Chemistry: An Atoms First Approach
Chemistry: An Atoms First Approach
2nd Edition
ISBN: 9781305079243
Author: Steven S. Zumdahl, Susan A. Zumdahl
Publisher: Cengage Learning
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Chapter 14, Problem 61E

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 61E

  • 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: An Atoms First Approach, Chapter 14, Problem 61E

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 14 Solutions

Chemistry: An Atoms First Approach

Ch. 14 - What are the major species in solution after...Ch. 14 - Prob. 2ALQCh. 14 - Prob. 3ALQCh. 14 - Prob. 4ALQCh. 14 - Sketch two pH curves, one for the titration of a...Ch. 14 - Prob. 6ALQCh. 14 - Prob. 7ALQCh. 14 - You have a solution of the weak acid HA and add...Ch. 14 - The common ion effect for weak acids is to...Ch. 14 - Prob. 10QCh. 14 - Prob. 11QCh. 14 - Consider the following pH curves for 100.0 mL of...Ch. 14 - An acid is titrated with NaOH. The following...Ch. 14 - Consider the following four titrations. i. 100.0...Ch. 14 - Prob. 15QCh. 14 - Prob. 16QCh. 14 - How many of the following are buffered solutions?...Ch. 14 - Which of the following can be classified as buffer...Ch. 14 - A certain buffer is made by dissolving NaHCO3 and...Ch. 14 - Prob. 20ECh. 14 - Calculate the pH of each of the following...Ch. 14 - Calculate the pH of each of the following...Ch. 14 - Prob. 23ECh. 14 - Compare the percent ionization of the base in...Ch. 14 - Prob. 25ECh. 14 - Calculate the pH after 0.020 mole of HCl is added...Ch. 14 - Calculate the pH after 0.020 mole of NaOH is added...Ch. 14 - Calculate the pH after 0.020 mole of NaOH is added...Ch. 14 - Which of the solutions in Exercise 21 shows the...Ch. 14 - Prob. 30ECh. 14 - Calculate the pH of a solution that is 1.00 M HNO2...Ch. 14 - Calculate the pH of a solution that is 0.60 M HF...Ch. 14 - Calculate the pH after 0.10 mole of NaOH is added...Ch. 14 - Calculate the pH after 0.10 mole of NaOH is added...Ch. 14 - Calculate the pH of each of the following buffered...Ch. 14 - Prob. 36ECh. 14 - Calculate the pH of a buffered solution prepared...Ch. 14 - A buffered solution is made by adding 50.0 g NH4Cl...Ch. 14 - Prob. 39ECh. 14 - An aqueous solution contains dissolved C6H5NH3Cl...Ch. 14 - Prob. 41ECh. 14 - Prob. 42ECh. 14 - Consider a solution that contains both C5H5N and...Ch. 14 - Calculate the ratio [NH3]/[NH4+] in...Ch. 14 - Prob. 45ECh. 14 - Prob. 46ECh. 14 - Prob. 47ECh. 14 - Prob. 48ECh. 14 - Calculate the pH of a solution that is 0.40 M...Ch. 14 - Calculate the pH of a solution that is 0.20 M HOCl...Ch. 14 - Which of the following mixtures would result in...Ch. 14 - Prob. 52ECh. 14 - Prob. 53ECh. 14 - Calculate the number of moles of HCl(g) that must...Ch. 14 - Consider the titration of a generic weak acid HA...Ch. 14 - Sketch the titration curve for the titration of a...Ch. 14 - Consider the titration of 40.0 mL of 0.200 M HClO4...Ch. 14 - Consider the titration of 80.0 mL of 0.100 M...Ch. 14 - Consider the titration of 100.0 mL of 0.200 M...Ch. 14 - Prob. 60ECh. 14 - Lactic acid is a common by-product of cellular...Ch. 14 - Repeat the procedure in Exercise 61, but for the...Ch. 14 - Repeat the procedure in Exercise 61, but for the...Ch. 14 - Repeat the procedure in Exercise 61, but for the...Ch. 14 - Prob. 65ECh. 14 - In the titration of 50.0 mL of 1.0 M methylamine,...Ch. 14 - You have 75.0 mL of 0.10 M HA. After adding 30.0...Ch. 14 - A student dissolves 0.0100 mole of an unknown weak...Ch. 14 - Prob. 69ECh. 14 - Prob. 70ECh. 14 - Potassium hydrogen phthalate, known as KHP (molar...Ch. 14 - A certain indicator HIn has a pKa of 3.00 and a...Ch. 14 - Prob. 73ECh. 14 - Prob. 74ECh. 14 - Prob. 75ECh. 14 - Prob. 76ECh. 14 - Prob. 77ECh. 14 - Estimate the pH of a solution in which crystal...Ch. 14 - Prob. 79ECh. 14 - Prob. 80ECh. 14 - Prob. 81AECh. 14 - Prob. 82AECh. 14 - Tris(hydroxymethyl)aminomethane, commonly called...Ch. 14 - Prob. 84AECh. 14 - You have the following reagents on hand: Solids...Ch. 14 - Prob. 86AECh. 14 - Prob. 87AECh. 14 - What quantity (moles) of HCl(g) must be added to...Ch. 14 - Calculate the value of the equilibrium constant...Ch. 14 - The following plot shows the pH curves for the...Ch. 14 - Calculate the volume of 1.50 102 M NaOH that must...Ch. 14 - Prob. 92AECh. 14 - A certain acetic acid solution has pH = 2.68....Ch. 14 - A 0.210-g sample of an acid (molar mass = 192...Ch. 14 - The active ingredient in aspirin is...Ch. 14 - One method for determining the purity of aspirin...Ch. 14 - A student intends to titrate a solution of a weak...Ch. 14 - Prob. 98AECh. 14 - Prob. 99AECh. 14 - Consider 1.0 L of a solution that is 0.85 M HOC6H5...Ch. 14 - Prob. 101CWPCh. 14 - Consider the following acids and bases: HCO2H Ka =...Ch. 14 - Prob. 103CWPCh. 14 - Prob. 104CWPCh. 14 - Consider the titration of 100.0 mL of 0.100 M HCN...Ch. 14 - Consider the titration of 100.0 mL of 0.200 M...Ch. 14 - Prob. 107CWPCh. 14 - Prob. 108CPCh. 14 - A buffer is made using 45.0 mL of 0.750 M HC3H5O2...Ch. 14 - A 0.400-M solution of ammonia was titrated with...Ch. 14 - Prob. 111CPCh. 14 - Consider a solution formed by mixing 50.0 mL of...Ch. 14 - When a diprotic acid, H2A, is titrated with NaOH,...Ch. 14 - Consider the following two acids: In two separate...Ch. 14 - The titration of Na2CO3 with HCl bas the following...Ch. 14 - Prob. 116CPCh. 14 - A few drops of each of the indicators shown in the...Ch. 14 - Malonic acid (HO2CCH2CO2H) is a diprotic acid. In...Ch. 14 - A buffer solution is prepared by mixing 75.0 mL of...Ch. 14 - A 10.00-g sample of the ionic compound NaA, where...Ch. 14 - Prob. 121IPCh. 14 - Prob. 122MP
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