lab report (vinegar)

ABSTRACT/ SUMMARY

The purposes of this experiment are to determine the percent by mass of acetic acid

in vinegar by titration with standardized sodium hydroxide (NaOH) solution. Vinegar is a

dilute solution of acetic acid. Since vinegar is an acid, it can be titrated with a base. In this

experiment, the titration comprised of two stages. The first stage is to standardize NaOH

solution with KHP. 0.6 M NaOH was used in this standardization. The second stage is the

titration between NaOH and commercial vinegar sample. The purpose of titration is to

determine the equivalence point of the reaction. At the equivalence point is occurs when an

equal number of moles of acid and base are mixed, the pH of the solution will change

drastically, as the solution is being flooded with excess base (NaOH). The percent of acetic

acid in vinegar and the percentage error for the experiment are 5.6983 % and 42.457 %

error. The conclusion is the experimental percentage of CH3COOH in the vinegar sample is

higher than the true value which is 4.00%.

1

INTRODUCTION

According to Stanton et. al., 2009, concentration of solution is the amount of solute (species

dissolved) in a given amount solvent (dissolving agent). A concentrated solution contains a

relatively great quantity of solute in a given amount of solvent. In another hand, dilute

solution contains relatively little solute in a given amount of solvent. Chemist uses specific

terms to express the concentration of solution. Two of these terms are molarity and percent

by mass:

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

Molarity (M) = moles of soluteliter of solution

(Equation 1-1)

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

Percent solute = gramsof solutegrams of solution

× 100% (Equation 1-2)

Vinegar or French for sour wine is formed by aerobic bacteria oxidizing grain alcohol to

acetic acid and water (Scharf and Malerich, 2010). The smell of vinegar attributed to acetic

acid (an organic acid). The molecular formula for acetic acid is CH3COOH. According to the

U.S. Food and Drug Administration (FDA), for a manufacturer of vinegar to use the word

“vinegar” in its labeling of the product, it must contain a minimum of 4.0 grams of acetic acid

per 100 mL of solution, which corresponds to a concentration of 0.67 M. The rest of the

solution is primarily water, with trace amounts of alcohol, phosphoric acid, sugar, and

glycerol present. Since vinegar is an acid, it can be titrated with a base to determine the

molarity and percent by mass of acetic acid in a vinegar solution. Titration is the process of

adding a known amount of a solution of known concentration to a known amount of solution

of unknown concentration. The more accurately the concentration of the solution of known

concentration is known, the more accurately the concentration of the unknown solution can

be determined. The concentration of unknown solution will be determined by knowing the

volume of known solution needed to complete the titration. The purpose of titration is to

determine the equivalence point of the reaction. The equivalence point of a titration is the

point at which the acid has completely reacted with or been neutralized by the base. At this

point, the pH changes rapidly with a small addition of acid or base (Chang, 2007).

OBJECTIVES

2

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

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

THEORY

In this experiment, the equivalence point is reach when the added quantity of base is the

exact amount necessary for stoichiometric reaction with acid (Stanton et. al., 2009). For

example, the stoichiometric amount of 1 mole of the strong base, sodium hydroxide (NaOH),

is necessary to neutralize 1 mole of the weak acid, acetic acid (CH3CO2H), as shown in

Equation 1-3:

NaOH(aq) + CH3CO2H(aq) NaCH3CO2(aq) + H2O(l) (Equation 1-3)

At the equivalence point of titration, the pH of the solution will change drastically, as the

solution is being flooded with excess strong base or strong acid (depending on what is being

added). pH in an aqueous solution is related to its hydrogen ion concentration. Symbolically,

the hydrogen ion concentration is written as [H3O+]. The molar concentration of hydrogen

ions in aqueous solution is frequently very small. For convenience, therefore the pH of a

solution is defined as the negative logarithm of the hydrogen ion concentration (Brown et. al.,

2009).

