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1.0Abstract

Vinegar can be defined as a solution composed of acetic acid and water. This experiment was

conducted to determine the molarity of the solution and the percent by mass of acetic acid in

vinegar by titration with the standardized sodium hydroxide solution. This is experiment is

done in two parts which is the part 1 is the standardization of sodium hydroxide and the second

part is done to determine the molarity of acetic acid in vinegar. In standardizing the NaOH

solution, approximately 250 mL of 0.6M sodium hydroxide from NaOH solid was prepared.

KHP of 250 mL solution is then titrated with the NaOH. The pH of the solution is recorded for

every 1 mL. The experiment was repeated to achieve accurate result. Then, the graph of pHagainst volume of NaOH was constructed. Based on the graph, the average molarity of NaOH

can be determined by plotting the graph. For the second experiment, vinegar of 10 mL was

titrated with the NaOH and the pH were recorded for every 2 mL of NaOH solution. The

titration process was also repeated to get accurate result. For experiment 1, the volume of

NaOH need to neutralize the acid is 11.03 mL for titration 1 and 11.103 mL for titration 2. The

molarity for titration 1 is 0.67 M and titration 2 is 0.66 M. The average molarity is 0.665 M. For

experiment 2, the the volume of NaOH need to neutralize the acid is 34.29 mL and 23.19 mL

for titration 1 and for titration 2. The molarity for is 2.28 M and 1.87 M for titration 1and

titration 2 is 0.66 M. The average molarity is 2.075 M. The percent by mass of acetic acid in

vinegar solution for titration 1 is 13.69% while the percent by mass in titration 2 is 11.23 %.

Hence the average percent by mass calculated is 12.46%. Based on the result, it can be

conclude that the higher the concentration, the higher the molarity and the higher volume

needed to neutralize the acid.

2.0 Introduction

Vinegar has been used since ancient times as an important household item. Vinegar can be

defined as a solution composed of acetic acid and water. The molecular for the acetic acid is

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CH3COOH. The acidic properties of vinegar can be titrated with a base by using a titration

method to determine the molarity and the percent by mass of the solution. Titration is processes

in an amount of a solution of known concentration are added to a solution of unknown

concentration until the stoichiometry of the solution is achieved. Concentration of solution is

the amount of solute in a given amount of solvent. The concentrated solution contain large

amount of solute in the given solvent while it is vice versa in diluted solution where the

solutions contain little amount of solute in a given solution. Molarity and the percent by mass

are the specific terms that can be used in expressing concentration.

Molarity is the number of moles of solute per liter of solution.

(Equation 2-1)

Percent by mass is the mass in grams of solute per 100 grams of solution.

(Equation 2-2)

In this experiment, the sodium hydroxide was standardized and the molarity and the percent by

mass were done by using the titration method. The titration process is done to determine theequivalent point of reaction. The equivalence point is reach when the added quantity of the

reactant is the exact amount necessary for stoichiometric reaction with another reactant which

in this experiment the NaOH will turn the solution basic.

3.0 Objectives

The objectives of this experiment is to determine the molarity of the solution and the percent by

mass of acetic acid in vinegar by titration with the standardized sodium hydroxide solution.

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4.0 Theory

Vinegar is a dilute solution of acetic acid. Since vinegar is an acid, it can be titrated with a base.NaOH is used as a base to neutralize the acidic properties. Titration is the process of adding a

known amount of a solution of known concentration to a known amount of solution of unknown

concentration. In this experiment, the equivalence point occur when the amount of 1 moles of

sodium hydroxide ( Base ) is necessary to neutralize the1 mole acetic acid ( acid ). The reaction is :

CH3COOH(aq) OH-(aq) H2O(l) CH3CO2

-(aq)

The solution is said to reached equivalent when there are sudden change in the pH value of the

solution. pH is defined as the negative of the logarithm of the hydrogen ion concentration.

pH = -log[HO]

pH scale indicate the acidity or the basicity of a solution. pH 7 is said to be natural. The value

of pH which lower than 7 is term as acid and above the value of 7 is called basic.

The experiment was first started by taking the pH of KHP before titrated with the standardize

sodium hydroxide solution. When the NaOH is added . the hydrogen ion will be neutralized and

the concentration ion will decreases.

