Vinegar Lab 1

8/10/2019 Vinegar Lab 1

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1. Abstract

Acetic acid, CH3COOH is an organic compound that is in form of colorless solution and

classified as a weak acid. Acetic acid is the main component ofvinegar apart from water. In this

experiment, the molarity of a solution and the percent by mass of acetic acid in vinegar is

determined by using titration with a standardized sodium hydroxide, NaOH solution. The

experiment is divided into two parts which are standardizing the NaOH solution is the first part

and the second part is proceeded with the determining the molarity of a solution and the percent

by mass of acetic acid in vinegar. In standardizing the NaOH solution, 250 mL of distilled water

is used to dilute NaOH solid in order to prepare 0.6 M NaOH solution. This NaOH solution is

then titrated with potassium hydrogen phthalate, KHC8H4O4 or KHP solution which has been

prepared by diluting 6.0 g of KHP granules in 30 mL of distilled water. The experiment is then

preceded to the second part of the experiment which standardized NaOH solution is titrated the

with 10 mL vinegar that has been diluted with approximate 100 mL of distilled water. Both

titration for part 1 and 2 are repeated thrice to get more accurate results. Based on results, it can

be conclude that the greater the mass of solute in the acid solution, the more concentrated the

solution becomes thus, the higher the molarity and more volume of NaOH needed to neutralize

the acid. The experiment is completed and successfully conducted.

2. Introduction

Concentration of solution is the amount of solute in a given amount of solvent. A concentrated

solution contains a relatively large quantity of solute in a given amount of solvent. Dilute

solutions contains a relatively little solute in a given amount of solvent. There are two specific

terms to express concentration, namely molarity and percent by mass:

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

Molarity (M) = moles of solute

Litre of solution (Equation 21)

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

Percent solute = grams of solute X 100%

Grams of solution (Equation 21)
http://en.wikipedia.org/wiki/Organic_compoundhttp://en.wikipedia.org/wiki/Vinegarhttp://en.wikipedia.org/wiki/Vinegarhttp://en.wikipedia.org/wiki/Organic_compound


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Vinegar is a dilute solution of acetic acid. The molecular formula for acetic acid is CH 3COOH.

Both molarity and percent by mass of acetic acid in the vinegar solution can be determined by

performing a titration. A titration is 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 that reaction is attained. Knowing the quantity of the known solution required

to complete the titration enables calculation of the unknown solution concentration. 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.

3. Objectives

To :

(a)Determine the morality of a solution and the percent by mass of acetic acid in vinegar

by titration with the standardized sodium hydroxide solution.

4. Theory

In the titration process, a burette is used to dispense a small, quantifiable increment of solution of

known concentration. A typical burette has the smallest calibration unit of 0.1 mL, therefore the

volume dispensed from the burette should be estimated to the nearest 0.05 mL. In this

experiment, the equivalent point occurs when the moles of acid in the solution equals the moles

of base added in the titration. For example, the stoichiometric amount of 1 mole of strong base,

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

(CH3CO2H), as indicated in equation 41:

NaOH (aq) + CH3CO2H (aq) NaCH3CO2(aq) + H2O (l) (Equation 41)

The sudden change in the solution pH shows that the titration has reached the equivalence point.

pH in an aqueous solution is related to its hydrogen in concentration. Symbolically, the hydrogen

ion concentration is written as [H3O+]. pH is defined as the negative of the logarithm of the

hydrogen ion concentration.

pH =log10[H3O+] (Equation 42)


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pH scale is a method of expressing the acidity or basicity of a solution. Solutions with pH < 7 are

acidic, pH = 7 are neutral, and pH > 7 are basic as shown in Figure 4.2. For example,, a solution

having an H3O+ concentration of 2.35 x 10

2 M would have a pH of 1.629 and is acidic. pH

electrodes will be used in this experiment.

