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1. Summary

This experiment is conducted to determine the molarity of a solution and the percent by

mass of acetic acid (CH3COOH) in vinegar by titration with the standardized sodium hydroxide

(NaOH) solution. First step is to obtain standardized NaOH solution by titrate it with hydrogen

phthalate (KHP) solution which the molarity is already known. Expressing the chemical reaction

for the reaction and identify the concentration of OH- in the NaOH solution. Thus, obtain the

molarity of NaOH. Then, calculate the moles of NaOH that reacted with CH3COOH. Only after

that, the moles of CH3COOH neutralized by NaOH can be obtained. Mass of CH3COOH in the

solution can be calculated from the moles obtained. Same goes to the percent by mass of

CH3COOH in solution.

2. Introduction

Amount of solute in a given amount of solvent is the concentration of solution. Large

quantity of solute in a given amount of solvent will make the solution more concentrated. While

for dilute solutions, it contained relatively small amount of solute in a given amount of solvent.

Concentration can be expressed in two ways, which is molarity and percent by mass.

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

( )

(Equation 1)

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

(Equation 2)

One example for dilute solution is vinegar which contained solvent of acetic acid (CH3COOH).

The concentration of acetic acid for both, molarity and percent by mass, can be determined by

titration process. A process in which small increments of a solution of known concentration are

added to a specific volume of a solution of unknown concentration until the stoichiometry for

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that reaction is attained. Then the equivalence point of the reaction is determined. The point

when the added amount of reactant is the exact amount necessary for stoichiometric reaction

with another reactant.

3. Objective

I) 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.

4. Theory

Burette is used to dispense a small, quantifiable increment of solution of known

concentration as in titration process. Common burette has the smallest calibration unit of 0.1 mL,

thus, volume dispense from the burette should be estimated to the nearest 0.01 mL.

Figure 1: Set-up for titration

When the moles of acid in the solution equals to the mole of base added in the titration, the

equivalence point achieved. As example, amount of one mole of the strong base, NaOH, is

necessary to neutralize one mole of weak acid, CH3COOH, in stoichiometric as shown in

equation 3.

NaOH (aq) + CH3COOH (aq) → NaCH3CO2 (aq) + H2O (l) (Equation 3)

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When the pH undergoes sudden changes, it shows that the titration process had achieved

equivalence point. pH indicates the concentration of hydronium ion concentration [H3O+]

and

defined as the negative logarithm of the hydronium ion concentration.

pH = -log [H3O+] (Equation 4)

pH scale is used to identify the acidity or basicity of one solution. If pH scale is larger than 7,

then the solution are basic, if the pH scale is lesser than 7, than the solution is acidic, and if pH

scale is 7, the solution are neutral. To measure the pH scale, in this experiment, pH electrode is

use.

In titration process, as NaOH is added, the concentration of hydronium ion will decrease

or been neutralize. When NaOH added is enough to neutralize the acid, the next drop of NaOH

will cause sharp increase in pH. This is where equivalence point of titration placed, thus, the

volume of base needed to completely neutralized the acid will be determine from this point of

equivalence.

Figure 2: Titration curve

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For this experiment, titration of vinegar with a standardized NaOH solution was done. Before

that, NaOH need to be standardized thus a primary standard acid solution is prepared. Primary

standard acid or bases have several common charactheristic:

they must be available in at least 99.9% purity

they must have a high molar mass to minimized error in weighing

they must stable upon heating

they must be soluble in the solvent of interest

In this experiment, NaOH will be titrated with hydrogen phthalate (KHC8H4O4) or KHP. The

equation for this reaction was like this.

KHC8H4O4(aq) + NaOH(aq) → KNaC8H4O4(aq) +H2O(l) (Equation 5)

When the NaOH has been standardized, it will be used to titrated 10.00 mL aliquots of vinegar.

The equation of reaction of NaOH with CH3COOH is as below.

CH3COOH (aq) + NaOH (aq) → NaCH3COO (aq) + H2O (l) (Equation 6)

By using equation 5 and 6, we can determine the molarity ad percent by mass of acetic acid in

the vinegar.

