Determination of The Concentration of Acetic Acid in
Vinegar
Name
:
Muhammad Haziq bin Abd Shukur
Matrix Number
:
2021132209
Lecturer’s Name
:
Dr Nik Raikhan Nik Him
Date
:
3/11/2021
Group
:
EH2202
Submit to
: Dr Nik Raikhan Nik Him
Student’s Name
: Muhammad Haziq bin Abd Shukur
Student ID
: 2021132209
No
1
2
3
4
5
6
7
8
9
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11
12
Title
Abstract/ Summary
Introduction
Objectives
Theory/ Literature review
Apparatus/ Material
Procedure
Result
Calculation
Discussion
Conclusion
Recommendation
Reference/ Appendix
Total Marks
Remarks
:
Checked By
:
…………………………………….
Date By
:
Allocated Marks (%)
5
10
5
10
5
10
10
10
20
5
5
5
100
Marks
Determination of the concentration of acetic acid in vinegar
1.0 Summary
The purpose of the experiment was to obtain the molarity and the percent by mass
of acetic acid in vinegar. The experiment carried out using appropriate procedure
of titration with the standardized sodium hydroxide, (NaOH) solution. The
experiment was start with standardization sodium hydroxide solution through
titration with a potassium hydrogen phthalate (KHP). Next, the standardized NaOH
solution needed to proceed with the second part of the experiment to titrate 10 mL
aliquots of vinegar. The graph of a acid-base titration was successfully plotted in
the orientation of pH versus volume of sodium hydroxide. The equivalent point of
the represented graph was utilized to obtain the pH and volume of NaOH to
neutralize vinegar solution. Also, the equivalence points were used to determine
the molarity of NaOH, molarity of acetic acid in vinegar and the percent by mass
of the acetic acid in vinegar. Throughout the experiment, it revealed that the
greater the mass of the solute in the acid solution, the concentration of the solution
increase. Resulting in higher molarity and more volume of NaOH needed to
neutralize the acid. Based on theory, As sodium hydroxide, NaOH, is incrementally
added to the acid solution, some of the hydrogen ions will be neutralized. There
were few possible errors which was common titration errors, air bubbles in the tip
and imprecision reading caused by wrong eye positioned ie not perpendicular to
the meniscus.
2.0 Introduction
There are 2 basic terms for the expression of concentration which is molarity
and mass percentage. Firstly, molarity is the amount of solution moles per litre.
Formula below shows how to determine molarity and percent by mass.
𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦, 𝑀 =
𝑀𝑜𝑙𝑒𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑜𝑙𝑢𝑡𝑒(𝑛)
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛(𝐿)
𝑃𝑒𝑟𝑐𝑒𝑛𝑡 𝑠𝑜𝑙𝑢𝑡𝑒, % =
𝑀𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒(𝑔)
× 100%
𝑀𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛(𝑔)
It should be noted, solution concentration is the solvent level for each solvent
quantity. Meaning, in any solvent volume, a concentrated solution contains a
relatively large amount of solvent in the particular solution. In the other hand, dilute
solutions contain relatively small amount of solvent. The vinegar is a dilute liquid
from acetic acid solution. In chemistry, the molecular formula of acetic acid is
CH3COOH. The main question raised in this particular experiment what are the
exact molarity and the percent by mass of acetic acid in vinegar can be achieve at
the end of the experiment. Next, the main aim of this experiment is to find out the
molarity of the solution and the percent by mass of acetic acid in vinegar by apply
the appropriate procedure of titration method. Besides, the overview of this
experiment is about titration of a vinegar sample with standardized sodium
hydroxide, NaOH solution.
3.0 Objective
The uttermost objective of this experiment is to determine the molarity of the
solution and the percent by mass of acetic acid in vinegar by using titration with
the standardized sodium hydroxide solution. In the relation to that, there are few
sub-objectives need to be determined. Firstly, the volume of sodium hydroxide
solution required to neutralized vinegar in titration. Secondly, the volume of sodium
hydroxide solution required to neutralize the KHP solution. Lastly, the equivalent
point of the reaction gained from the titration curve of pH against volume of NaOH.
4.0 Theory
In this experiment, the equivalence point happens when the acid moles in the
solution are equal to the base moles added to the titration. For example, the
stoichiometric amount of 1 mole of a strong base, sodium hydroxide (NaOH), is
required to neutralize 1 mole of weak acid, acetic acid (CH3CO2H), as indicated
at below equation.
