UNIVERSITI TEKNOLOGI MARA FAKULTI KEJURUTERAAN KIMIA

ENGINEERING CHEMISTRY LABORATORY (CHE485)

No. Title Allocated Marks (%) Marks

1 Abstract/Summary 5  2 Introduction 10  3 Aims 5  4 Theory 10  5 Apparatus 5  6 Methodology/Procedure 10  7 Results 10  8 Calculations 10  9 Discussion 20  10 Conclusion 5  11 Recommendations 5  12 Reference / Appendix 5  

TOTAL MARKS 100  

Remarks:

Checked by :

---------------------------

Date :

NAME : NUR NAJWA BINTI YUNUSSTUDENT NO. : 2011207298GROUP : EH 222 1AEXPERIMENT : CONCENTRATION OF ACETIC ACID DATE PERFORMED : 26TH SEPTEMBER 2011SEMESTER : 1PROGRAMME / CODE : BACHELOR OF ENGINEERING IN

CHEMICAL AND BIOPROCESSSUBMIT TO : CIK NURUL ASYIKIN MD ZAKI

TABLE OF CONTENTS

Abstract…………………………………………………………………………. 1

Introduction…………………………………………………………………….. 2

Aims…………………………………………………………………………….. 3

Theory…………………………………………………………………………... 4

Methodology……………………………………………………………………. 5

Results…………………………………………………………………………... 8

Calculations……………………………………………………………………... 15

Discussion………………………………………………………………………. 17

Conclusion………………………………………………………………………. 19

Recommendations………………………………………………………………. 20

References……………………………………………………………………….. 21

Appendices………………………………………………………………………. 22

ABSTRACT

Vinegar is a dilute solution of acetic acid and popular for its domestic use. This experiment was

conducted in order to determine the percent by mass of acetic acid in vinegar by titration with a

standardized sodium hydroxide solution. The process that plays major role in this experiment

was titration whereby the required acid was titrated with sodium hydroxide (NaOH). In this

experiment, there were two acids used which were potassium hydrogen phthalate (KHP) as a

primary standard acid and vinegar itself. Both of the solutions were titrated until a sudden change

of the pH is noted. The sudden change was referred as the equivalence point. This point indicated

the amount of NaOH required to neutralize the acids. The results of both experiment was used to

calculate the mass of acetic acid in the vinegar. The average percent of mass of acetic acid is

5.0146 %. Therefore, the experiment was completely and successfully done.

1

INTRODUCTION

The term of concentration is used by scientist to estimate the amount of solute dissolved

in a given quantity of solvent or quantity of solution. The greater the amount of solute dissolved

in a particular amount of solvent, the more concentrated the resulting solution. Two specific

terms are used to express the concentration, which are molarity and percent by percent by mass.

Stoichiometry with the relative quantities of reactants and products in chemical

reactions. After balancing the chemical equation, the relations between quantites of reactants and

products form a ratio of whole numbers. Stoichiometry wil be used to calculate quantities such as

mass, number of moles, volume and percent yield. Stoichiometry is related to the law of

conservation of mass and with a few calculations, we can predict how elements and components

diluted in a standard solution react in certain experimental conditions and thus, able to determine

the molarity.

Molarity (M) expresses the concentration of a solution as the number of moles of solute

in a liter of a solution. Therefore, the formula is as below:

Molarity (M )= moles solutevolumeof solution∈liters

[ Eq .1]

whereby percent by mass is the mass in grams of solute per 100 grams of solution and the

formula is as below:

Percent solute= grams of solutegrams of solution

× 100 %[ Eq .2]

Acetic acid, CH3COOH is the main component of vinegar other than water. Typically,

vinegar consist of 4 to 18% of acetic acid by mass. Vinegar is always used for medical purposes,

culinary and agricultural purposes. Therefore, some country strictly limited the percent of acetic

acid inside commercial vinegar for safety. Though acetic acid is considered as weak acid, higher

concentration of acetic acid may results to the permanent eye damage, skin burns and irritation to

mucous membrane.

