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 of vinegar 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 approximately 6 g of 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 1.5 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 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 (Equation 2 – 1)
Litre of solution Percent by mass is the mass in grams of solute per 100 grams of solution Percent solute = grams of solute Grams of solution
X 100% (Equation 2 – 1)
Vinegar is a dilute solution of acetic acid. The molecular formula for acetic acid is CH3COOH. 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
1
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. Aim Aim of this experiment is to determine the molarity of a solution and the percent by mass of acetic acid in vinegar by titration with a 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 as shown in Figure 4.1, therefore the volume dispensed from the burette should be estimated to the nearest 0.05 mL.
Figure 4.1: Depicts a typical 50 mL burette with indication of smallest calibration unit (0.1 mL) 2
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 4 – 1: NaOH (aq) + CH3CO2H (aq) → NaCH3CO2 (aq) + H2O (l)
(Equation 4 – 1)
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 4 – 2)
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 3 – 5). 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.
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Figure 4.3: Acid – base titration curve of weak acid titrated with NaOH
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
Potassium hydrogen phthalate, KHC8H4O4, 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: 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)
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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.
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.
5
6. Procedure 6.1
Standardization of sodium hydroxide, NaOH solution
1. Weight of NaOH solid needed to prepare the 0.6 M NaOH solution is calculated and confirmed by the lecturer. 2. 250 mL of approximately 0.6 M NaOH solution is prepared from the NaOH solid. 3. A beaker is placed on the weighing balance and the beaker’s weight is tarred and recorded. 4. Then, 1.5 g of KHP granules is added into the beaker and its weight is recorded to the nearest 0.001 g. 5. Next, 30 mL of distilled water is added into the beaker and the solution is stirred until the KHP granules completely dissolved in the solution. 6. The pH value of the KHP solution without the addition of NaOH solution is recorded until pH meter reading stabilized. 7. This solution is then titrated with 2 mL of NaOH solution. 8. Then, the pH value of the solution is recorded after the reading on pH meter is stabilized. 9. Step 7 and 8 are repeated until the pH values of the solution achieve its stable state. 10. After that, the titration process is repeated twice for NaOH standardization. 11. The graph of pH versus NaOH for all titration processes are plotted and the volume of NaOH needed to neutralize the KHP solution is determined from all three graphs. 12. Next, the molarity of NaOH solution for titration 1, 2, and 3 are calculated. 13. Lastly, the average molarity of the NaOH solution is calculated. 6.2
Molarity of acetic acid and mass percent in vinegar
1. 10 mL of vinegar is transferred into a clean and dry beaker. 2. 100 mL of distilled water is then added into the beaker. 3. Next, pH value of the vinegar solution is recorded after the pH meter reading stabilized. 4. Then, 2 mL of NaOH solution is added into the vinegar solution and its pH value is recorded after the pH meter reading stabilized. 5. Step 4 is repeated until the pH values of the solution achieve its stable state. 6. After that, step 1 until 5 are repeated twice more for standardization.
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7. The graph of pH versus NaOH for all titration processes are plotted and the volume of NaOH needed to neutralize the vinegar solution is determined from all three graphs. 8. Next, the molarity and the average molarity of acetic acid solution for titration 1, 2, and 3 are calculated. 9. Lastly, the percent by mass and the average percent by mass of acetic acid solution in vinegar for titration 1, 2, and 3 are calculated.