A burette, as shown in appendix 1, is used in the titration process to dispense
a small, quantifiable increase of solution of known concentration. Because the
lowest calibration unit for the standard burette is 0.1mL, the volume discharging
from the burette must be calculated to the nearest 0.01mL.
Fig 1. a) Represents a typical 50 mL burette.b) Specifies the smallest calibration unit,
0.1mL on a typical 50 mL burette
Fig 2. Represents a standard 50mL burette with calibration unit 0.1mL
The rapid shift in the pH of the solution indicates that the titration has reached
its equivalence point. The concentration of hydrogen ions in an aqueous solution
is related to its pH. The concentration of hydrogen ion is symbolically expressed
as [H3O+]. The negative value of the hydrogen ion concentration logarithm,
represented as pH = - log [H30+], is referred to as pH. The pH scale is a method
of displaying a solution's acidity or basicity. Acidic solutions have a pH less than
7, neutral solutions have a pH of 7, and basic solutions have a pH more than 7. In
this experiment, the pH electrode will be used. Titration begins with the placement
of a pH electrode in an acid solution beaker with a pH value within 3 to 5.
Some hydrogen ions will be neutralised as sodium hydroxide, NaOH, is
gradually added to the acid solution. The pH of the solution will gradually climb as
the concentration of hydrogen ion drops. If enough NaOH is added to entirely
neutralise the acid, the pH will rise sharply with the next drop of NaOH added. At
the equivalence point of the titration, as indicated in appendix3, compute the
volume of base required to completely neutralise the acid.
Fig 3: The graph of pH versus volume of NaOH and the graph of the titration
between weak acid and strong base.
The vinegar sample will be titrated against a standardised sodium hydroxide
solution in this experiment. To standardise the sodium hydroxide solution, a
primary standard acid solution is first prepared. Primary standard solutions are
made by dissolving a known volume of pure acid or base in a weighed quantity of
solution. Primary standard acids and bases must have at least 99.9% purity, have
a high molar mass to reduce weight error, be heat stable, and be soluble in the
solvent of interest.
Primary standard acids were potassium hydrogen phthalate KHC8H4O4 and
oxalic acid (COOH)2. The most popular is sodium carbonate (Na2CO3). As
primary standards, most acids and bases, such as HCl, CH3COOH, NaOH, and
KOH, are easily accessible. It is necessary to titrate one of these acids or based
solutions with the primary standard in order to standardise it. In this experiment, a
NaOH solution will be titrated with potassium hydrogen phthalate (KHP). This
reaction's equation is as follows:
KHC8H4O4 (aq) + NaOH (aq)→KNaC8H4O4 (aq) + H2O (l)
10.00 mL vinegar aliquots will be titrated only after sodium hydroxide solution
has been standardised. The reaction of vinegar with NaOH is described by the
equation below.
CH3COOH (aq) + NaOH(aq)→NaCH3COO (aq) + H2O(l)
The molarity and percentage of acetic acid in the vinegar solution can be
calculated using the following equation and the standard concentration of NaOH.
The titration curve of 1.523 gram of KHP diluted in 20.0 mL of NaOH titrated
distilled water is shown in Appendix 3.
Fig 4: Titration curve of potassium hydrogen phthalate , KHP with Sodium
Hydroxide, NaOH. The volume of Sodium Hydroxide, NaOH used at the
equivalent point is 15.3mL
To determine the molarity of NaOH the moles of the potassium hydrogen
phthalate, KHP used should be calculated as below calculations:
1.523𝑔𝐾𝐻𝐶8𝐻404
1 𝑚𝑜𝑙 𝐾𝐻𝐶8𝐻404
= 0.007458 𝑚𝑜𝑙 𝐾𝐻𝐶8𝐻404
204.2𝑔 𝐾𝐻𝐶8𝐻404
Then, based on below eqaution , the moles of Na OH required neutralizing the
moles of KHP can be calculated.
KHC8H4O4 (aq) + NaOH (aq)→KNaC8H4O4 (aq) + H2O (l)
0.007458𝑚𝑜𝑙𝐾𝐻𝑃 ×
1𝑚𝑜𝑙𝑁𝑎𝑂𝐻
= 0.007458𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
1𝑚𝑜𝑙𝐾𝐻𝑃
After that the molarity of NaOH is determined.