Titration is a common laboratory method to determine the unknown concentration of an

identified analyte. Titration method is the process involved in neutralization of acid-base. The

base is added to acid until sudden change of the pH of the solution. The changes of the pH is

very sudden. Therefore, by using titration method, the volume of base poured into the acid can

2

be controlled and by small increments. Through this method, the equivalence point in which the

solutions are neutralize can be determined.

OBJECTIVE

The objective of this experiment is to:

apply the process of titration

determine the molarity of a solution

determine the percent by mass of acetic acid in vinegar by titration with a standardized

sodium hydroxide solution.

3

THEORY

In this experiment, titration is the major process involved to get the most accurate result. In the

titration process, a burette is used to dispense solution in a small amount. Most of the burette has

the smallest calibration unit of 0.1mL.

Figure 1: a) Shows a 50-mL burette. b) Smallest calibration unit, 0.1-mL

While conducting this experiment, the equivalence point will occurs when the moles of acid in

the solution equals the moles of base added in the titration. For example, amount of 1 mole of

sodium hydroxide (NaOH), is necessary to neutralize 1 mole of acetic acid (CH3CO2H), as

shown below:

NaOH(aq) + CH3CO2H(aq) → NaCH3CO2(aq) + H2O(l) [Eq. 3]

The titration has reached the equivalence point if there is a sudden change of pH in the solution .

pH is the negative logarithm in base 10 of [H+] and related to the hydrogen ion concentration.

pH = -log10[H+]

pH scale varies according to the basicity and acidity of a solution. The pH of a neutral solution is

7.00 at 25⁰C. If the pH is below 7.00 the solution is considered as an acid whereas if it is above

7.00, it is referred as a basic solution. The pH of a solution can be measured quickly and

accurately with a pH meter. The electrode of the pH meter was inserted into a beaker with

required acid while the solution was titrated with NaOH solution. The hydrogen ions will be

neutralized as NaOH is incrementally added to the acid solution in which resulting to the

increased of the pH of that particular acid. After a sufficient amount of NaOH is added, the next

drop of NaOH will cause a sudden sharp increase in pH. The equivalence point of titration

indicates the volume of based required to completely neutralized the acid.

4

Figure 2 Titration curve of weak acid titrated with NaOH

During this experiment was conducted, titration of a vinegar sample with a standardized NaOH

solution will be performed. In order to standardize the NaOH solution, a primary standard acid

solution is initially prepared. Potassium hydrogen phthalate (KHC8H4O4) and oxalic acid

(COOH)2 were commonly used as a primary standard acid while sodium carbonate, Na2CO3, is

the commonly used base. Potassium hydrogen phthalate (KHP) was used throughout this

experiment as a primary standard acid. The KHP was produced by dissolving around 1.500g with

30 mL of distilled water. In order to choose the standard acid or even bases,they must have at

least these characteristics:

available in at least 99.9 purity

have high molar mass

stable upon heating

must be soluble in the required solvent

The reaction equation of this reaction between KHP and NaOH solution is shown below:

KHC8H4O4(aq) + NaOH(aq) → KNaC8H4O4(aq) + H2O(l) [Eq. 4]

After the NaOH solution was standardized, the process of titration with 10.0 mL aliquots of

vinegar began. The equation of the reaction is:

CH3COOH(aq) + NaOH(aq) → NaCH3COO(aq) + H2O(l) [Eq. 5]

By referring to the equation above, the molarity and the percent by mass of acetic acid can be

determined.

5

METHODOLOGY

Material and Apparatus

Solid NaOH

Distilled water

KHP

Beakers

Measuring cylinder

Magnetic stirrer

pH meter

Vinegar

10 mL volumetric pipette

Analytical Balance

Prodecure

Experiment A

Standardization of Sodium Hydroxide Solution

1. 250 mL of approximately 0.6 M sodium hydroxide solution was prepared from NaOH solid.

The solution was prepared in a beaker after the calculation was checked with the laboratory

instructor. The calculation was recorded.

