Download - 92849518 Acetic Acid in Vinegar
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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)