [H+] is calculated by the formula.

[H+]=[1.0×1014][OH]

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

[H+]=[1.0×1014][5.6×103][H+]=1.78×1012

The value of pH is calculated by the formula.

pH=log[H+]

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

pH=log[1.78×1012]pH=11.74_

At 30mL addition of NaOH to the 25mLof0.100Mlacticacid, the concentration of lactate ion at equilibrium is calculated by using reaction stoichiometric coefficients.

HC3H5O3+NaOHC3H5O3NaNumberofMolesbeforethereaction0.00250.00300NumberofMolesafterthereaction00.00050.0025

The [OH] value =0.00050.025+0.028=9.09×103M

The value of [H+] is calculated by the formula.

[H+]=[1.0×1014][OH]

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

[H+]=[1.0×1014][9.09×103][H+]=1.1×1012

The value of pH is calculated by the formula.

pH=log[H+]

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

pH=log[1.1×1012]pH=11.95_

The values of pH corresponding to the different volumes of NaOH added are shown in the following table.

VolumeofNaOHadded pH
0 2.43
4 3.06
8 3.53
12.5 3.86
20 4.46
24.5 4.24
25 8.27
25.1 10.29
26 11.29
28 11.74
30 11.95

Table 1

The graph between the volume of NaOH added and the corresponding pH values is stated as follows.

Chemistry, Chapter 15, Problem 67E

Conclusion

Conclusion

  • The pH when 0.0mLof0.100MNaOH is added to 25mLof0.100MHC3H5O3 is 2.43_.
  • The pH when 4.0mLof0.100MNaOH is added to 25mLof0.100MHC3H5O3 is 3.06_.
  • The pH when 8.0mLof0.100MNaOH is added to 25mLof0.100MHC3H5O3 is 3.53_.
  • The pH when 12.5mLof0.100MNaOH is added to 25mLof0.100MHC3H5O3 is 3.86_.
  • The pH when 20.0mLof0.100MNaOH is added to 25mLof0.100MHC3H5O3 is 4.46_.
  • The pH when 24.5mLof0.100MNaOH is added to 25mLof0.100MHC3H5O3 is 4.24_.
  • The pH when 25.0mLof0.100MNaOH is added to 25mLof0.100MHC3H5O3 is 8.27_.
  • The pH when 25.1mLof0.100MNaOH is added to 25mLof0.100MHC3H5O3 is 10.29_.
  • The pH when 26.0mLof0.100MNaOH is added to 25mLof0.100MHC3H5O3 is 11.29_.
  • The pH when 28.0mLof0.100MNaOH is added to 25mLof0.100MHC3H5O3 is 11.74_.
  • The pH when 30.0mLof0.100MNaOH is added to 25mLof0.100MHC3H5O3 is 11.95_

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Chapter 15 Solutions

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