Figure 4.2: pH scale

The titration is initiated by inserting a pH electrode into a beaker containing the acid solution

(pH 35). As sodium hydroxide, NaOH, is incrementally added to the acid solutions, some of

the hydrogen ions will be neutralized. As the hydrogen ion concentrated decreases, the pH of the

solution will gradually increase. When sufficient NaOH is added to completely neutralize the

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

a sudden sharp increase in pH as shown in Figure 4.3. The volume of based required to

completely neutralized the acid is determined at the equivalent point of titration.

Figure 4.3: Acidbase titration curve of weak acid titrated with NaOH


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In this experiment, titration of a vinegar sample with standardized sodium hydroxide solution

will be performed. 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. Primary standard acid or

bases have several common characteristics:

They must be available in at least 99.9 purity

They must have a high molar mass to minimize error in weighing

They must be stable upon heating

They must be soluble in the solvent of interest

Vinegar is a dilute solution of acetic acid (HC2H3O2) which is one of the many weak acids

used in chemistry. In contrast, there are only seven common strong acids: HCl, HBr, HI, HNO3,

H2SO4, HClO3, and HClO4. The difference between a strong acid and a weak acid is the degree

of dissociation when placed in water. A strong acid is essentially 100% dissociated into H+ (or

H3O+) and an anion when placed in water; a weak acid is less than 100% dissociated. Whether

an acid is strong or weak does not affect its reaction with a base - strong and weak acids both

react swiftly and completely with a base. The concentration seen on a vinegar label is typically

expressed as a mass percent (mass of solute / mass of solution 100). The molarity of vinegar

can be converted to mass percent once the density of the vinegar solution is known. The molarity

of the vinegar provides the number of moles of acetic acid in 1 L vinegar the moles of acetic

acid can be converted to grams acetic acid (solute) while the 1 L vinegar (solution) can be

converted into grams of vinegar (solution). Potassium hydrogen phthalate, KHC 8H4O4, and

oxalic acid, (COOH)2, are common primary standard acids. Sodium carbonate, Na2CO3, is the

most commonly used base. Most acids and bases (e.g. HCL, CH3COOH, NaOH, and KOH) are

mostly available in primary standard form. To standardize one of these acidic or basic solutions,

titration of the solution with a primary standard should be performed. In this experiment, NaOH

solution will be titrated with potassium hydrogen phathalate (KHP). The reaction equation for

this is:


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KHC8H4O4(aq) + NaOH (aq) KNaC8H4O4(aq) + H2O (l) (Equation 4

3)

Once the sodium solution has been standardized it will be titrated with 10.00 mL aliquots of

vinegar. The reaction equation for vinegar with NaOH is:

CH3COOH (aq) + NaOH (aq) NaCHCOO (aq) + H2O (l) (Equation 4

4)

Knowing the standardized NaOH concentration and using Equation 4 4, the molarity and

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


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5. Apparatus

Apparatus used in this experiment including descriptions are listed in Table 5.1. Meanwhile,

materials used in this experiment are listed in Table 5.2.

Table 5.1: List of apparatus used in the experiment including its description

No. Apparatus Descriptions

1 Hot plate To assist the stirring of solution in the experiment.

2 Magnetic stirrer To stir the solution.

3 Retort stand To hold the burette.

4 Beaker There are two types of beaker used in this experiment:

I. 250mL beaker

II.

10mL beaker

All solutions used in this experiment were placed in these

beakers.

5 pH meter To measure changes of pH value of the solution.

6 Burette To place the NaOH solution for titration.

7 Weighing balance To weight sufficient amount of materials used in the

experiment.

8 Measuring cylinder To measure and transfer the right amount of solutions needed

from its actual container into the beaker.

Table 5.2: List of materials used the experiment.

No. Materials

1 Sodium hydroxide, NaOH solid.

2 Potassium hydrogen phthalate, KHC8H4O4 (KHP) granules.

3 Vinegar.

4 Distilled water.


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6. Procedure

Standardization of sodium hydroxide solution

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

solid. The solution is prepared in a beaker, and the calculation is checked with the

laboratory instructor prior to preparing the solution. The calculations are recorded.

2. A 250 mL beaker is weighed and the mass is recorded to the nearest 0.001g. 1.5

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

to the nearest 0.001 g. The mass of KHP by difference is calculated and the data is

recorded. 30mL of distilled water is added to the beaker. The solution is stirred until

the KHP has dissolved completely.