KHC8H4O4(aq)+NaOH(aq) KNaC8H4O4(aq)+H2O(l)

CH3COOH(aq)+NaOH(aq) NaCH3COO(aq)+H2O(l)

5.0 Procedures

5.1 Standardization of sodium hydroxide solution

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1. 250 mL of approximately 0.6 M sodium hydroxide solution is prepared from

NaOH solid in a beaker. The calculation was checked with the laboratory

instructor before preparing the solution. The calculation is recorded

2. 250 mL beakers was weighed and the mass was recorded to the nearest 0.001g.

1.5 grams of KHP is added to the beaker. The mass of the beaker and KHP are

recorded to the nearest 0.001g. Differences in mass of KHP are calculated and

the data are recorded. 30 mL of distilled water is then added to the beaker and

the solution was stirred until the KHP dissolved completely.

3. The solution was then titrated with Naoh and the pH was recorded with the

addition of 1 mL NaOH solution.

4.

Steps 1 to 3 were repeated and the second trial were performed to standardize

the NaOH solution.

5. Graph of pH versus NaOH was plotted and the volume of NaOH required to

neutralize the KHP solution in each solution were determined.

6.

The molarity for sodium hydroxide in titration 1 and 2 were calculated

7. The average molarity for sodium hydroxide in titration 1 and 2 were calculated.

The resulting sodium hydroxide concentration were used in the part B of the

experiment.

5.2 Molarity of acetic acid and percent of vinegar

1. 10.00 mL of vinegar was transferred to a clean, dry 250 mL beaker using a 10

mL volumetric pipette. Sufficient water in rang of 75 to 100 mL was added to

cover the pH electrode tip during titration.

2. 1 mL of NaOH was added to the vinegar solution and the pH were recorded.

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3. Step 12 were repeated once more

4. Graph of pH vs volume Naoh added were plotted. The volume of NaOH

required to neutralize vinegar in each titration were determined and the data

were recorded.

5. The molarity of the acetic acid in vinegar for titration 1 and 2 were calculated.

6. The average molarity of the acetic acid in vinegar for titration 1 and 2 were

calculated.

7.

The percent by mass of acetic acid in vinegar for titration 1 and 2 werecalculated

8. The percent by mass of acetic acid in vinegar was calculated.

6.0 Apparatus & Material

6.1 Apparatus

250 mL Beaker, Volumetric Flask, Weigh Balance, Magnetic Stirrer, Retort Stand, Ph

Meter, Pipette, Burette, Magnetic Rod,

6.2 Material

0.6 M Naoh, 1.5 Grams KHP, Distilled Water, Vinegar

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7.0 Result

Titration 1 Titration 2

Mass of beaker (g) 99.672 148.036

Mass of beaker + KHP (g) 101.181 149.543

Mass of KHP (g) 1.509 1.507

Volume of NaOH to neutralize the

KHP solution (mL)

Titration 1 (pH) Titration 2 (pH)

0 4.459 4.216

1 4.720 4.481

2 4.869 4.717

3 5.025 4.863

4 5.139 5.008

5 5.271 5.143

6 5.369 5.287

7 5.503 5.423

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8 5.697 5.587

9 5.873 5.770

10 6.177 6.024

11 6.827 6.417

12 12.468 12.070

13 12.923 12.732

14 13.072 12.978

(pH) pH of KHP after being titrate with NaOH

Volume of NaOH required to

neutralize vinegar

Titration 1 (pH) Titration 2 (pH)

0 2.444 2.655

2 3.220 3.380

4 3.588 3.717

6 3.846 3.974

8 4.056 4.159

10 4.223 4.338

12 4.381 4.490

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14 4.521 4.640

16 4.677 4.790

18 4.802 4.942

20 4.940 5.092

22 5.098 5.273

24 5.236 5.485

26 5.240 5.748

28 5.446 6.468

30 5.771 11.995

32 6.188 12.487

34 6.196 12.689

36 11.867 12.808

38 12.408 12.892

40 12.628 12.962

* (pH) pH of vinegar after being titrate with NaOH

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Figure 7.1 : Titration of KHP solutions with NaOH

Figure 7.2 : Titration of Vinegar solutions with NaOH

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

pH

Volume of NaOH

Graph of pH vs Volume of NaOH

Titration 1

Titration 211.103 mL

equivalent point

0123456789

1011121314

0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38

pH

Volume of NaOH

Graph of pH vs Voume of Volume of NaOH

Titration 1

Titration 2

28.19 mL 34.29

mL

11.03 mL

equivalent point

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8.0 Sample of calculations

Standardization of sodium hydroxide solution

1)Calculations for preparing 250 mL of approximately 0.6 M sodium hydroxide solution.