(15.30𝑚𝐿 𝑁𝑎𝑂𝐻) ×
1𝐿
= 0.01530𝐿 𝑁𝑎𝑂𝐻
1000𝑚𝐿
𝑀 = 𝑚𝑜𝑙
𝑁𝑎𝑂𝐻
0.007458 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
=
= 0.4875𝑀 𝑁𝑎𝑂𝐻
𝐿 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
0.01530𝐿 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
Then ,to determinw the acetic acid concentration in vinegar by titration with
standard base, the molarity and the percent by mass of CH3COOH in the solution
is calculated.
A 10.00mL aliquot of vinegar requires 16.95 mL of 0.4875 m NaOH
standardized solution to reach the equivalence point of tittration. The density of the
solution of vinegar is 1.00g/mL. To find out molarity the moles of NaOH reacted is
calculated:
1𝐿
= 0.01695𝐿 𝑁𝑎𝑂𝐻
100𝑚𝐿
0.4875𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
0.01695𝐿 𝑁𝑎𝑂𝐻 ×
= 0.008263𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
1𝐿 𝑁𝑎𝑂𝐻 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
16.95𝑚𝐿 𝑁𝑎𝑂𝐻 ×
Then, the moles of CH3COOH which that has been neutralized by the moles of
NaOH is determined by using below equation.
CH3COOH (aq) + NaOH(aq)→NaCH3COO (aq) + H2O(l)
0.008263𝑚𝑜𝑙 𝑁𝑎𝑂𝐻 ×
1𝑚𝑜𝑙𝐶𝐻3𝐶𝑂𝑂𝐻
= 0.008263𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
1𝑚𝑜𝑙𝑁𝑎𝑂𝐻
After that the molarity of the CH3COOH solution is determined.
1𝐿
= 0.010𝐿 𝐶𝐻3𝐶𝑂𝑂𝐻 𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛
1000𝑚𝐿
𝐶𝐻3𝐶𝑂𝑂𝐻
0.008263 𝑚𝑜𝑙 𝐶𝐻3𝐶𝑂𝑂𝐻 0.8263𝑚𝑜𝑙 𝐶𝐻3𝐶𝑂𝑂𝐻
𝑀 = 𝑚𝑜𝑙
=
=
𝐿 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
0.01𝐿 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝐿 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
10𝑚𝑙 𝐶𝐻3𝐶𝑂𝑂𝐻 ×
= 0.8263𝑀 𝐶𝐻3𝐶𝑂𝑂𝐻
Next , to find the percent the mass of acetic acid, the first step is to find the
mass of the acetic acid in the solution:
1𝐿
= 0.010𝐿 𝐶𝐻3𝐶𝑂𝑂𝐻 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
1000𝑚𝐿
0.8263 𝑚𝑜𝑙 𝐶𝐻3𝐶𝑂𝑂𝐻 60.06𝑔 𝐶𝐻3𝐶𝑂𝑂𝐻
0.01𝐿 𝐶𝐻3𝐶𝑂𝑂𝐻 ×
×
1𝐿 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
1𝑚𝑜𝑙 𝐶𝐻3𝐶𝑂𝑂𝐻
10𝑚𝐿 𝐶𝐻3𝐶𝑂𝑂𝐻 ×
= 0.4963𝑔 𝐶𝐻3𝐶𝑂𝑂𝐻
Then , the mass of acetic acid solution is also calculated
10𝑚𝐿 𝐶𝐻3𝐶𝑂𝑂𝐻 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 ×
1𝑔 𝐶𝐻3𝐶𝑂𝑂𝐻 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
1𝑚𝐿𝐶𝐻3𝐶𝑂𝑂𝐻 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
= 10.00𝑔 𝐶𝐻3𝐶𝑂𝑂𝐻 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
Finally the percent by mass of acetic acid solution in the solution can be
determined by below equation:
%𝑎𝑐𝑒𝑡𝑖𝑐 𝑎𝑐𝑖𝑑 𝑖𝑛 𝑣𝑖𝑛𝑒𝑔𝑎𝑟 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 =
𝑝𝑒𝑟𝑐𝑒𝑛𝑡 𝑚𝑎𝑠𝑠 𝐶𝐻3𝐶𝑂𝑂𝐻 =
𝑝𝑒𝑟𝑐𝑒𝑛𝑡 𝑚𝑎𝑠𝑠 𝐶𝐻3𝐶𝑂𝑂𝐻 =
𝑔𝑟𝑎𝑚𝑠 𝑜𝑓 𝑎𝑐𝑒𝑡𝑖𝑐 𝑎𝑐𝑖𝑑
× 100
𝑔𝑟𝑎𝑚𝑠 𝑜𝑓 𝑣𝑖𝑛𝑒𝑔𝑎𝑟 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝑔𝐶𝐻3𝐶𝑂𝑂𝐻
× 100%
𝑔𝐶𝐻3𝐶𝑂𝑂𝐻 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
0.4693𝑔𝐶𝐻3𝐶00𝐻
× 100% = 1.963%𝐶𝐻3𝐶𝑂𝑂𝐻
10.00𝑔𝐶𝐻3𝐶𝑂𝑂𝐻
5.0 Methodology
Fig 5. Experimental set up for titration of NaOH and KHP
Fig 6. Experimental set up for tittration of NaOH and vinegar
5.1 Apparatus
1. Burette
2. Retord stand
3. 10mL volumetric pipette
4. Vinegar solution