3. This solution is titrated with NaOH and the pH with 1 ml additions of NaOH

solution is recorded.

4. Steps 1 to 3 are repeated to perform a second and third trial to standardize the NaOH

solution.

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

the KHP solution in each titration is determined from the plots.

6. The molarity of sodium hydroxide for titration 1 and 2 is calculated.

7. The average morality of sodium hydroxide solution for titration 1 and 2 is calculated.

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

experiment.

Molarity of acetic acid and percent of vinegar

1. 10.00mL of vinegar is transfered to a clean, dry 250 mL beaker using a 10mL

volumetric pipette. Sufficient water is added, 75 to 100 mL, to cover the pH

electrode tip during the titration.

2. 1 ml of NaOH is added to the vinegar solution and the pH is recorded.

3. The steps above is repeated twice more.

4. The graph of pH vs volume NaOH added is plotted. The volume of NaOH required to

neutralize vinegar in each titration is determined from the plots. Data are recorded.

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

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


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7. The percent by mass of acetic acid in vinegar for titration 1 and 2 is calculated.

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

7. Results

Standardization of NaOH Solution (KHP)

Titration 1 Titration 2

Mass of beaker (g) 98.4076 g 96.8620 g

Mass of beaker (g) + KHP (g) 99.9080 g 98.3615 g

Mass of KHP (g) 1.5004 g 1.4995 g

Volume of NaOH to neutralize

the KHP solution (ml)

14.1 ml 14.1 ml

Average volume = 14.1 ml

Standardization of NaOH Solution (Vinegar)

Titration 1 Titration 2

Volume of NaOH to neutralize

the vinegar (ml)

22.1 ml 34.1 ml

Average volume = 28.1 ml


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Titration of NaOH to KHP

Volume of

NaOH/ml Titration 1 KPH/ph Titration 2 KPH/ph

1 4.27 4.28

2 4.46 4.41

3 4.68 4.56

4 4.80 4.65

5 4.83 4.77

6 4.93 4.89

7 5.04 5.01

8 5.15 5.12

9 5.28 5.24

10 5.39 5.39

11 5.55 5.54

12 5.71 5.7213 6.00 5.98

14 6.43 6.62

15 12.00 12.11

16 12.55 12.60

17 12.74 12.77

18 12.88 12.91

19 12.98 12.98

20 13.02 13.04


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Graph 1: Graph of pH versus volume of NaOH for titration 1

0

2

4

6

8

10

12

14

0 5 10 15 20 25

KPH/pH

Volume of NaOH/ml

Titration 1 KPH/ph

Titration 1 KPH/ph


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Graph 2: Graph of pH versus volume of NaOH for titration 2

0

2

4

6

8

10

12

14

0 5 10 15 20 25

KPH/pH

Volume of NaOH/ml

Titration 2 KPH/ph

Titration 2 KPH/ph


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Titration of NaOH to KHP

Volume of

NaOH/ml

Titration 1

Vinegar/pH

Titration 2

Vinegar/pH

1 3.37 3.102 3.63 3.42

3 3.82 3.60

4 3.98 3.73

5 4.10 3.85

6 4.22 3.95

7 4.32 4.04

8 4.42 4.12

9 4.5 4.20

10 4.59 4.26

11 4.70 4.34

12 4.76 4.4413 4.90 4.49

14 4.95 4.51

15 5.09 4.55

16 5.15 4.60

17 5.27 4.65

18 5.35 4.70

19 5.48 4.74

20 5.67 4.80

21 6.03 4.85

22 6.61 4.90

23 11.74 4.97

24 12.23 5.02

25 12.44 5.08

26 5.15

27 5.23

28 5.30

29 5.41

30 5.53

31 5.66

32 5.82

33 6.1334 6.51

35 11.47

36 12.04

37 12.28

38 12.43

39 12.55

40 12.62


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Graph 3: Graph of pH versus volume of vinegar for titration 1