250 mL NaOH

= 0.25 L NaOH

Molarity, M =

0.6 M NaOH =

Moles of NaOH = 0.15 mol

Mass NaOH = moles of NaOH molar mass NaOH

=0.15 mol NaOH 40 g/mol

=6 g solid NaOH

2)

Titration 1 Titration 2

Mass of beaker (g) 99.672 148.036

Mass of beaker + KHP (g) 101.181 149.543

Mass of KHP (g) 1.509 1.507

Volume of NaOH to

neutralize the KHP solution

(mL)

11.03 11.103

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3)Calculation of molarity of sodium hydroxide for each titration

Titration 1 :

Mol KHP =

=

= 0.00738 mol

1 mol KHP = 1 mol NaOH

Molarity NaOH =

=

= 0.67 M

Titration 2 :

Mol KHP =

=

= 0.00740 mol

1 mol KHP = 1 mol NaOH

Molarity NaOH =

=

= 0.66M

4)Calculation of the average molarity of sodium hydroxide for each titration.

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Average molarity of NaOH =

=0.665 M

Molarity of acetic acid and percent of vinegar

1)

Titration 1 Titration 2

Volume of NaOH required

to neutralize vinegar (mL)

34.29 28.19

2)Calculation of molarity of acetic acid in each titration 1 and 2.

Titration 1:

Volume of NaOH = 34.29 mL

Mol of NaOH = 0.03429 L 0.665 mol/L

= 0.0228 mol

From the equation :

CH3COOH(aq) + NaOH(aq)NaCH3COOH(aq)+ H2O(l)

1 mol NaOH = 1 mol CH3COOH

Thus 0.0228 mol NaOH = 0.0228 mol CH3COOH

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Molarity of CH3COOH =

=

= 2.28M

Titration 2:

Volume of NaOH = 28.19 mL

Mol of NaOH = 0.02819 L 0.665 M

= 0.0187 mol

From the equation :

CH3COOH(aq) + NaOH(aq)NaCH3COOH(aq)+ H2O(l)

1 mol NaOH = 1 mol CH3COOH

Thus 0.0198 mol NaOH = 0.0198 mol CH3COOH

Molarity of CH3COOH =

=

= 1.87 M

3)Calculation of average molarity of acetic acid

Average molarity of acetic acid =

= 2.075 M

4)Calculation of percent by mass of acetic acid in vinegar

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Titration 1 :

Mass of CH3COOH =0.01 L 2.28 M CH3COOH 60.06 g/mol CH3COOH

=1.369 g CH3COOH

Percentage by mass

x 100%

x 100%

= 13.69 %

Titration 2 :

Mass of CH3COOH = 0.01 L 1.87 M CH3COOH 60.06 g/mol CH3COOH

=1.123 g CH3COOH

Percentage by mass

x 100%

x 100%

= 11.23 %

5)Calculation of average percent by mass of acetic acid in vinegar

average percent by mass=

= 12.46 %

9.0 Discussions

The objectives of this experiment are to determine the molarity of the solution and the percent

by mass of acetic acid in vinegar by titration with the standardized sodium hydroxide solution.

The experiment is divided into two parts. In the first experiment, KHP solution is titrated with

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NaOH. The figure 7.1 illustrates the graph of titration of the solution to neutralize the acid. The

pH is taken 1 mL of NaOH each time being titrated. The pH value of KHP started to increase

drastically at 12 mL of NaOH solution where pH value was increase from 6.827 to 12.468. The

solution was tending to be alkali onwards. From the graph, the pH value was plotted and the

value of NaOH to neutralize the KHP solution at pH 7 is at 11.03 mL. The process was

repeated for the second time to get accurate result. For the second titration, the pH value of

KHP started to increase drastically also at range 12 mL of NaOH solution where pH value was

increase from 6.417 to 12.070.. From the graph, the value of NaOH to neutralize the KHP

solution at pH 7 is at 11.103 mL.