5. Weighing scales
6. Magnetic stirrer
7. pH meter
8. Filter funnel
9. pH electrode
5.2 Materials
1. 0.6M Sodium Hydroxide, NaOH Solid
2. 250mL Beaker
3. Potassium Hydrogen Phthalate, KHP
4. Distilled water
5.3 Procedure
Standardization of Sodium Hydroxide Solution
1. 250 mL of 0.6 M Sodium Hydroxide, NaOH was prepared from 6 g of Sodium
Hydroxide, NaOH solid in a beaker.
2. The 250 mL 0.6 NaOH was transferred to burette by using filter funnel. Drain
the solution into the sink through the tip of the burette. Fill the burette with
NaOH solution to the top of the initial calibration line. Make sure the tip of the
burette is filled with the sodium hydroxide solution.
3. Then, the 250 mL beaker was placed on the weighing balance and its weight
was recorded in table 1 to the nearest 0.001g.
4. 1.5 g of Potassium Hydrogen Phthalate, KHP granules was added into the
250 mL beaker and the weight was recorded to the nearest 0.001 g.
5. After that, 30 mL of distilled water was added into the beaker consist of KHP
and stirred using magnetic stirrer.
6. The solution was stirred with magnetic stirrer until the KHP granules
dissolved completely in the solution.
7. The pH value of the KHP solution without addition of the NaOH solution was
recorded until pH meter reading stabilized.
8. Next, the solution consist of KHP was titrated with 250 mL of 0.6 Sodium
Hydroxide, NaOH prepared in step 1 and 2.
9. The pH was recorded every 1 mL additions of NaOH until the pH values of
the solution achieved the stable state. The pH value was recorded by pH
electrode and pH meter
10. Steps 1 until 10 was repeated to perform second titration to standardize the
NaOH solution.
11. The graph of pH versus volume NaOH for all titration was plotted. From the
plots, the volume of NaOH required neutralizing the KHP solution in each
titration was determined.
12. The molarity of sodium hydroxide for titration 1 and 2 were calculated.
13. The average molarity of sodium hydroxide solution was calculated for titration
1 and 2.
14. The resulting sodium hydroxide concentration was used in part B of the
experiment.
Molarity of acetic acid and percent of vinegar
1. 250 mL of beaker was rinsed and cleaned with distilled water
2. Then, 10 mL of volumetric pipette was rinsed with vinegar
3. 10 mL of vinegar was transferred to a clean 250 mL beaker by using a 10
mL of volumetric pipette. The sufficient water was added from 75 to 100
mL to cover the pH electrode tip during the titration.
4. The resulting NaOH in a burette from part A of the experiment was used
and transfer to the burette and ready for titration.
5. Every 1 mL of NaOH was titrated to the vinegar solution and the pH value
was recorded by inserting pH in the beaker and connected to pH meter.
6. Steps 1 until 5 was repeated to perform second titration to standardize
the NaOH solution.
7. The graph of pH vs volume NaOH added was plotted. The volume of
NaOH required to neutralized vinegar in each titration was determined.
