Chemical Principles
Chemical Principles
8th Edition
ISBN: 9781305581982
Author: Steven S. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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Acid Base Titration
An acid-base titration is the method of quantitative analysis for determining the concentration of an acid and base by exactly neutralizing it with a standard solution of base or acid having known concentration.
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Chapter 8, Problem 67E
Interpretation Introduction

Interpretation:

The pH values after the addition of each proportion of the base to the acid is to be determined. Also, the titration curve needs to be drawn.

Concept introduction:

Titration curve is drawn to determine the change in pH of an acid or base with respect to the added volume of base or acid to it.

The titration curve can be drawn between a strong/weak acid and strong/weak base. The change in pH shows different patterns for different combinations of acids and bases.

Expert Solution & Answer
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Explanation of Solution

Initial pH of the analyte solution can be calculated as follows:

Lactic acid is a weak acid that forms an equilibrium mixture when dissolved in water. The equilibrium is as follows.

  HC3H5O3+H2O⇌C3H5O3−+H3O+

The initial molarity of lactic acid is 0.1 M.

The amount of lactic acid at the beginning can be calculated from. By constructing an ICE table, the concentration of lactate ion in the solution after the acid dissociation can be determined.

    Reaction Lactic acidLactateH+
    Initial 0.100
    Change -x+x+x
    Equilibrium (0.1-x)xx

The acid dissociation constant can be represented as follows:

  Ka=[C5H5O3−][H+][HC5H5O3]1.38×10−4=[x][x][0.1−x]=x2[0.1−x]1.38×10−5−1.38×10−4x=x20=x2+1.38×10−4x−1.38×10−5

Solving this quadratic equation gives the amount of hydrogen ions in the solution.

  0=x2+1.38×10−4x−1.38×10−5x=−b±b2−4ac2ax=−(1.38×10−4)±(1.38×10−4)2−4(1)(−1.38×10−5)2(1)x=−(1.38×10−4)±(1.90×10−8+1.38×10−5)2=−(1.38×10−4)±0.000013822=−(1.38×10−4)±0.003722=0.00372=0.00185

Thus, the concentration of hydrogen ion is 0.00185 and pH of the solution can be calculated as follows:

  pH=−log[H+]=−log[0.00185]=2.73

Addition of 4.0mLof base:

Total amount of lactic acid to be neutralized can be calculated from its molarity and volume as follows:

  n=M×V

Or,

  n=0.100mol/L×25.0×10−3L=2.5×10−3mol

Now, the amount of base added can be calculated as follows:

  n=0.100mol/L×4.0×10−3L=4×10−4mol

Then the ICE table after the addition of base is created in order to determine the pH of the solution using Henderson-Hasselbalch equation.

    Reaction Lactic acidOH-LactateH+
    Initial 0.0025000
    Add00.0004
    Change -0.0004-0.00040.00040.0004
    Equilibrium 0.002100.00040.0004

Concentration of lactic acid after addition of base =0.0021mol(25.0+4.0)×10−.3L=0.072mol/L

Concentration of lactate ion =0.0004mol(25.0+4.0)×10−.3L=0.014mol/L

Applying the Henderson-Hasselbalch equation,

  pH=pKa+log[lactate][lactic acid]pH=3.86+log[0.014][0.072]=3.14

Addition of 8.0mLof base:

Total amount of lactic acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×8.0×10−3L=8×10−4mol

Then the ICE table after the addition of base is created in order to determine the pH of the solution using Henderson-Hasselbalch equation.

    Reaction Lactic acidOH-LactateH+
    Initial 0.0025000
    Add00.0008
    Change -0.0008-0.00080.00080.0008
    Equilibrium 0.001700.00080.0008

Concentration of lactic acid after addition of base =0.0017mol(25.0+8.0)×10−.3L=0.051mol/L

Concentration of lactate ion =0.0008mol(25.0+8.0)×10−.3L=0.024mol/L

Applying the Henderson-Hasselbalch equation,

  pH=pKa+log[lactate][lactic acid]pH=3.86+log[0.024][0.051]=3.53

Addition of 12.5mLof base:

Total amount of lactic acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×12.5×10−3L=12.5×10−4mol

Then the ICE table after the addition of base is created in order to determine the pH of the solution using Henderson-Hasselbalch equation.