Next, the figure 7.2 shows the graph about the ph value against the volume of NaOHsolution to neutralize the vinegar. Same as the experiment 1, the titration process was done two

times to ensure the data recorded are accurate. In the first titration, it can be stated that the

volume needed for the NaOH to neutralize vinegar was 34.29 mL and amount of 23.19 mL of

NaOH is needed to neutralize vinegar at titration 2. .

In this experiment, the average molarity and the percent by mass were also needed to be

calculated and determined. The molarity of NaOH for the first titration is 0.67 M and 0.66 M

for second titration. The average molarity for both titration is 0.665 M. the percent by mass ofthe acetic acid in vinegar are 13.69 % and 11.23 % for titration 1 and 2. The average mass

percent of this experiment is 12.46 %

10.0 Conclusions

Based on the result, it can be concluded that the experiment achieved its objectives. However,

the results obtained are not very precise. The figure 7.1 and 7.2 illustrate the pattern of the

solution to achieve base pH value. The pattern shows that the data are not consistent. In the

experiment 1, it required an amount of 11.03 mL of NaOH to neutralize the acid in titration 1

while it required an amount of 11.103 mL of NaOH to neutralize the acid in titration 2. For the

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molarity, the value of molarity in titration 1 is 0.67M and 0.66 M in titration 2. The average

molarity for both titration is 0.665 M.

In experiment 2, the result shown that the amount of volume of NaOH to neutralize the vinegar in

titration 1 is 34.29 mL. On the other hand, the volume need by NaOH to neutralize the vinegar in

titration 2 is 23.19 mL. The molarity in titration 1 is 2.28 M and the molarity in titration 2 is 1.87 M.

The average molarity is 2.075M. The percent by mass of acetic acid in vinegar solution for

titration 1 is 13.69% while the percent by mass in titration 2 is 11.23 %. Hence the average

percent by mass calculated is 12.46%. Thus, it can be concluded that, the higher the molarity of

solution the more volume of NaOH needed to neutralize the acid.

11.0 Recommendations

1. Strictly make sure that the eye level are parallel to the instrument to avoid parallax error

2. Make sure that there are no air bubbles trapped in the burette before conducting the

experiment.

3. The experiment should be repeated up to three or four times to get the accurate readings.

4. Use indicator solution so that changes in the solution can be observe

12.0 References

Lab manual of Engineering Chemistry Laboratory (CHE 485) Uitm Shah Alam

Chemistry in water. (n.d.). Retrieved October 16, 2014, from

http://linus.chem.ku.edu/genchemlab/184FA06/Download184_Labs/Vinegar%20Download

.htm

http://linus.chem.ku.edu/genchemlab/184FA06/Download184_Labs/Vinegar%20Download.htmhttp://linus.chem.ku.edu/genchemlab/184FA06/Download184_Labs/Vinegar%20Download.htmhttp://linus.chem.ku.edu/genchemlab/184FA06/Download184_Labs/Vinegar%20Download.htmhttp://linus.chem.ku.edu/genchemlab/184FA06/Download184_Labs/Vinegar%20Download.htm

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Science Buddies Staff : Measuring the Amount of Acid in Vinegar by Titration with an

Indicator Solution (2014,). Retrieved October 16, 2014 from

http://www.sciencebuddies.org/science-fair-projects/project_ideas/Chem_p045.shtml

R. T. Morison & R. N. Boyd (1973). Fundamentals of General, Organic And Biological

Chemistry, Prentice Hall, Inc. Upper Saddle River, New Jersey.

13.0 Appendices

http://www.sciencebuddies.org/science-fair-projects/project_ideas/Chem_p045.shtmlhttp://www.sciencebuddies.org/science-fair-projects/project_ideas/Chem_p045.shtmlhttp://www.sciencebuddies.org/science-fair-projects/project_ideas/Chem_p045.shtml