All data were recorded in table 2.
8. The molarity of acetic acid in vinegar for titration1 and 2 was calculated.
9. The average molarity of acetic acid for each titration was calculated.
10. The percent by mass of acetic acid in vinegar for titration 1 and 2 was
calculated
11. The percent by mass of acetic acid in vinegar was calculated.
6.0 Results
Table 1- Standardization of Sodium Hydroxide Solution Using KHP Solution
Titration 1
Titration 2
Mass of Beaker (g)
106.485
106.485
Mass of Beaker + KHP (g)
107.996
107.998
1.511
1.513
12.5
12.5
30.0
30.0
Mass of KHP (g)
Volume of NaOH to neutralize
KHP Solution (mL)
Volume of Distilled Water (mL)
Table 3- Titration between Sodium Hydroxide Solution and KHP Solution
Volume of NaOH (mL)
Titration 1
Titration 2
0
pH Volume
4.18
pH Volume
4.68
1
4.73
4.75
2
5.15
5.00
3
5.16
5.05
4
5.24
5.17
5
5.31
5.41
6
5.49
5.58
7
5.60
5.60
8
5.83
5.74
9
5.90
5.93
10
6.13
6.27
11
6.49
6.55
12
7.34
6.98
13
12.65
12.42
14
12.88
12.84
15
12.99
12.97
Fig 7. Titration Curve pH Versus volume of NaOH (mL) in titration 1. The volume of
NaOH used at the equivalence point is 12.5mL
Fig 8.Titration Curve pH Versus volume of NaOH (mL) in titration 2. The volume of
NaOH used at the equivalence point is 12.5mL
Table 2- Standardazation of Sodium Hydroxide with vinegar solution
Titration 1
Titration 2
Mass of Beaker (g)
106.485
106.485
Mass of Beaker + Vinegar (g)
116.498
116.500
10
10
14.5
17.5
100
100
Volume of Vinegar (mL)
Volume of NaOH to neutralize
Vinegar solution (mL)
Volume of Distilled Water (mL)
Table 4- Titration between Sodium Hydroxide with vinegar solution
Volume of NaOH
(mL)
Titration
1
pH Volume
Titration 2
pH Volume
0
4.18
4.68
1
4.73
4.75
2
5.15
5.00
3
5.16
5.05
4
5.24
5.17
5
5.31
5.41
6
5.49
5.58
7
5.60
5.60
8
5.83
5.74
9
5.90
5.93
10
6.13
6.27
11
6.49
6.55
12
7.34
6.98
13
12.65
12.42
14
12.88
12.84
15
12.99
12.97
Fig 9. Titration Curve pH Versus volume of NaOH (mL) in titration 1. The volume of
NaOH used at the equivalence point is 14.5ML
Fig 10. Titration Curve pH Versus volume of NaOH (mL) in titration 2. The volume of
NaOH used at the equivalence point is 17.5ML
7.0 Calculations
Standardization of Sodium Hydroxide solution using KHP solution
i)
Calculations for preparing 250mL of approximately 0.6M sodium hydroxide solution.
𝑉 = 250𝑚𝐿 = 0.25𝐿
𝑀 = 0.60𝑀
𝑀𝑜𝑙𝑎𝑟 𝑀𝑎𝑠𝑠 𝑁𝑎𝑂𝐻 = 40
𝑔
𝑚𝑜𝑙
𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 = 𝑀𝑜𝑙/𝑉𝑜𝑙𝑢𝑚𝑒
𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 =
ii)
0.60𝑚𝑜𝑙 40𝑔
×
× 0.25𝐿 = 𝟔𝒈 𝑵𝒂𝑶𝑯
𝐿
𝑚𝑜𝑙
Calculations for preparing 150mL of approximately 0.6M sodium hydroxide solution.