pH = -log10 [H3O+] (Equation 1-4)

pH scale is used to express the acidity or basicity of solution. Solutions with pH < 7 are

acidic, pH = 7 are neutral and pH > 7 are basic. For example a solution having [H 3O+]

concentration of 4.23 × 10-3 would have a pH of 2.374 and is acidic. There are a couple of

methods that can be used to monitor the pH change in a solution. In this experiment, pH

electrodes will be used to constantly measure and record the pH of the solution. The titration

is initiated by using a pH electrode to measure the pH of acid solution (pH within 3-5). As

sodium hydroxide, NaOH, is incrementally added to the acid solution, some of the hydrogen

ions will be neutralized. The pH of the solution will gradually increase as the hydrogen ion

concentration decreases. When sufficient NaOH is added to completely neutralize the acid

(most of the H3O+ ions are removed from the solution), the next drop of NaOH added will

cause sudden change in pH (increase sharply). This shows that the equivalence point of

titration is reached. The volume of based needed to completely neutralized the acid is

obtained from the equivalence point of titration. From the graph of pH versus the volume of

3

base added plotted, the equivalence point is the middle of the vertical part of the curve

(Chang, 2007).

Titration of vinegar sample with a standardized sodium hydroxide solution will be performed

in this experiment. To standardize the sodium hydroxide solution, a primary standard acid

solution is initially prepared. In general, primary standard solutions are produce by dissolving

a weighed quantity of pure acid or base in a known volume of solution.

Sodium hydroxide (NaOH) is one of the bases commonly uses in laboratory. However, it is

difficult to obtain solid sodium hydroxide in a pure form because it has tendency to absorb

moisture from air, its solution react with carbon dioxide and often appears wet. For these

reasons, a solution of sodium hydroxide must be standardized before it can be used in

accurate analytical work. The sodium hydroxide solution can be standardized by titrating it

against an acid solution of accurately known concentration. The acid chosen for this

experiment is a monoprotic acid called potassium hydrogen phthalate (KHP), for which the

molecular formula is KHC8H4O4 (Chang, 2007). The equation for the reaction of potassium

hydrogen phthalate with sodium hydroxide is:

KHC8H4O4(aq) + NaOH(aq) KNaC8H4O4 + H2O(l) (Equation 1-5)

Once the sodium hydroxide solution has been standardized, it will then used to titrate 10.0

mL aliquots of vinegar. The equation for the reaction between vinegar and sodium hydroxide

is:

CH3COOH(aq) + NaOH(aq) NaCH3COO(aq) + H2O(l) (Equation 1-6)

By knowing the standardized NaOH concentration and using Equation 1-6, the molarity and

mass percentage of acetic acid in the vinegar solution can be determined.

APPARATUS/ MATERIALS

4

pH meter

Vinegar

Magnetic stirrer

Weighing balance

Stir bar

Burette and burette clamp

Beaker (100, 250 mL)

Measuring cylinder

Conical flask

Retort stand

10 mL volumetric pipette

0.6 M NaOH solution

1.5 g potassium hydrogen phthalate (KHP)

Diagram of titration

.

PROCEDURES

5

Standardization of sodium hydroxide solution

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

solid. The solution was initially prepared in a beaker to make it easier to transfer into

the burette for titration.

2. A beaker was placed on the balance and tare. 1.5 grams of KHP was added to the

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

water was added to the beaker. The solution was stirred until the KHP had dissolved

completely.

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

solution was recorded.

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. From the plotted graph, the volume of

NaOH required to neutralize the KHP solution in each titration was determined.

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

7. The average molarity of the sodium hydroxide solution was calculated. The result of

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

Molarity of acetic acid and mass percent in vinegar

1. A 10 mL volumetric pipette was used to transfer 10.0 mL of vinegar to a clean, dry

100 mL beaker. Sufficient water, 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 more.

4. The graph of pH versus NaOH volume added was plotted. The volume of NaOH

required to neutralize the vinegar in each titration was determined from the plotted

graph and the data was recorded.