2. The beaker was placed and tarred on the balance. 1.5 grams of KHP was added to the beaker.

The mass of the KHP was recorded to the nearest 0.001 g. 30 mL of distilled water was

added. The solution was stirred with magnetic stirrer until the KHP has dissolved completely.

3. The solution was titrated with NaOH and the pH was recorded with 1 mL additions of NaOH

solution.

4. Steps 1 to 3 were repeated and two more solutions for NaOH standardization were prepared.

5. The graph of pH versus NaOH was plotted. The volume of NaOH required to neutralize the

KHP solution in each titration were determined from the plots

6. The molarity of sodium hydroxide for titrations 1, 2 and 3 was calculated.

7. The average molarity of sodium hydroxide solution was calculated. The resulting sodium

hydroxide concentration was used in part B of the experiment.6

Experiment B

Molarity of acetic acid and mass percent in vinegar

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

pipette. 90.0 mL of water was added in order to cover the pH electrode tip during the

titration.

2. 1.0 mL of NaOH was added to the vinegar solution and the pH was recorded.

3. The above steps was repeated twice more.

4. The graph of pH vs NaOH was plotted and from the plots, the volume of NaOH required to

neutralize the vinegar in each titration. The data was recorded.

5. The molarity of acetic acid was calculated for titrations 1, 2 and 3.

6. The average molarity of acetic acid for each titration was calculated.

7. The percent by mass of acetic acid in vinegar was calculated for titrations 1, 2 and 3.

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

7

RESULTS

Experiment A

Graph 1 - Titration 1

8

0.00 2.00 4.00 6.00 8.00 10.00 12.00 14.00 16.00 18.000.00

2.00

4.00

6.00

8.00

10.00

12.00

14.00

KHP Titrated With NaOH

Volume of NaOH (mL)

pH

12.50

0.00 2.00 4.00 6.00 8.00 10.00 12.00 14.00 16.00 18.000.00

2.00

4.00

6.00

8.00

10.00

12.00

14.00

KHP Titrated With NaOH

Volume of NaOH (mL)

12.50

Graph 2 - Titration 2

9

0.00 2.00 4.00 6.00 8.00 10.00 12.00 14.00 16.00 18.000.00

2.00

4.00

6.00

8.00

10.00

12.00

14.00KHP Titrated With NaOH

Volume of NaOH (mL)

pH

12.00

Graph 3 - Titration 3

10

Experiment B

11

Graph 5 - Titration 1

0.00 5.00 10.00 15.00 20.00 25.000.00

2.00

4.00

6.00

8.00

10.00

12.00

Vinegar Titrated With NaOH

mL of NaOH

pH

14.00

Graph 4 - Titration 1

12

Graph 7 - Titration 2

0.00 5.00 10.00 15.00 20.00 25.000.00

2.00

4.00

6.00

8.00

10.00

12.00

Vinegar Titrated With NaOH

mL of NaOH

pH

12.00

Graph 6 - TItration 2

Graph 8 - Titration 3

13

0.00 5.00 10.00 15.00 20.00 25.000.00

2.00

4.00

6.00

8.00

10.00

12.00

14.00

Vinegar Titrated With NaOH

mL of NaOH

pH

16.00

Experiment A

Titration 1 Titration 2 Titration 3

Mass of KHP (g) 1.5018 1.5011 1.5015

Volume of NaOH to

neutralize the KHP

solution (mL)

12.50 12.50 12.00

Molarity of NaOH (M) 0.5884 0.5881 0.6128

Average molarity (M) 0.5964

Table 1 - Tabulation data for Experiment A

Experiment B

Titration 1 Titration 2 Titration 3

Volume of NaOH to

neutralize the vinegar

solution (mL)

14.00 12.00 16.00

Molarity of Acetic

Acid (M)0.8350 0.7157 0.9542

Average molarity (M) 0.8350

% by mass of acid in

vinegar (%)5.0150 4.2980 5.0146

Average percent by

mass of acetic acid in

vinegar (%)

5.0146

Table 2 - Tabulation Data of Experiment B

CALCULATIONS

14

Experiment A

Calculation of the moles of KHP used in the first titration:

1.5018 gof KH C8 H 4O 4×1mol of KH C8 H 4 O4

204.2 gof K H C8 H 4O 4

=0.007355 mol KH C8 H 4 O4

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

From equation above, the moles of NaOH is required to neutralize the moles of KHP can be

calculated.