    Reaction Lactic acidOH-LactateH+
    Initial 0.0025000
    Add00.00125
    Change -0.00125-0.001250.001250.00125
    Equilibrium 0.0012500.001250.00125

Concentration of lactic acid after addition of base =0.00125mol(25.0+12.5)×10−.3L=0.033mol/L

Concentration of lactate ion =0.00125mol(25.0+4.0)×10−.3L=0.033mol/L

Applying the Henderson-Hasselbalch equation,

  pH=pKa+log[lactate][lactic acid]pH=3.86+log[0.033][0.033]=3.86

Addition of 20.0mLof base:

Total amount of lactic acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×20.0×10−3L=20×10−4mol

Then the ICE table after the addition of base is created in order to determine the pH of the solution using Henderson-Hasselbalch equation.

    Reaction Lactic acidOH-LactateH+
    Initial 0.0025000
    Add00.002
    Change -0.002-0.0020.0020.002
    Equilibrium 0.000500.0020.002

Concentration of lactic acid after addition of base =0.0005mol(25.0+20.0)×10−.3L=0.011mol/L

Concentration of lactate ion =0.002mol(25.0+20.0)×10−.3L=0.044mol/L

Applying the Henderson-Hasselbalch equation,

  pH=pKa+log[lactate][lactic acid]pH=3.86+log[0.044][0.011]=4.46

Addition of 24.0mLof base:

Total amount of lactic acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×24.0×10−3L=24×10−4mol

Then the ICE table after the addition of base is created in order to determine the pH of the solution using Henderson-Hasselbalch equation.

    Reaction Lactic acidOH-LactateH+
    Initial 0.0025000
    Add00.0024
    Change -0.0024-0.00240.00240.0024
    Equilibrium 0.000100.00240.0024

Concentration of lactic acid after addition of base =0.0001mol(25.0+24.0)×10−.3L=0.002mol/L

Concentration of lactate ion =0.0024mol(25.0+24.0)×10−.3L=0.049mol/L

Applying the Henderson-Hasselbalch equation,

  pH=pKa+log[lactate][lactic acid]pH=3.86+log[0.049][0.002]=5.24

Addition of 24.5mLof base:

Total amount of lactic acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×24.5×10−3L=24.5×10−4mol

Then the ICE table after the addition of base is created in order to determine the pH of the solution using Henderson-Hasselbalch equation.

    Reaction Lactic acidOH-LactateH+
    Initial 0.0025000
    Add00.00245
    Change -0.00245-0.00245-0.00245-0.00245
    Equilibrium 0.000050-0.00245-0.00245

Concentration of lactic acid after addition of base =0.00005mol(25.0+24.5)×10−.3L=0.0010mol/L

Concentration of lactate ion =0.00245mol(25.0+24.5)×10−.3L=0.049mol/L

Applying the Henderson-Hasselbalch equation,

  pH=pKa+log[lactate][lactic acid]pH=3.86+log[0.049][0.0010]=5.55≈5.6

Addition of 24.9mLof base:

Total amount of lactic acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×24.9×10−3L=24.9×10−4mol

Then the ICE table after the addition of base is created in order to determine the pH of the solution using Henderson-Hasselbalch equation.

    Reaction Lactic acidOH-LactateH+
    Initial 0.0025000
    Add00.00249
    Change -0.00249-0.00249-0.00249-0.00249
    Equilibrium 0.000010-0.00249-0.00249

Concentration of lactic acid after addition of base =0.00001mol(25.0+24.9)×10−.3L=0.0002mol/L

Concentration of lactate ion =0.00249mol(25.0+24.9)×10−.3L=0.049mol/L

Applying the Henderson-Hasselbalch equation,

  pH=pKa+log[lactate][lactic acid]pH=3.86+log[0.049][0.0002]=6.3

Addition of 25.0mLof base:

Total amount of lactic acid to be neutralized =0.100mol/L×25.0×10−3L=2.5×10−3mol

Amount of base added =0.100mol/L×25.0×10−3L=25×10−4mol

Then the ICE table after the addition of base is created in order to determine the pH of the solution using Henderson-Hasselbalch equation.