7. Results 7.1
Standardization of sodium hydroxide, NaOH solution
7.1.1
Results for titration 1 Titration 1
Volume
0
2
4
6
8
10
12
14
4.668
4.931
5.192
5.522
6.220
16
18
of NaOH (mL) pH
3.976 4.363
12.147 12.528 12.689
value
Graph of pH versus volume of NaOH for titration 1 14 12
pH = 9.5
10
pH
8
Equivalence Point (13.11 mL)
6 4 2 0 0
5
10
15
20
Volume of NaOH (mL) Based on the graph, molarity of NaOH solution for titration 1 = 0.5715 M
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7.1.2
Results for titration 2 Titration 2
Volume
0
2
4
6
8
10
12
14
4.659
4.918
5.174
5.513
6.220
16
18
of NaOH (mL) pH
3.972 4.343
12.196 12.538 12.685
value
Graph of pH versus volume of NaOH for titration 2 14 12 10
pH = 9.4
pH
8
Equivalence Point (13.06 mL)
6 4 2 0 0
5
10
15
20
Volume of NaOH (mL) Based on the graph, molarity of NaOH solution for titration 2 = 0.5625 M 7.1.3
Results for titration 3 Titration 3
Volume
0
2
4
6
8
10
12
4.683
4.933
5.176
5.512
6.184
14
16
18
of NaOH (mL) pH
3.960 4.340
12.201 12.535 12.679
value
8
Graph of pH versus volume of NaOH for titration 3 14 12 10
pH = 9.3
pH
8
Equivalence Point (13.04 mL)
6 4 2 0 0
5
10
15
20
Volume of NaOH (mL) Based on the graph, molarity of NaOH solution for titration 3 = 0.5643 M
7.1.4
Results for average value of titration Average
Volume
0
2
4
6
8
10
12
4.670
4.927
5.181
5.516
6.208
14
16
18
of NaOH (mL) pH
3.969 4.349
12.181 12.534 12.684
value
9
Graph of pH versus volume of NaOH for average value of titration 14 12 10
pH = 9.4
pH
8
Equivalence Point (13.07 mL)
6 4 2 0 0
5
10
15
20
Volume of NaOH (mL) Based on the graph, molarity of NaOH solution for average value of titration = 0.5661 M
7.1.5
Overall results Titration 1
Titration 2
Titration 3
Mass of KHP (g)
1.5298
1.5000
1.5027
Volume of NaOH to
13.11
13.06
13.04
neutralize the KHP solution (mL)
7.2
Molarity of acetic acid and mass percent in vinegar
7.2.1
Results for titration 1 Titration 1
Vol.
0
2
4
6
8
10
12
14
16
18
20
22
of NaOH (mL) pH
2.545 3.417 3.808 4.351 4.453 4.593 4.860 5.252 7.136 11.619 11.911 12.050
10
Graph of pH versus volume of NaOH for titration 1 14 12 10
pH = 8.8
pH
8
Equivalence Point (16.74 mL)
6 4 2 0 0
5
10
15
20
25
Volume of NaOH (mL) Based on the graph; Molarity of acetic acid, CH3COOH in vinegar solution for titration 1 = 0.9477 M Percent of acetic acid, CH3COOH in vinegar solution for titration 1 = 5.962 % 7.2.2
Results for titration 2 Titration 2
Vol.
0
2
4
6
8
10
12
14
16
18
20
22
of NaOH (mL) pH
2.751 3.508 3.899 4.178 4.423 4.669 4.937 5.333 6.849 11.699 11.966 12.039
11
Graph of pH versus volume of NaOH for titration 2 14 12 10
pH = 8.8 pH
8 6
Equivalence Point (16.80 mL)
4 2 0 0
5
10
15
20
25
Volume of NaOH (mL) Based on the graph; Molarity of acetic acid, CH3COOH in vinegar solution for titration 2 = 0.9510 M Percent of acetic acid, CH3COOH in vinegar solution for titration 2 = 5.712 % 7.2.3
Results for titration 3 Titration 3
Vol.
0
2
4
6
8
10
12
14
16
18
20
22
of NaOH (mL) pH
2.850 3.608 3.984 4.263 4.491 4.729 5.030 5.414 7.260 11.748 12.008 12.078
12
Graph of pH versus volume of NaOH for titration 3 14 12 10
pH = 8.9
pH
8
Equivalence Point (16.73 mL)
6 4 2 0 0
5
10
15
20
25
Volume of NaOH (mL) Based on the graph; Molarity of acetic acid, CH3COOH in vinegar solution for titration 3 = 0.9471 M Percent of acetic acid, CH3COOH in vinegar solution for titration 3 = 5.688 % 7.2.4
Results for average value of titration Average
Vol.