0.007355 mol KHP ×1 mol NaOH1mol KHP

=0.007355 mol NaOH

Thus, the molarity of the NaOH can be calculated like below:

12.50 mL NaOH ×1 L

1000 mL=0.0125 L NaOH

M=mol NaOH1 L

=0.007355 mol NaOH0.0125 L solution

=0.05884 mol NaOH1 L

=0.5884 M NaOH

The average molarity of NaOH:

Titration1+Titration 2+Titration 33

=0.5884 M +0.5881 M+0.61283

═ 0.5964 M

Experiment 2

Calculation of the moles of NaOH that reacted:

14.00 mL NaOH ×1 L

1000 mL=0.0140 L NaOH

0.0140 L NaOH ×0.5964 mol NaOH1 L NaOH solution

=0.008350 mol NaOH

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

Calculation of the moles of CH3COOH neutralized by the moles of NaOH by using the

equation above:

0.008350 mol NaOH ×1mol C H 3 COOH

1 mol NaOH=0.008350 mol C H 3 COOH

Calculation of the molarity of the CH3COOH solution:

10 mLC H 3 COOH ×1 L

1000 mL=0.010 LC H 3COOH solution

15

M=mol C H 3 COOH

L of solution=

0.008350 mol of C H 3COOH

0.01 L solution=0.8350 M C H 3 COOH

Average molarity of acetic acid for each titration:

Averagemolarity=0.8350 M +0.7157 M +0.9542 M3

= 0.8350 M CH3COOH

Calculation of the mass of acetic acid in the solution:

10 mLC H 3 COOH ×1L

1000 mL=0.010 LC H 3COOH solution

0.01 LC H 3COOH ×0.8350 mol C H 3 COOH

1 L solution×

60.06 gC H 3COOH

1 molC H 3 COOH=0.5015 gC H 3 COOH

Calculation of the mass of the acetic acid solution:

10 mLC H 3 COOH ×1 gC H 3COOH solution

1 mLC H 3COOH solution=10.00 g C H 3COOH solution

Calculation of the percent by mass of acetic acid in the solution:

percent by mass C H 3 COOH=g C H 3 COOH

gC H 3 COOH solution×100 %

percent mass C H 3 COOH=0.5015 gC H 3COOH

10.00 gC H 3COOH× 100%=5.0150 %C H 3 COOH

Average percentage by mass of acetic acid in vinegar

Average% of acetic acid=5.0150 %+4.2980 %+5.7309 %3

= 5.0146 %

16

DISCUSSION

Acetic acid (CH3COOH) or ethanoic acid as in IUPAC name is the major component of vinegar,

other than water. This acid is classified as a type of weak acid. However, concentrated acetic

acid is corrosive and harmful to the skin. Vinegar is produces by fermentation of ethanol .

The aim of this experiment was to determine the molarity of a solution and the percent by

mass of acetic acid in vinegar by titration with a standardized sodium hydroxide (NaOH)

solution. The reaction involved can be considered the neutralization of the vinegar and NaOH

solution whereby an acid and a base are combined until certain point. The most suitable process

to get the most accurate results for this experiment is via titration process. Titration is a process

where small increments of a solution are added to a specific volume of solution until the

stoichiometry is obtained. Therefore, the equivalence point of the reaction can be determined. By

knowing the equivalence point, the molarity, mass and percent by mass of the acetic acid

(CH3COOH) can also be determined.