𝑉 = 150𝑚𝐿 = 0.15𝐿
𝑀 = 0.60𝑀
𝑀𝑜𝑙𝑎𝑟 𝑀𝑎𝑠𝑠 𝑁𝑎𝑂𝐻 =
40𝑔
𝑚𝑜𝑙
𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 = 𝑀𝑜𝑙/𝑉𝑜𝑙𝑢𝑚𝑒
𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 =
iii)
0.60𝑚𝑜𝑙 40𝑔
×
× 0.15𝐿 = 𝟑. 𝟔𝒈 𝑵𝒂𝑶𝑯
𝐿
𝑚𝑜𝑙
The molarity of sodium hydroxide for each titration 1 and 2
KHC8H4O4 (aq) + NaOH (aq) → KNaC8H4O4 (aq) + H2O (l)
Titration 1
Mw Potassium Hydrogen Phthalate, KHP (C8H5KO4) =204.22g/mol
1.5𝑔
𝑀𝑜𝑙 =
0.0073 𝑚𝑜𝑙
204.22𝑔
𝑚𝑜𝑙
𝑉𝑜𝑙𝑢𝑚𝑒 𝑁𝑎𝑂𝐻 = 12.5𝑚𝑙 = 0.0125𝐿
𝑛
𝑀=
𝑣
0.0073
𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑁𝑎𝑂𝐻 =
= 0.584𝑀
0.0125
Titration 2
𝑉𝑜𝑙𝑢𝑚𝑒 𝑁𝑎𝑂𝐻 = 12.5𝑚𝑙 = 0.0125𝐿
𝑛
𝑀=
𝑣
0.0073
𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑁𝑎𝑂𝐻 =
= 0.584𝑀
0.0125
Average Molarity of NaOH
0.584+0.584
2
= 𝟎. 𝟓𝟖𝟒𝑴
Standardization of Sodium Hydroxide with vinegar solution
i)
Molarity of acetic acid in vinegar for titration 1 and 2
Titration 1
Reacted NaOH in Moles
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑁𝑎𝑂𝐻 𝑟𝑒𝑎𝑐𝑡𝑒𝑑 = 14.5𝑚𝐿 = 0.0145𝐿
𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 = 0.584
𝑚𝑜𝑙
𝑣𝑜𝑙𝑢𝑚𝑒
𝑀𝑜𝑙𝑒(𝑛) = 𝑀𝑉
𝑀𝑜𝑙𝑒 𝑜𝑓 𝑁𝑎𝑂𝐻 𝑟𝑒𝑎𝑐𝑡𝑒𝑑 = (0.584)(0.0145) = 0.0085 𝑚𝑜𝑙𝑒
1 𝑚𝑜𝑙𝑒 𝑜𝑓 𝑁𝑎𝑂𝐻 = 1 𝑚𝑜𝑙𝑒 𝑜𝑓 𝐶𝐻3𝐶𝑂𝑂𝐻
𝑀𝑜𝑙𝑒 𝑜𝑓 𝐶𝐻3𝐶𝑂𝑂𝐻 = 0.0085 𝑚𝑜𝑙𝑒
Molarity of the CH3COOH solution
𝑉𝑜𝑙𝑢𝑚𝑒 𝑎𝑐𝑒𝑡𝑖𝑐 𝑎𝑐𝑖𝑑 = 10𝑚𝐿 = 0.01𝐿
𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 𝑎𝑐𝑒𝑡𝑖𝑐 𝑎𝑐𝑖𝑑 =
=
𝑚𝑜𝑙𝑒
𝑣𝑜𝑙𝑢𝑚𝑒
0.0085
= 0.850𝑀
0.01
Titration 2
Reacted NaOH in Moles
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑁𝑎𝑂𝐻 𝑟𝑒𝑎𝑐𝑡𝑒𝑑 = 17.5𝑚𝐿 = 0.0175𝐿
𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 = 0.584
𝑚𝑜𝑙
𝑣𝑜𝑙𝑢𝑚𝑒
𝑀𝑜𝑙𝑒(𝑛) = 𝑀𝑉
𝑀𝑜𝑙𝑒 𝑜𝑓 𝑁𝑎𝑂𝐻 𝑟𝑒𝑎𝑐𝑡𝑒𝑑 = (0.584)(0.0175) = 0.0102 𝑚𝑜𝑙𝑒
1 𝑚𝑜𝑙𝑒 𝑜𝑓 𝑁𝑎𝑂𝐻 = 1 𝑚𝑜𝑙𝑒 𝑜𝑓 𝐶𝐻3𝐶𝑂𝑂𝐻
𝑀𝑜𝑙𝑒 𝑜𝑓 𝐶𝐻3𝐶𝑂𝑂𝐻 = 0.0102 𝑚𝑜𝑙𝑒
Molarity of the CH3COOH solution
𝑉𝑜𝑙𝑢𝑚𝑒 𝑎𝑐𝑒𝑡𝑖𝑐 𝑎𝑐𝑖𝑑 = 10𝑚𝐿 = 0.01𝐿
𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 𝑎𝑐𝑒𝑡𝑖𝑐 𝑎𝑐𝑖𝑑 =
=
Average molarity of acetic acid
𝑚𝑜𝑙𝑒
𝑣𝑜𝑙𝑢𝑚𝑒
0.0102
= 1.02𝑀
0.01
0.850+1.02
2
= 𝟎. 𝟗𝟑𝟔𝑴
ii)
The percentage,% by mass of acetic acid in vinegar for titration 1 and 2
Titration 1
Mass of acetic acid in the solution
𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑎𝑐𝑒𝑡𝑖𝑐 𝑎𝑐𝑖𝑑 (𝐶𝐻3𝐶𝑂𝑂𝐻) = 60𝑔/𝑚𝑜𝑙
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑎𝑐𝑒𝑡𝑖𝑐 𝑎𝑐𝑖𝑑 = 10𝑚𝐿 = 0.01𝐿
𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 𝑜𝑓 𝑎𝑐𝑒𝑡𝑖𝑐 𝑎𝑐𝑖𝑑 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = 0.850𝑀
