Determination of Acetic Acid in Vinegar

Abstract

The Arrhenius definition defines acids as substances that increase the concentration of hydrogen ions (H+) when dissolved in water while base is defined as substances that produce hydroxide ions (OH−) in an aqueous solution. In this experiment, a technique known as a titration will be used to determine the concentration of acetic acid in vinegar. The concentration of acetic acid in vinegar can be determined when the equivalence point between NaOH solution and acetic acid is achieved. At the end of experiment, the molarity of acetic acid is 0.458 M and its mass percentage is 2.7485%.

Introduction

Vinegar is a dilute solution of acetic acid. The molecular formula for acetic acid is CH3COOH. In this experiment, vinegar is used to determine the molar concentration of acetic acid by titrating it with a standard solution of NaOH.

Titration is a process in which slow addition of one solution of a known concentration to a known volume of another solution of unknown concentration until the stoichiometry for the reaction is attained. The purpose of titration is to determine the equivalence point of the reaction. The equivalence point is reach when the added quantity of one reactant is the exact amount necessary for stoichiometric reaction with another reactant.

CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l)

By adding the sodium hydroxide, which is a basic solution gradually in small amounts from a burette to the acetic acid, which is an acidic solution, a neutralization reaction will occurs. Burette is a device that allows the precise delivery of a specific volume of a solution. The

amount of NaOH used to standardize the vinegar can then be used to determine the molarity and percentage by mass of acetic acid in the vinegar.

Objectives

The objectives of this experiment are to: 

Determine the molarity of a solution and the percent by mass of acetic acid in vinegar by titration with the standardized sodium hydroxide solution

Theory A burette is a device that allows the precise delivery of a specific volume of a solution. A typical burette has the smallest calibration unit of 0.1mL. Therefore, volume dispense from the burette should be estimated to the nearest 0.01mL. The NaOH will be added to the vinegar sample until all the acetic acid in the vinegar has been exactly consumed (reacted away). At this point the reaction is completed, and no more NaOH is required. This is called the equivalence point of the titration. NaOH (aq) + CH3COOH(aq)

NaCH3COO (aq) + H2O (l)

In this experiment, acid and base react, forming salt and water (neutral) . Acid + base

salt + water

In order to express the acidity or basicity, pH scale method is used. The pH value determine either the solution is basic or acidic, solutions having a pH < 7 are acidic , pH = 7 are neutral while pH > 7 are basic. Also pH electrode will be used in this experiment. When a pH electrode is inserting into a beaker containing the acid solution (pH within 3-5), the titration is initiated.. Some of the hydrogen ions will be neutralized as sodium hydroxide, NaOH, is incrementally added to the acid solution. When the concentration of hydrogen ion decreases, the pH of the solution will increase gradually. A sudden sharp increase in pH will occur when the next drop of NaOH is added after a sufficient NaOH is added to completely neutralize the acid (most of the H3O+ ions are removed from the solution). The equivalence point of titration is used to determine the volume of based required to completely neutralized the acid.

Procedures

A. Standardization of sodium hydroxide solution 1. Sodium hydroxide solution of 250 ml of approximately 0.6M was prepared from NaOH solid. The solution was prepared in a beaker and the calculations were checked by the laboratory instructor prior to preparing the solution. The calculation then was recorded. 2. A 250 ml beaker was weighed and the mass was recorded to the nearest 0.001g. 1.5 grams of KHP was added to the beaker and the mass of both KHP and the beaker was recorded to the nearest 0.001 g. the mass of KHP was calculated by thedifference recorded in the data. Distilled water of 30 ml was added to the beaker and the solution was stirred until the KHP dissolved completely. 3. This solution was titrated with NaOH and the pH value were recorded with 1 ml additions of NaOH solution. 4. Steps 1 to 3 were repeated to perform a second and third trial to standardize the NaOH solution. 5. The graph of pH versus NaOH was plotted and from the plots, the volume of NaOH required to neutralize the KHP solution in each titration were determined. 6. The molarity of sodium hydroxide for titration 1 and 2 were calculated. 7. The average molarity of sodium hydroxide solution for titration 1 and 2 were calculated. The resulting sodium hydroxide concentration were used in part B of the experiment.