5. Procedure

5.1 Standardization of Sodium Hydroxide Solution. (Part A)

i. 250 mL of approximately 0.6 M NaOH was prepared from NaOH solid. The

calculation for preparing the solution is checked with laboratory instructor prior and

recorded.

ii. 250 mL beaker was weighted to the nearest 0.001g and recorded. 1.5g of KHP

was added to the empty beaker. Mass of beaker and the KHP recorded to nearest 0.001g.

The mass of KHP added is calculated by difference and was recorded. 30 mL of distilled

water was added into the beaker. The solution is stirred until KHP has dissolved

completely.

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iii. KHP been titrated with NaOH that has been prepared. The initial pH reading of

KHP solution was taken. 1 mL of NaOH was dropped one at time while recording the pH

change of KHP solution. The titration is stopped after 2 mL after the equivalent point

achieve. Repeat the process twice.

iv. Graph of pH versus NaOH was plotted. From the graph, the volume of NaOH

needed to neutralize KHP solution is determined.

v. Molarity of NaOH for both titration and average molarity are calculated.

vi. Result from Part A will be used in Part B.

5.2 Titration of Vinegar. (Part B)

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

mL volumetric pipette. Sufficient water was added around 90 mL, enough to cover the

pH electrode tip during titration.

ii. The solution was titrated with NaOH same ways in Part A.

iii. The step i and ii is repeated twice.

iv. Graph of pH values versus volume of NaOH added was plotted. The equivalent

point is determined.

v. Molarity of acetic acid in titration 1 and 2 and average molarity are calculated.

vi. Percent by mass of acetic acid is calculated.

6. Apparatus and Material

6.1 Apparatus

i. Retort Stand

ii. 250 mL Beaker

iii. 50 mL Burette

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iv. 10 mL Pipette

6.2 Meterial

i. Solid NaOH

ii. Solid KHP

iii. Vinegar

7. Result

Part A

Titration 1 Titration 2

Mass Beaker (g) 101.46 175.19

Mass Beaker + KHP (g) 102.99 176.72

Mass of KHP 1.53 1.58

Volume of NaOH to neutralize

the KHP solution (mL)

12.14 12.18

Part B

Titration 1 Titration 2

Volume of NaOH required to

neutralized vinegar

29.5 29.5

8. Sample of Calculation

Part A

Mass needed to prepare 250 mL of approximately 0.6 M NaOH solution.

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Molarity of NaOH for titration 1(a):

Moles of KHP used in titration:

Moles of NaOH required to neutralize the moles of KHP according equation 5:

Molarity of NaOH solution:

Molarity of NaOH in titration 2(a):

Moles of KHP used in titration:

Moles of NaOH required to neutralize the moles of KHP according equation 5:

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Molarity of NaOH solution:

Average molarity of NaOH used:

Part B

Molarity of acetic acid (a.a) in titration 1(b) and 2(b):

Moles of NaOH reacted:

Moles of a.a neutralized by the moles of NaOH according to equation 6:

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Molarity of a.a solution:

Mass of a.a in the solution:

Mass of the a.a solution:

Percent by mass of a.a in the solution:

9. Sample of Calculation Error

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10. Discussion

In the experiment conducted, result in finding the molarity of the solution and percent by

mass of acetic acid in vinegar was slightly higher from expected value. This is because the

experiment did not been conducted properly.

The molarity of standardize NaOH solution obtain was 0.62626 M in average. The actual

data was 0.4875 M. There are two probabilities why this error occur. First, there was error in

weighting solid KHP or solid NaOH by the student. Second one is the volume of distilled water

added to produce NaOH and KHP solution was not taken properly.

The molarity of acetic acid in vinegar obtain was 1.8475 M but in the actual data was

0.8263 M. This error happen maybe because the calculations in molarity of standardize NaOH

solution obtains was different. Also, it could be that the beaker used for storing the vinegar has

contaminated by other strong acid or mixing of vinegar with KHP when the used beaker for KHP

not properly washed then been used to produce the vinegar-water solution. There also may occur

due to systematic error which the instrument used give false reading such as the pipette.