5. The molarity of acetic acid in vinegar for titrations 1, 2 and 3 were calculated.

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

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

calculated.

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

RESULTS

6


TITRATION 1 TITRATION 2 TITRATION 2

Mass of KHP (g) 1.529 1.5047 1.5057

Volume of NaOH to

neutralize the KHP

solution (mL)

12.27 11.82 12.29

Molarity of NaOH 0.6102 M 0.6234 M 0.6000 M

Average molarity of

NaOH for each titration0.6112 M


TITRATION 1 TITRATION 2 TITRATION 3

Volume of NaOH to

neutarlize the vinegar

solution (mL)

15.54 15.48 15.55

Molarity of acetic acid in

vinegar0.9498 M 0.9461 M 0.9504 M

Average molarity of

acetic acid for each

titration

0.9488 M

% by mass of acetic acid

in vinegar5.7045 % 5.6823 % 5.7081 %

Average percent by

mass of acetic acid in

vinegar

5.6983 %

% error 42.457 % error


7

0 2 4 6 8 10 12 14 160

2

4

6

8

10

12

14

KHP TITRATED WITH NaOH (TITRATION 1)

VOLUME OF NaOH (mL)

pH

0 2 4 6 8 10 12 14 160

2

4

6

8

10

12

14


VOLUME OF NaOH (mL)

pH

8

12.27 mL

11.82 mL

0 2 4 6 8 10 12 14 160

2

4

6

8

10

12

14


VOLUME OF NaOH (mL)

pH


0 2 4 6 8 10 12 14 16 180

2

4

6

8

10

12

14

VINEGAR TITRATED WIYH NaOH (TITRATION1)

VOLUME OF NaOH (mL)

pH

9

12.29 mL

15.54 mL

0 2 4 6 8 10 12 14 16 180

2

4

6

8

10

12

14

VINEGAR TITRATED WITH NaOH (TITRATION 2)

VOLUME OF NaOH (mL)

pH

0 2 4 6 8 10 12 14 16 180

2

4

6

8

10

12

14

VINEGAR TITRATED WITH NaOH (TITRATION 3)

VOLUME OF NaOH (mL)

pH

10

15.48 mL

15.55 mL

CALCULATION TO PREPARE 0.6 M NaOH

0.6molL

× 250 mL × 40 g NaOHmol

× 1 L

1000mL = 6 g NaOH solid

SAMPLE OF CALCULATION FOR STANDARDIZING BASED WITH KHP

Titration 1

Calculate the moles of KHP used in the titration.

1.529 g KHC8H4O4 × 1mol KHC8H 4O 4

204.2g KHC 8H 4O 4 = 0.007488 mol KHC8H4O4

Calculate the moles of NaOH required neutralizing the moles of KHP.

0.007488 mol KHC8H4O4 × 1mol NaOH1mol KHP

= 0.007488 mol NaOH

Calculate the molarity of NaOH solution.

12.27 mL NaOH × 1 L

1000mL = 0.01227 L NaOH

M1 = molesof NaOHLof solution

= 0.007488mol NaOH

0.01227 LNaOH = 0.6102 M NaOH

* The same calculation was used for titrations 2 and 3 to find the molarity of NaOH solution

(Appendix 1).

Calculate average molarity of sodium hydroxide for each titration.

Mav = M 1+M 2+M 3

3 =

0.6102+0.6234+0.60003

= 0.6112 M NaOH

11

SAMPLE OF CALCULATIONS FOR DETERMINING THE ACETIC ACID

CONCENTRATION IN VINEGAR BY TITRATION WITH STANDARD BASE

Titration 1

Calculate the moles of NaOH that reacted.

15.54 mL NaOH × 1 L

1000mL = 0.01554 L NaOH

0.01554 L NaOH × 0.6112mol NaOH1L NaOH solution

= 0.009498 mol NaOH

Calculate the moles of CH3COOH neutralized by the moles of NaOH.