In Experiment A, NaOH solution was initially prepared for both of the experiment. In

order to standardize the NaOH solution, a primary standard acid solution is initially prepared. In

this case, potassium hydrogen phthalate, KHC8H4O4 (KHP) was chosen as they have high molar

mass. The KHP was measured at 1.5 g and added with 30 mL of distilled water into a beaker.

The solution was stirred until the KHP dissolved completely. After the preparation was done, the

KHP solution was titrated with the NaOH solution until sudden change in the pH reading. The

experiment was repeated two more times and recorded.

After the readings were recorded, the graphs were plotted and thus, the equivalence point

can be obtained. By using the Eq. 1, molarity of NaOH solution can be calculated for all the

titration. All of the molarity are added and divided with three to get the average molarity of

NaOH. The molarity of NaOH are 0.5884 M, 0.5881 M and 0.6128 M for Titration 1, 2 and 3

respectively. The average molarity of NaOH are 0.5964 M.

Meanwhile, in Experiment B, the KHP solution was replaced with vinegar. The solution

was titrated with NaOH solution that was prepared earlier 1 mL at one time. The solution was

titrated until a sudden change in the pH reading noted. The reading were recorded and the

17

experiment was repeated twice more. The graphs are plotted and the equivalence points were

determined. With these equivalence points, the molarity of acetic acid for each titration were

calculated. The molarity of acetic acid in the first titration was 0.8350 M. On the other hand, the

molarity of acetic acid in titration 2 and 3 were 0.7157 M and 0.9542 M respectively. All of the

molarity reading were added and divided with three to get the average molarity of acetic acid

which is 0.8350 M. Last but not least, by using the molarity of acetic acid, the percent by mass of

acetic acid in vinegar for each titration can be calculated, and the average percent by mass of

acetic acid in vinegar was 5.0146 %.

According to a webpage vinegar contains of about 4.0 % of acetic acid [1]. The percentage

error was high which is 25.37 %. Therefore, it can be concluded that some errors had occurred.

In order to minimize the error, the experiment must be repeated until several times and the

average is taken. Other than that, the mass of solid KHP and NaOH must be taken carefully and

precisely. The beaker must be tarred to get the exact mass of the solid. In order to have the

slightest error, the solid must be placed inside the beaker outside the range of the balance. This is

to ensure that there are no solid fall onto the balance and affect the reading. In addition, the

reading of the burette must be taken perpendicularly to the meniscus in order to avoid parallex

error. Magnetic stirrer and analytical balance must be used throughout this experiment to stir

those acids. This is to ensure uniform concentration of the solution inside the beaker. Last but not

least, the electrode of the pH meter must be dipped thoroughly to ensure no disturbance in the pH

reading.

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CONCLUSION

From this experiment, it can be concluded that the molarity of a solution and percent by

mass of acetic acid in vinegar by titration with a NaOH solution can be determined by the graphs

of the results. The graphs contributes a lot to determine the point where both of the solutions are

neutralize completely. After the equivalence point was reached, the molarity of NaOH and acetic

acid can be calculated. The percent by mass of acetic acid in vinegar for each titration is

5.0146%. The experiment was a success.

19

RECOMMENDATIONS

There are a few recommendations in order to minimize the error. First of all, the

experiment must be repeated until several times and the average is taken. Other than that, the

mass of solid KHP and NaOH must be taken accurately. The solid must be placed inside the

beaker outside the range of the balance. This is to ensure that there are no solid fall onto the

balance and affect the reading. Other than that, the reading of the burette must be taken

perpendicularly to the meniscus in order to avoid parallex error. Magnetic stirrer and analytical

balance can be used throughout this experiment to stir those acids as a safety measures. This is to

ensure uniform concentration of the solution inside the beaker. Last but not least, the electrode of

the pH meter must be dipped thoroughly to ensure no disturbance in the pH reading.

20

REFERENCES

[1]

Acetic Acid. (2011, October 8). Retrieved October 9, 2011, from Wikipedia: http://en.wikipedia.org/wiki/Acetic_acid[2]Brown, T. L., LeMay, H. E., Bursten, B. E., & Murphy, C. J. (2009). Chemistry The Central

Science. NJ: Pearson Education, Inc.