B. Molarity of acetic acid and percent of vinegar

1. Vinegar of 10.00 ml was transferred to a clean, dry 250 ml beaker using a 10 ml volumetric pipette. Sufficient water of 75 to 100 ml was added to cover the pH electrode tip during the titration. 2. 1 ml of NaOH was added to the vinegar solution and the pH was recorded. 3. The above steps were repeated twice. 4. The graph of pH versus volume of NaOH added were plotted and from the plots, the volume of NaOH required to neutralized vinegar in each titration were determined. The data was recorded. 5. The molarity of acetic acid in vinegar for titration 1 and 2 werecalculated. 6. The average molarity of acetic acid for each titration were calculated. 7. The percent by mass of acetic acid in vinegar for titration 1 and 2 were calculated. 8. The percent by mass of acetic acid in vinegar was calculated.

Apparatus and Materials       

0.6 M sodium hydroxide solid 1.5 g KHP Water 30 ml distilled water 10 ml volumetric pipette Burette pH meter

   

stirrer bar analytical balance 250 ml beaker 10 ml vinegar

Results and Calculations

A. Standardization of sodium hydroxide solution

Volume of NaOH (ml)

Titration 1

Titration 2

Average

1 2

4.08 4.31

4.22 4.52

4.15 4.42

3 4 5 6 7 8 9 10 11 12

4.57 4.77 4.93 5.09 5.28 5.39 5.62 5.94 6.64 12.08

4.67 4.8 4.95 5.08 5.22 5.38 5.57 5.88 6.53 12.12

4.62 4.79 4.94 5.09 5.25 5.39 5.6 5.91 6.59 12.1

1. Preparation of 150 ml of approximately 0.6M sodium hydroxide solution. Molarity of NaOH = 0.6M Volume = 250 ml No of mole =

=

MV 1000 0.6(250) 1000

= 0.15 mol Mass = no of mol × molar mass = 0.15 (23+16+1) =6g

Titration 1

Titration 2

Mass of beaker (g)

114.096

112.214

Mass of beaker + KHP (g)

115.653

113.722

1.557

1.508

11.25

11.25

2.

Mass of KHP (g) Volume of NaOH to neutralize the KHP solution (mL)

3. Calculation for molarity of sodium hydroxide for titration 1 and 2:

Titration 1: Mol KHP =

=

mass KHP Molar mass 1.557 204.2 −3

= 7.62× 10

mol

1 mol KHP  1 mol NaOH −3

7.62× 10

−3

mol KHP  7.62× 10

Molarity of NaOH =

=

mol of solute volume of solution 7.62 ×10−3 0.01125

= 0.677 M

Titration 2: Mol KHP =

=

mass KHP Molar mass 1.508 204.2

mol NaOH

−3 = 7.38 ×10 mol

1 mol KHP  1 mol NaOH 7.38 ×10−3 mol KHP  7.38 ×10−3 mol NaOH mol of solute volume of solution

Molarity of NaOH =

=

7.38 ×10−3 0.01125

= 0.656 M 4. Calculation for average molarity of sodium hydroxide for titration 1 and 2: Average molarity =

=

molarity for titration 1+molarity for titration2 2

0.677+0.656 2

= 0.667 M

B. Standardization of sodium hydroxide solution

Volume of NaOH

Titration 1

Titration 2

Average

0 1 2 3 4 5 6 7 8

2.85 3.73 4.11 4.36 4.6 4.83 5.03 5.48 9.41

2.8 3.6 4.09 4.11 4.56 4.93 5.03 5.46 7.8

2.83 3.67 4.1 4.24 4.58 4.88 5.03 5.47 8.61

1.

Titration 1

Titration 2

7.62

7.65

Volume of NaOH required to neutralize vinegar (mL)