Equivalence point was determined when one mole of H3O+ is equal to one mole OH

-. As

graph plotted from obtaining data (refer graph in appendices), it can be seen that at the beginning

of the titration, the pH value of the solution (KHP or vinegar) was increase slowly. This is where

the OH- ion in the base solution reacted with H3O

+ ion in acidic solution to produce water thus

decreasing the concentration of H3O+ ion. When the acidic is fully neutralized, another drop of

NaOH will turn the solution into basic solution since the concentration of OH- ion increase

dramatically. This is where the equivalent point placed. After the sharp increase of pH has

occurred, addition of NaOH solution will gradually increase the pH value.

For average titration graph of KHP to NaOH (refer graph average(a) in appendices), the

initial pH value of KHP solution is 3.910. The pH value is gradually increased when NaOH is

added until it reached pH 6.150 with 12 mL NaOH. At this point, the KHP solution has been

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fully neutralized by NaOH. Another addition of NaOH has make the pH value rocketed from

6.15 to 11.590 with 12 mL to 13 mL NaOH. The equivalent point occurs which the concentration

of H3O+ equal to OH

-. Addition of NaOH of after that will increase the pH value slightly again.

Thus this form a sigmoid curve for the graph pH values versus volume of NaOH added.

Average titration graph for vinegar to NaOH (refer graph average(b) in appendices),

initial pH value of vinegar solution was 2.565. The addition of NaOH slowly increase the pH

value of vinegar solution until it reached pH 5.945 with 28 mL NaOH. Drops after that makes

the pH reading jumped from 5.945 to 11.300 with 28 mL to 30 mL of NaOH. This where the

equivalent point occurs which the concentration of H3O+ was equal to OH

-. The addition of

NaOH after that makes the pH solution increase slightly. This will also form a sigmoid curve

graph.

11. Conclusion

As a conclusion, since the result of the experiment has more than 5% error, the

experiment has fail to identify accurate data for the objective given which to find the molarity of

vinegar and percent by mass of acetic acid in vinegar. The actual result is 0.8263 M for molarity

of vinegar and 1.963% by mass of acetic acid in vinegar. Still, the experiment has given students

brilliant idea how to standardize base or acidic solution and also to identify the equivalent point.

12. Recommendations

To overcome the error during the experiment, student must:

Using a proper instrument to weighting material which has high sensitivity at least to

nearest 0.001 g.

Avoid parallax error when taking measurement by makes sure that eye position is

perpendicular to the reading meter.

Wash properly every each instrument used to measured volume and the beaker with

distilled water to minimize the contamination.

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13. References

Books:

Steven S. Zumdahl, Susan, Titration: Ph Indicator, Thermometric Titration, Nonaqueous

Titration, Equivalence Point, Acid-Base Titration, Amperometric Titration, 2010,

General Books LLC

Brown, LeMay, Bursten, Murphy, Chemistry: The Central Science,2009, Pearson

Education Inc.

John McMurry, Organic Chemistry, 2008, Cengage Learning

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14. Appendices

Part A

Graph 1: 1st titration of KHP with NaOH

Graph 2: 2nd

titration of KHP with NaOH

0

2

4

6

8

10

12

14

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

pH

val

ue

Volume of NaOH (mL)

Titration 1(a)

Titration 1(a)

0

2

4

6

8

10

12

14

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

pH

val

ue

Volume of NaOH (mL)

Titration 2(a)

Titration 2(a)

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Graph 3: Average titration of KHP with NaOH

Part B

Graph 4: 1st titration of vinegar with NaOH

0

2

4

6

8

10

12

14

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

pH

val

ue

Volume of NaOH (mL)

Average (a)

Average (a)

0

2

4

6

8

10

12

14

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

pH

val

ue

Volume of NaOH (mL)

Titration 1(b)

Titration 1(b)

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Graph 5: 2nd

titration of vinegar with NaOH

Graph 6: Average titration of vinegar with NaOH

0

2

4

6

8

10

12

14

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

pH

val

ue

Volume of NaOH (mL)

Titration 2(b)

Titration 2(b)

0

2

4

6

8

10

12

14

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

pH

val

ue

Volume of NaOH (mL)

Average (b)

Average (b)