0.009498 mol NaOH × 1molCH 3COOH

1mol NaOH = 0.009498 mol CH3COOH

Calculate the molarity of the CH3COOH solution.

10 mL CH3COOH × 1 L

1000mL = 0.010 L CH3COOH solution

M1 = molesof CH 3COOH

Lof solution =

0.009498molCH 3COOH0.010 Lsolution

= 0.9498 M CH3COOH

Calculate the mass of acetic acid in the solution.

10 mL CH3COOH × 1 L

1000mL = 0.010 L CH3COOH solution

0.010 L CH3COOH × 0.9498molCH 3COOH

1 Lsolution ×

60.06gCH 3COOH1molCH 3COOH

= 0.57045 g

CH3COOH

Calculate the mass of acetic acid solution.

10 mL CH3COOH solution × 1 gCH 3COOH

1mLCH 3COOH = 10.0 g CH3COOH solution

Calculate the percent by mass of acetic acid in the solution.

Percent mass CH3COOH = gCH 3COOH

gCH 3COOH solution × 100%

Percent mass CH3COOH = 0.57045gCH 3COOH

10.0gCH 3COOH solution × 100% = 5.7045 % CH3COOH

* The same calculation was used for titrations 2 and 3 to find the molarity of NaOH solution

12

(Appendix 1).

Calculate average molarity of acetic acid for each titration.

Mav = M 1+M 2+M 3

3 =

0.9498+0.9461+0.95043

= 0.9488 M CH3COOH

Calculate average molarity of acetic acid for each titration.

Percent mass CH3COOH av = %1+%2+%3

3 =

5.7045 %+5.6823 %+5.7081 %3

= 5.6983

% CH3COOH

SAMPLE ERROR CALCULATIOS

% error = %massdata−%mass theory%mass theory

= 5.6983 %−4.00 %

4.00 % × 100% = 42.457 % error

DISCUSSION

13

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 a standardization sodium hydroxide

(NaOH) solution. For the first experiment, the purposes are to standardize the sodium

hydroxide solution and to determine the molarity of sodium hydroxide. Sodium hydroxide

was needed to standardize because it can easily contaminated by the absorption of carbon

dioxide or water from the air as it absorbs moisture from the air and usually appears wet.

Thus if a solution of sodium hydroxide is prepared by weighing the sodium hydroxide, the

concentration for the solution may be not the approximately the desired concentration. On

another hand, potassium hydrogen phthalate (KHP) has a lesser tendency to absorb water

from the air and when dried will remain dry for a reasonable period of time. Potassium

hydrogen phthalate is a white, soluble solid that is commercially available in highly pure

form. Potassium hydrogen phthalate is a primary standard. Besides that, KHP also must be

dissolved completely and carefully at the beginning of the titration to get the accurate

concentration of NaOH. This means that more accurately the concentration of the solution of

known concentration is known (KHP), the more accurately the concentration of the unknown

solution can be determined (NaOH).

pH is a method of expressing the acidity or basicity of a solution. Solution with pH<7

are acidic, pH=7 are neutral and pH>7 are basic. In this experiment pH was used to

determine the equivalence point of the solutions. Equivalence point occurs when chemically

equivalent amounts of acid and base are present. At this point, the pH will rapidly increase

with a small addition of NaOH. This means that equivalence point in this experiment is the

point at which the volume of NaOH required to neutralize the acid solution.