2. Calculation for molarity of acetic acid in vinegar for titration 1 and 2:

Titration 1: Volume of vinegar solution = 10 ml Volume of NaOH = 7.62 ml

Mol of NaOH = MV = 0.6 × 0.00762 = 0.004572 mol

1 mol NaOH  1 mol

CH 3 COOH mol CH 3 COOH

4.572 ×10−3 mol NaOH  4.572 ×10−3

Molarity of acetic acid =

=

mol of solute volume of solution

4.572× 10−3 0.01

= 0.457 M

Titration 2: Volume of vinegar solution = 10 ml Volume of NaOH = 7.65 ml

Mol of NaOH = MV = 0.6 × 0.00765 = 0.00459 mol 1 mol NaOH  1 mol 4.59 ×10−3

CH 3 COOH

−3 mol NaOH  4.59 ×10

Molarity of acetic acid =

=

mol of solute volume of solution

4.59 ×10−3 0.01

= 0.459 M

mol CH 3 COOH

3. Calculation for average molarity of acetic acid for titration 1 and 2:

Average molarity =

=

molarity for titration 1+molarity for titration 2 2

0.457+0.459 2

= 0.458 M

4. Calculation for percentage by mass of acetic acid in vinegar for titration 1 and 2:

Titration 1: mol of CH 3 COOH =4.572× 10−3 mol

Mass of

CH 3 COOH

= no of mol × molar mass

−3 = 4.572 ×10 ×60

= 0.27432 g Percentage by mass =

0.27432 ×100 10

= 2.7432%

Titration 2: −3

mol of CH 3 COOH =4.59× 10

Mass of

CH 3 COOH

mol

= no of mol × molar mass

−3 = 4.59 ×10 ×60

= 0.2754 g Percentage by mass =

0.2754 × 100 10

= 2.754%

5. Calculation for average percent by mass of acetic acid in vinegar: Average % by mass = average by mass for titration 1+average by mass for titration 2 2

=

2.743+ 2.754 2

= 2.7485%

Discussion

This experiment is carried out in order to determine the molarity of a solution and the percentage by mass of acetic acid in vinegar by titrating standardized sodium hydroxide solution, NaOH with the acetic acid. In part A of the experiment, 11.25 ml of NaOH was used to titrate the potassium hydrogen phthalate, KHP. The titration process was repeated twice in order to get an average reading of NaOH solution used. The molarity of the NaOH used is 0.6 M and the volume obtained from both of the trial is 11.25 ml. Equivalence point can be attained when the pH value is equal to 7. Thus, the average molarity of NaOH after it has been standardized was 0.667 M. In part B, 0.004572 mol of NaOH is required to react with 0.004572 mol of

CH 3 COOH

for the first trial while for the second trial, 0.00459 mol is required to react with 0.00459 mol of

CH 3 COOH

. The molarity for first and second trial are 0.457 M and 0.459 M

respectively. The average molarity of the acetic acid is 0.0458 M. The percentage by mass of acetic acid in 10 g of vinegar for both trial are 2.74% and 2.75%, making the average percentage by mass is 2.7485%.

Conclusions

As a conclusion, the molarity of acetic acid in a sample of vinegar is 0.458 M and its percentage of mass is 2.75%. the value obtained in this experiment is not the same with the theoretical value due to some errors that occurred during the experiment.

Recommendations

There are many factors that have contributed in causing error in the results of this experiment. One major factor that affected the result of this experiment was the strength of the sodium hydroxide. If either of these substances is left open in the atmosphere, they begin to lose their strength. During the experiment, the sodium and sodium hydroxide were both left open to interact with the environment for some time. Thus, the final answer did not match the theoretical value accurately because the strength was weakened, meaning that the numbers used to calculate the molar concentration were not as accurate. In order to prevent this error from affecting the results of the experiment, we shold keep the sodium hydroxide in an enclosed environment at all times, thus limiting the time it has to interact with the atmosphere. In addition, the equipment used could have also contributed to the error as all pieces of apparatus have an uncertainty attached to it. These uncertainties are then applied to calculations in order to keep up the amount of uncertainty associated with the amount of material used. These uncertainties can be reduced by using more accurate equipments, for example a more accurate mass balance. Also, limiting the transfer of solution from one container to another will also reduce the amount of error. Human judgment also accounts for some of the error in this experiment as the person performing the experiment was required to read off many measurements from the pipette and

burette. This error can be reduced by always ensuring that readings are always made at eye level and that the same person taking the readings is constant as judgment varies with each person.

References

1. http://infohost.nmt.edu/~jaltig/Vinegar.pdf 2. http://www.smc.edu/projects/28/Chemistry_10_Experiments/Ch10_Titration.pdf 3. http://schoolworkhelper.net/titration-of-vinegar-lab-answers/