[3]Chang, R. (2010). Chemistry Tenth Edition (10th Edition ed.). Singapore: McGraw-Hill.

[4] Freddy Thomas, Eric Jamin, 2H NMR and 13C-IRMS analyses of acetic acid from vinegar,

18O-IRMS analysis of water in vinegar: International collaborative study report, Analytica

Chimica Acta, Volume 649, Issue 1, 1 September 2009, Pages 98-105, ISSN 0003-2670,

10.1016/j.aca.2009.07.014.

[5] A. Caligiani, D. Acquotti, G. Palla, V. Bocchi, Identification and quantification of the main

organic components of vinegars by high resolution 1H NMR spectroscopy, Analytica Chimica

Acta, Volume 585, Issue 1, 28 February 2007, Pages 110-119, ISSN 0003-2670,

10.1016/j.aca.2006.12.016. [6] Åke Olin, Bo Wallén, A note on the calculation of acid-base

titration curves and their

equivalence points, Talanta, Volume 25, Issues 11-12, November-December 1978, Pages 720-

721, ISSN 0039-9140, 10.1016/0039-9140(78)80186-6.

21

APPENDICES

Sample Calculation of Standardizing NaOH with KHP

The figure below presents the titration curve of 1.523 g of KHP dissolved in 20.0 mL of distilled

water.

Figure 3 The volume of NaOH used at the equivalence point is 15.3 mL

Calculation of the moles of KHP used in the titration:

1.523 gof KH C8 H 4O 4×1 mol of KH C8 H 4 O4

204.2 g of KH C8 H 4O 4

=0.007458 mol KH C8 H 4 O4

From Eq. 5, the moles of NaOH is required to neutralize the moles of KHP can be calculated.

0.007458 mol KHP ×1mol NaOH1mol KHP

=0.007458 mol NaOH

Thus, the molarity of the NaOH can be calculated like below:

15.30 mL NaOH ×1 L

1000 mL=0.01530 L NaOH

M=mol NaOH1 L

=0.007458 mol NaOH0.01530 L solution

=0.04875 mol NaOH1 L

=0.4875 M NaOH

Sample calculations to determine the concentration of acetic acid in vinegar

Calculation of the moles of NaOH that reacted:

16.95 mL NaOH ×1 L

1000 mL=0.01695 L NaOH

0.01695 L NaOH ×0.4875 mol NaOH1 L NaOH solution

=0.008263 mol NaOH

Calculation of the moles of CH3COOH neutralized by the moles of NaOH:

22

0.008263 mol NaOH ×1 mol C H 3 COOH

1 mol NaOH=0.008263 mol C H 3 COOH

Calculation of the molarity of the CH3COOH solution:

10 mLC H 3 COOH ×1L

1000 mL=0.010 LC H 3COOH solution

M=mol C H 3 COOH

L of solution=

0.008263 mol of C H 3COOH

0.01 L solution=0.8263 M C H 3 COOH

Calculation of the mass of acetic acid in the solution:

10 mLC H 3 COOH ×1L

1000 mL=0.010 LC H 3COOH solution

0.01 LC H 3COOH ×0.8263 mol C H 3 COOH

1 L solution×

60.06 gC H 3COOH

1 molC H 3 COOH=0.4963 gC H 3 COOH

Calculation of the mass of the acetic acid solution:

10 mLC H 3 COOH ×1 gC H 3COOH solution

1 mLC H 3COOH solu tion=10.00 g C H 3COOH solution

Calculation of the percent by mass of acetic acid in the solution:

percent by mass C H 3 COOH=g C H 3 COOH

gC H 3 COOH solution×100 %

percent mass C H 3 COOH=0.4693 gC H 3COOH

10.00 gC H 3COOH× 100%=4.963 %C H 3COOH

Calculation of error in this experiment

5.0146 %−4.0 %4.0

×100 %=25.37 %

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