0

2

4

6

8

10

12

14

0 5 10 15 20 25 30

Vinegar/pH

Volume of vinegar/ml

Titration 1 Vinegar/pH



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Graph 4: Graph of pH versus volume of vinegar for titration 2

0.00

2.00

4.00

6.00

8.00

10.00

12.00

14.00

0 10 20 30 40 50

Vinegar/pH

Volume of vinegar/ml




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8. Calculations

8.1 Standardization of sodium hydroxide, NaOH solution

8.1.1 Calculation for preparing 150 mL of approximately 0.6 M NaOH solution

Molarity (M) = moles of solute

Litre of solution

0.6 = moles of solute

(250 X 103

)L

Moles of solute = (0.6) (0.25)

= 0.15 mol

No. of moles = mass

Molecular weight of NaOH

Mass = (no. of moles) (MW NaOH)

= (0.15) (22.99 + 16.00 + 1.01g)

= 6.00 g


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8.1.2.2Molarity of NaOH

Moles of KHP:

1.5004 g KHC8H4O4 X 1 mol KHC8H4O4= 0.007348 mol HC8H4O4

204.2 g KHC8H4O4

Moles of NaOH required neutralizing moles of KHP:

0.007348 mol KHC8H4O4 X 1 mol NaOH = 0.007348 mol NaOH

1 mol KHC8H4O4

Molarity of NaOH solution:

14.5 mL NaOH X 1L = 0.0145 L NaOH

1000 mL

M = mol of NaOH = 0.007348 mol NaOH = 0.5068 M NaOH

L of solution 0.0145 L NaOH

8.1.2.3Average molarity of NaOH

Mave

= (M1+ M

2) / 2

= (0.5068 + 0.5068) / 2

= 0.5068 M NaOH


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8.2 Molarity of acetic acid and mass percent in vinegar

8.2.1 Sample calculation for average percent by mass of acetic acid if vinegar using data from

titration 1

8.2.1.1Volume of NaOH needed to neutralize the vinegar

From the graph, volume of NaOH needed to neutralize the vinegar is 22.5 mL

8.2.1.2Molarity of acetic acid in vinegar

Moles of NaOH that reacted:

22.5 mL of NaOH X 1L = 0.0225 L NaOH

1000 mL

0.0225 L NaOH X 0.5068 mol NaOH = 0.011403 mol NaOH

1L NaOH solution

Moles of CH3COOH neutralized by moles of NaOH:

0.011403 mol NaOH X 1 mol CH3COOH = 0.011403 mol CH3COOH

Mol NaOH

Molarity of CH3COOH:

22.1 mL CH3COOH X 1L = 0.0221 L CH3COOH

1000 mL

M = mol of CH3COOH = 0.011403 mol CH3COOH = 0.5160 M CH3COOH

L of solution 0.0221 L CH3COOH


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8.2.1.3Average molarity of acetic acid in vinegar

Mave= (M1+ M2) / 2

= (0.5160 + 0.3344) / 2

= 0.4252 M CH3COOH

8.2.1.4Percent by mass of acetic acid in vinegar

Mass of acetic acid in the solution:

10 mL CH3COOH X 1L = 0.010 L CH3COOH

1000 mL

0.010 L CH3COOH X 0.5160 mol CH3COOH X 60.06 g CH3COOH

1 L solution 1 mol CH3COOH

= 0.3099 g CH3COOH

Mass of acetic acid solution:

22.1 mL CH3COOH X 1 g CH3COOH solution = 22.1 g CH3COOH solution

1 mL CH3COOH solution

Percent by mass of acetic acid in the solution = g CH3COOH X 100 %

g CH3COOH solution

= 0.3099 X 100%

22.1

= 1.4023 %


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9. Discussion

Sodium hydroxide (a base) is typically used as a standard to determine the concentration of

acids. When an acid and base react, they form salt and water.

acid + base salt + water

The decrease in the concentration of H3O + ions will cause the equilibrium system to shift to the

right and as such more of the weak acid (hydrogen phthalate ion, HP - ) will ionize.

Successive additions of NaOH will continually remove H3O + ions, shift the equilibrium to the

right and force the weak acid into complete ionization.