𝑀𝑎𝑠𝑠 =
60𝑔 0.850𝑚𝑜𝑙
×
× 0.01𝐿 = 0.510𝑔
𝑚𝑜𝑙
𝐿
Mass of acetic acid solution
𝑉𝑜𝑙𝑢𝑚𝑒 𝑎𝑐𝑒𝑡𝑖𝑐 𝑎𝑐𝑖𝑑 = 10𝑚𝐿
1𝑔
𝑀𝑎𝑠𝑠 = 10𝑚𝐿 ×
= 10𝑔
1𝑚𝐿
Percent mass of acetic acid in the solution
𝑃𝑒𝑟𝑐𝑒𝑛𝑡 𝑚𝑎𝑠𝑠 % =
0.510𝑔
× 100% = 𝟓. 𝟏%
10𝑔
Titration 2
Mass of acetic acid in the solution
𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑎𝑐𝑒𝑡𝑖𝑐 𝑎𝑐𝑖𝑑 (𝐶𝐻3𝐶𝑂𝑂𝐻) = 60𝑔/𝑚𝑜𝑙
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑎𝑐𝑒𝑡𝑖𝑐 𝑎𝑐𝑖𝑑 = 10𝑚𝐿 = 0.01𝐿
𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 𝑜𝑓 𝑎𝑐𝑒𝑡𝑖𝑐 𝑎𝑐𝑖𝑑 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = 1.02𝑀
𝑀𝑎𝑠𝑠 =
60𝑔 1.02𝑚𝑜𝑙
×
× 0.01𝐿 = 0.612𝑔
𝑚𝑜𝑙
𝐿
Mass of acetic acid solution
𝑉𝑜𝑙𝑢𝑚𝑒 𝑎𝑐𝑒𝑡𝑖𝑐 𝑎𝑐𝑖𝑑 = 10𝑚𝐿
1𝑔
𝑀𝑎𝑠𝑠 = 10𝑚𝐿 ×
= 10𝑔
1𝑚𝐿
Percent mass of acetic acid in the solution
0.612𝑔
𝑃𝑒𝑟𝑐𝑒𝑛𝑡 𝑚𝑎𝑠𝑠 % =
× 100% = 𝟔. 𝟏𝟐%
10𝑔
iii)
The average percent by mass of acetic acid in vinegar
5.04 + 6.12
= 5.58%
2
8.0 Discussion
The main aim of this experiment is to find out the molarity of the solution and
the percent by mass of acetic acid in vinegar by apply the appropriate procedure of
titration method. The overview of this experiment is about titration of a vinegar sample
with standardized sodium hydroxide, NaOH solution. In the first part of the experiment,
a primary standard acid solution has been used to standardize the sodium hydroxide
solution. From the calculation result, 6.0 g of NaOH solid was weighed to the nearest
tenth of a gram to prepare the solution. Next, 250 mL of distilled water was added to
the same beaker. The titration of the solution with a primary standard has been
performed in order to standardized solutions for sodium hydroxide NaOH. To be exact,
Potassium hydrogen phthalate, KHP was the standard acid that has been used in the
first part of the experiment. In addition, table 1 indicated the pH of the titration for every
1 mL added to the sodium hydroxide, NaOH until it reached 15 mL. Also, pH electrode
has been inserted into the beaker to initiate the pH. There was few steps taken to
enhance the quality of data including performed the experiment twice to get the
accurate data and continuously measure and record the pH of the solution by
continuing to immerse the pH meter in the solution. Chemical equation below shows
the reaction of NaOH solution and KHP during the titration.