    Reaction Lactic acidOH-LactateH+
    Initial 0.0025000
    Add00.0025
    Change -0.0025-0.0025-0.0025-0.0025
    Equilibrium 0.00000-0.0025-0.0025

Concentration of lactic acid after addition of base =0

Concentration of lactate ion =0.0025mol(25.0+25.0)×10−.3L=0.05mol/L

At this point, there is no excess acid or base. Therefore, the only possible reaction here is the dissociation of the conjugate base of the lactic acid (that is lactate ion).

  C5H5O3−+H2O⇌HC5H5O3+OH−

Thereafter, by obtaining the Kb value for lactate ion, the amount of hydroxide ions in the solution can be determined to get the pH value at this point.

    Reaction Lactic acidLactateOH-
    Initial 0.0500
    Change -Xxx
    Equilibrium (0.05-x)xx

  Kb=KwKaKb=1.0×10−141.38×10−4=7.24×10−11

Then the pH can be calculated as follows:

  Kb=[HC5H5O3][OH−][C5H5O3−]7.24×10−11=[x][x][0.05−x]=x2[0.05−x]3.62×10−12−7.24×10−11x=x20=x2+7.24×10−11x−3.62×10−12

Thereafter, this quadratic equation can be solved to determine the hydroxide ion concentration, thereby, the pOH and the pH can be determined.

  0=x2+7.24×10−11x−3.62×10−12x=−b±b2−4ac2ax=−(7.24×10−11)±(7.24×10−11)2−4(1)(−3.62×10−12)2(1)x=−(7.24×10−11)±52.42×10−22+1.45×10−112x=−(7.24×10−11)±1.45×10−112x=−(7.24×10−11)±3.80×10−62x=3.80×10−62x=1.90×10−6

The calculated value of x is concentration of hydroxide ion. The pOH of the solution will be:

  pOH=−log[1.90×10−6]=5.72

  pH=14−pOH=14−5.72pH=8.3

Addition of 28.0mLof base:

Total amount of lactic acid to be neutralized =0

Amount of base added =0.100mol/L×28.0×10−3L=28×10−4mol

Then the ICE table after the addition of base is created in order to determine the pH of the solution using Henderson-Hasselbalch equation.

    Reaction Lactic acidOH-LactateH+
    Initial 0.0025000
    Add00.0028
    Change -0.00250.002500
    Equilibrium 00.000300

Concentration of hydroxide =0.0003mol(25.0+28.0)×10−3L=0.0056mol/L

  pOH=−log[0.0056]=2.25pH=14.00−2.25=11.75

Addition of 30.0mLof base:

Total amount of lactic acid to be neutralized =0

Amount of base added =0.100mol/L×30.0×10−3L=30×10−4mol

Then the ICE table after the addition of base is created in order to determine the pH of the solution using Henderson-Hasselbalch equation.

    Reaction Lactic acidOH-LactateH+
    Initial 0.0025000
    Add00.0030
    Change -0.00250.002500
    Equilibrium 00.000500

Concentration of hydroxide =0.0005mol(25.0+30.0)×10−3L=0.009mol/L

  pOH=−log[0.009]=2.04pH=14.00−2.04=11.96

Thus, the value of pH with respect to added volume of base is as follows:

    Volume (in mL)pH
    02.73
    43.14
    83.53
    12.53.86
    204.46
    245.24
    24.55.6
    24.96.3
    25.08.28
    25.110.3
    26.011.30
    28.011.75
    30.011.96

The titration curve can be drawn as follows:

  Chemical Principles, Chapter 8, Problem 67E

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Chapter 8 Solutions

Chemical Principles

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