0
2
4
6
8
10
12
14
16
18
20
22
of NaOH (mL) pH
2.715 3.511 3.897 4.264 4.456 4.664 4.942 5.333 7.082 11.689 11.962 12.056
13
Graph of pH versus volume of NaOH for average value of titration 14 12 10
pH = 8.8
pH
8
Equivalence Point (16.76 mL)
6 4 2 0 0
5
10
15
20
25
Volume of NaOH (mL) Based on the graph; Molarity of acetic acid, CH3COOH in vinegar solution for average value of titration = 0.9486 M Percent of acetic acid, CH3COOH in vinegar solution for average value of titration = 5.7873 %
7.2.5
Overall results
Volume of NaOH to
Titration 1
Titration 2
Titration 3
16.74
16.80
16.73
neutralize the KHP solution (mL)
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 10–3) L 14
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) =6g
8.1.2
Sample calculation for the molarity of NaOH using data from titration 1
8.1.2.1 Volume of NaOH needed to neutralize the KHP Perform interpolation: 9.500 – 6.220
= Vol – 12
12.147 – 6.220
14 – 12
3.280
= Vol – 12
5.927
2
6.560
= 5.927 Vol – 71.724
Vol
= 13.11 mL of NaOH
8.1.2.2 Molarity of NaOH Moles of KHP: 1.5298 g KHC8H4O4 X
1 mol KHC8H4O4
= 0.007492 mol HC8H4O4
204.2 g KHC8H4O4 Moles of NaOH required neutralizing moles of KHP: 0.007492 mol KHC8H4O4
X
1 mol NaOH
= 0.007492 mol NaOH
1 mol KHC8H4O4 Molarity of NaOH solution: 13.11 mL NaOH
X
1L
= 0.01311 L NaOH
1000 mL
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M = mol of NaOH = 0.007492 mol NaOH = 0.5715 M NaOH L of solution
0.01311 L NaOH
8.1.2.3 Average molarity of NaOH
Mave = (M1 + M2 + M3) / 3 = (0.5715 + 0.5625 + 0.5643) / 3 = 0.5661 M NaOH 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
8.2.1.1 Volume of NaOH needed to neutralize the vinegar Perform interpolation: 8.800 – 7.136
= Vol – 16
11.619 – 7.136
18 – 16
1.664
= Vol – 16
4.483
2
3.328
= 4.483 Vol – 71.728
Vol
= 16.74 mL of NaOH
8.2.1.2 Molarity of acetic acid in vinegar
Moles of NaOH that reacted: 16.74 mL of NaOH
X
1L
= 0.01674 L NaOH
1000 mL 0.01674 L NaOH
X 0.5661 mol NaOH = 0.009477 mol NaOH 1L NaOH solution
Moles of CH3COOH neutralized by moles of NaOH: 0.009477 mol NaOH X 1 mol CH3COOH = 0.009477 mol CH3COOH Mol NaOH
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Molarity of CH3COOH: 10 mL CH3COOH
X
1L
=
0.010 L CH3COOH
1000 mL M = mol of CH3COOH = 0.009477 mol CH3COOH = 0.9477 M CH3COOH L of solution
0.010 L CH3COOH
8.2.1.3 Average molarity of acetic acid in vinegar
Mave = (M1 + M2 + M3) / 3 = (0.9477 + 0.9510 + 0.9471) / 3 = 0.9486 M CH3COOH 8.2.1.4 Percent 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 0.9477 mol CH3COOH
X
60.06 g CH3COOH
1 L solution
1 mol CH3COOH
= 0.5692 g CH3COOH Mass of acetic acid solution: 10 mL CH3COOH
X 1 g CH3COOH solution
= 10 g CH3COOH solution
1 mL CH3COOH solution Percent by mass of acetic acid in the solution =
g CH3COOH
X 100 %
g CH3COOH solution =
0.5692
X 100%
10.0 = 5.692 %
8.2.1.4 Average percent by mass of acetic acid in vinegar: %ave = (%1 + %2 + %3) / 3 = (5.962 + 5.712 + 5.688 %) / 3 = 5.7873 %
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9. Discussion The purpose of this experiment is achieved as the molarity of the solution and the percent by mass of acetic acid in vinegar have been determined by using titration with a standardized sodium hydroxide solution. In the first part of this experiment in which to standardize the sodium hydroxide solution, a primary standard acid solution is initially prepared. Sodium hydroxide, NaOH is used as the base meanwhile potassium hydrogen phthalate, KHC8H4O4 or jotted as KHP is used as primary standard acid. In addition, pH meter is used to measure the pH of the solution. It is necessary to constantly measure and record the pH of the solution by keep on immersing the pH meter in the solution to ensure more accurate results.