In the first experiment, 6 g of NaOH was diluted with 250 mL of distilled water in

order to get 0.6 M NaOH. For this experiment the titration of KHP with NaOH was repeated

for 3 times to get the average. At first titration, the pH was slowly increase from 1 to 12 mL

NaOH added which are from pH 4.06 to 6.77. After 0.5 mL of NaOH was added, the pH

sharply increases to 12.16. The same goes to the titration 2 and 3 at which the pH sudden

change after 12 mL of NaOH was added to the KHP solution. This means that the pH of the

solution had changed from acid to base. The equivalence points are 12.27 mL for the first

titration, 11.82 mL for the second titration and 12.29 mL for the third titration. The molarity of

NaOH for first titration is 0.6102 M, second titration, 0.6234 M and lastly third titration,

0.6000 M. The average molarity of NaOH for these three titrations is 0.6112 M.

14

In second experiment, 10 mL of vinegar was used as acid. The titration of vinegar

with NaOH solution was also repeated for three times in this experiment. For the first

titration, the pH of the solution was slowly increased when NaOH was added from 1 to 15

mL which are from pH 3.09 to 6.06. After 1 mL of NaOH was added, the pH drastically

increases to 11.89. The same goes to the titration 2 and 3 at which the pH sudden change

after 15 mL of NaOH was added to the vinegar solution. This means that the pH of the

solution had changed from acid to base. The equivalence points for each titration is 15.54mL

for the first titration, 15.48mL for the second titration and 15.55mL for the third titration. The

molarity of acetic acid in vinegar for first titration is 0.9498 M, second titration, 0.9461 M and

lastly third titration, 0.9504 M. The average molarity and percent by mass of acetic acid in

vinegar for these three titrations is 0.9488 M and 5.6983 %. The percentage error for the

experiment is 42.457 %.

CONCLUSION

The purpose of this experiment was to determine the molarity of a solution and the

percent by mass of acetic acid, CH3COOH, in a commercial vinegar sample by titration with

a standardized sodium hydroxide solution. The percentage of CH3COOH was found to be

5.6983 %. The exact value was 4.00 % from the label on the commercial vinegar sample.

The deviation of the experimental value from the true value was calculated to be 1.6983.

Using the true value and experimental value, a percent error was calculated. The percent

error was found to be 42.457 %. This shown that the objective was achieved because for a

manufacturer of vinegar to use the word “vinegar” in its labeling of the product, it must

contain a minimum of 4.00 % of CH3COOH.

Based on this experiment, we can conclude that the experimental percentage of

CH3COOH in the vinegar sample is higher than the true value. One possible source of error

to account for this different could be the titration past the true equivalence point of the

reaction. Addition of more NaOH than desired volume to reach the end point would give a

larger value for the number of moles of NaOH used. An error in the molarity of NaOH used

would result in a larger value for the number of moles of CH3COOH in the sample. A larger

number for the moles of CH3COOH would result in a larger mass of CH3COOH in the

sample. The larger mass of acetic acid would result in a higher percentage of acid in the

sample.

15

RECOMMENDATIONS

1. Student should be given more than one weak acid. Thus, they will able to compare

and study the efficiency of each weak acid.

2. The wet burette must be rinsed with NaOH before filling to make sure that the

concentration of the NaOH was not contaminated with other particles.

3. Potassium hydrogen phthalate solutions must be carefully prepared to get the

accurate concentration of NaOH.

4. The tip of the burette must be filled with NaOH before initial volume was recorded in

order to get the accurate volume that required neutralizing the acid solution.

5. An indicator such as phenolphthalein must be used to monitor the pH change. An

indicator changes color when the pH reaches a certain point. So, we will be aware

with the pH changes and when it reaches the equivalence point.

16

REFERENCES

Brown, T. L.,LeMay, H. E.,Bursten, B. E.and Murphy, C. J. (2009). "CHEMISTRY: the central

science," 11th edition/Ed. Prentice Hall. 673 - 675.

Chang, R. (2007). "Chemistry," 9th edition/Ed. McGraw-Hill. 150 - 709.

Stanton;, B.,Zhu;, L.and Atwood, C. H. (2009). "EXPERIMENTS IN GENERAL

CHEMISTRY: measureNet," 2nd edition/Ed. Cengage Learning. 155 - 157.

Scharf, W.and Malerich, C. (2010). Determination of Acetic Acid Content of Vinegar. Baruch

College.

17

lab report (vinegar)

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