The aim of this experiment has been successfully achieved as the percent by mass of acetic acid

in vinegar and the molarity of the solution and has been determined by using the titration method

with a standardized sodium hydroxide solution prepared. For the first part of this experiment, to

standardize the sodium hydroxide solution, a primary standard acid solution is initially prepared

beforehand. Sodium hydroxide, NaOH, is used as the base. Potassium hydrogen phthalate,

KHC8H4O4or normally known as KHP, is used as the primary standard acid. In addition to that,

an electronic pH meter was used to obtain the value of pH of the solution. To ensure more

accurate results are obtained, it is necessary to constantly measure the pH of the solution and

record the pH of the solution by keep on immersing the pH meter in the solution.

Based on graph that was plotted and result calculated, it is revealed that the average volume of

NaOH needed to neutralize the primary standard acid is 14.1 mL at average pH of 9.29. As

mention in the theory, solutions with pH less than 7 are acidic, pH equal to 7 are neutral, and pH

greater than 7 are basic. However, the result in this experiment does not exactly parallel with the

theory as the solution gains its equivalence point at average pH of 9.29. In other words, the

solution started to neutralize from pH of 6.525 up to 12.055. The reason being is that because of

some of the hydrogen ions are gradually neutralized with the increment volume of NaOH. Thus,

an immediate sharp escalation in pH occurred as sufficient volume of NaOH is added into the

acid solution. In addition, the pH at the endpoint of a weak acidstrong base titration is always


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greater than 7 because strong base will always allow hydrogen ions in weak acid to neutralize

more easily.

At the second part of this experiment with using vinegar as a substitute of KHP, there is also a

sudden fluctuation in the pH. In the second the experiment using vinegar, the molarity of acetic

acid and the mass percent in vinegar are determined. The molarity of acetic acid and the mass

percent in vinegar are calculated using the average volume of NaOH resulted from the first part

of the experiment and with the help of graph plotted based on results from the second part of the

experiment. Additionally, it is essential to dilute the vinegar in order to avoid a very small titre,

which would reduce the accuracy of the experiment.

The significance of percent by mass and molarity of solution in this experiment is that it tells

whether the solution is either diluted or concentrated solution. Hence, the acetic acid in the

vinegar is a dilute solution as its percent by mass and molarity are relatively small.

10. Conclusion

From the experiment conducted, the results of the first part of this experiment showed that when

the mass of KHP is 1.5004 g, volume of NaOH required to neutralize the acid is 14.1 mL and the

molarity of NaOH solution for titration 1 is 0.5068 M. While for titration 2, when the mass of

KHP is 1.4995 g, volume of NaOH required to neutralize the acid is 14.1 mL and the molarity of

NaOH solution for titration 2 is 0.5068 M.

Meanwhile, the results of the second part of the experiment showed that molarity of acetic acid

in vinegar solution for titration 1 = 0.5160 M, percent of acetic acid in vinegar solution for

titration 1 = 1.4023 % and the volume of NaOH required to neutralize the solution is 22.5 mL.

For titration 2, molarity of acetic acid in vinegar solution is 0.3344 M, percent of acetic acid invinegar is 0.5890 % and the volume of NaOH required to neutralize the solution is 34.1 mL.

As a result, it can be established that, the bigger the mass of solute in the acid solution, the more

concentrated the solution becomes. Henceforth, the higher the molarity and more volume of

NaOH needed to neutralize the acid.


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11. Recommendation

The swirling of the solution should be constant while adding the NaOH. The reason is to ensure

that the NaOH is totally dispersed and mixed well before taking every reading using the pH

meter. Secondly, make sure that the position of eye is directly perpendicular to the meniscus

when reading the volume of solution. The reason is to avoid inaccuracy in the volume entered in

the reaction. Third, it is better to carry out three or more accurate titration. The reason is that the

experimental error is reduces by calculating the average value. Fourth, ensure that the tip of the

burette is filled completely with NaOH so that no air bubbles are present in the tip. The presence

of air bubbles will affect the accuracy of the experiment.

Vinegar Lab 1

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