Based on graph and calculated result, it is showed that the average volume of NaOH needed to neutralize the primary standard acid is 13.07 mL at average pH of 9.4. 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 9.4. In other words, the solution started to neutralize from pH of 6.208 up to 12.181. This is because of some of the hydrogen ions are gradually neutralized with the increment volume of NaOH. Thus, a sudden sharp increase in pH occurred as sufficient volume of NaOH is added into the acid solution. Furthermore, the pH at the endpoint of a weak acid–strong base titration is always greater than 7 because strong base allows hydrogen ions in weak acid to neutralize more easily.
The sudden fluctuation in the pH also occurred at the second part of this experiment. In the second part of the experiment, 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 by 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. The average molarity of acetic acid in vinegar is 0.9486 and its average percent by mass is 5.7873 %. In addition, it is important 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.
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10. Conclusion The results of the first part of this experiment showed that when the mass of KHP is 1.5298 g, volume of NaOH required to neutralize the acid is 13.11 mL and the molarity of NaOH solution for titration 1 is 0.5715 M. While for titration 2, when the mass of KHP is 1.500 g, volume of NaOH required to neutralize the acid is 13.06 mL and the molarity of NaOH solution for titration 2 is 0.5625 M. For titration 3, when the mass of KHP is 1.5027 g, volume of NaOH required to neutralize the acid is 13.04 mL and the molarity of NaOH solution for titration 3 is 0.5643 M. Meanwhile, the results of the second part of the experiment showed that molarity of acetic acid in vinegar solution for titration 1 = 0.9477 M, percent of acetic acid in vinegar solution for titration 1 = 5.962 % and the volume of NaOH required to neutralize the solution is 16.74 mL. For titration 2, molarity of acetic acid in vinegar solution is 0.9510 M, percent of acetic acid in vinegar is 5.712 % and the volume of NaOH required to neutralize the solution is 16.80 mL. Lastly, molarity of acetic acid in vinegar solution for titration 3 = 0.9471 M, percent of acetic acid in vinegar solution for titration 3 = 5.688 % and the volume of NaOH required to neutralize the solution is 16.73 mL.
Thus, it can be concluded that, the greater the mass of solute in the acid solution, the more concentrated the solution becomes. Hence, the higher the molarity and more volume of NaOH needed to neutralize the acid.
11. Recommendation 1. It is better to use an indicator such as phenolphthalein as it colour will change at a pH of ~9.0 rather than using a pH meter as using a pH meter is fairly tedious and it will turns out unnecessary. 2. The swirling of the solution should be constant while adding the NaOH in order to ensure that the NaOH is totally dispersed. 3. Ensure that the position of eye is directly perpendicular to the meniscus when reading the volume of solution to avoid inaccuracy. 4. It is better to carry out three accurate titration so that the experimental error is reduces by calculating the average value. 5. Ensure that the tip of the burette is filled with NaOH so that no air bubbles are present in the tip.
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6. When the solution is being prepared, wear safety glasses and gloves as solid sodium hydroxide is corrosive, and can cause severe burns to eyes and skin while sodium hydroxide solution irritates the eyes.
12. References/Appendix Sample figure of equipment’s used in this experiment.
Figure 12.1: Hot plate
Figure 12.2: pH meter
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