   Chapter 15, Problem 61E

Chapter
Section
Textbook Problem

# 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.

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.

Explanation

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

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

#### The bladder straddles the cardiovascular system and filters the blood. T F

Nutrition: Concepts and Controversies - Standalone book (MindTap Course List)

#### True or false? Animal cells do not have walls.

Biology: The Unity and Diversity of Life (MindTap Course List)

#### Which elements make up about 99% of the mass of living things?

Oceanography: An Invitation To Marine Science, Loose-leaf Versin 