KHC8H4O4 (aq) + NaOH (aq) → KNaC8H4O4 (aq) + H2O (l)
Furthermore, the data documented in the Table 1 was used to produce the
graph of the titration curve pH versus volume of NaOH, mL in the titration 1 and 2 as
shown at figure 1 and figure 2. Along with that, the volume and pH of NaOH to
neutralize KHP solution were determined by captured the equivalent point at both
graph of titration 1 and titration 2. In particular, the volume of NaOH at the equivalent
point in both titration is 12.5mL. While the pH obtained in both titration is 10 and 9.7
respectively. Conversely, from pH 7.34 up to 12.65 for titration1 began to neutralize
and from pH 6.98 up to 12.42 for titration 2 the solution began to neutralize. Based on
the theory the increment amount of the NaOH is finally neutralized by the hydrogen
ions. Thus, there was a rapid rise in pH of the solution.
Indeed, the pH scale is often said to range from 0 to 14 and the most solutions
do fall within this range. Anything below 7.0 is acidic and anything above 7.0 is alkaline.
In titration 1 and 2 the solutions has outreached at the basic state of the equivalence
point since the pH was 10 and 9.7 respectively. From the set data collection, the
molarity and percentage of acetic acid in the vinegar solution were determined. Next,
the concentration of the sodium hydroxide can be expressed by molarity and percent
by mass. As a result, the average molarity of sodium hydroxide for each titration 1 and
2 is 0.584 M
Subsequently, the experiment was followed by the standardization of sodium
hydroxide solution with vinegar solution. 10.00 mL of vinegar aliquots has been titrated
during the standardization. Chemical equation below shows the reaction of vinegar
and NaOH.
The balanced chemical equation portrayed a stoichiometric amount of 1 mole
of a strong base of NaOH was needed to neutralize q mole of weak acid CH3COOH.
From the set data collection, the molarity and percentage of acetic acid in the vinegar
solution were determined. As a result, the molarity of the acetic acid in vinegar in
titration 1 and 2 during the second part of the experiment are 0.850 M and 1.020 M
respectively. Next the average molarity of both titrations is 1.025 M. Apart from that
the percentage of acetic acid in vinegar was 5.1% in titration 1 and 6.12% in titration
2. Lastly but not least, the average percent by mass of acetic acid in vinegar in titration
1 and 2 is 5.58%.
9.0 Conclusion
To conclude, the main objective of the whole experiment is accomplished. As
well as that, the molarity of a solution and the percent by mass of acetic acid in vinegar
using titration with the standardized sodium hydroxide solution was determined. In the
first part of the experiment, standardization of sodium hydroxide solution using KHP
solution clearly revealed the molarity of a sodium hydroxide in each titration 1 and 2
was 0.584 M. Therefore, the molarity of the acetic acid in vinegar for titration 1 and 2
were achieved by a stoichiometric amount of 1 mole of a strong base of sodium
hydroxide. It was required to neutralize 1 mole of weak acid of acetic acid CH3COOH.
From the findings, the molarity of acetic acid in vinegar in titration 1 and 2 was 0.850
M and 1.020 M respectively. Besides, percent by mass of acetic acid in vinegar
successfully determined. The percent by mass o acetic acid in vinegar in titration 1
was 5.1% and titration 2 is 6.12%. The average of percent by mass of acetic acid in
vinegar was 5.58%. Thus it can be written that the higher the solute mass in the acid
solution, the more concentrated the solution becomes. Finally, the greater the molarity
required to neutralize the acid, the greater the volume of NaOH.
10.0 Recommendation
A few recommendations can be applied throughout the experiment. Firstly,
focused and put high attention during the titration process because there might be
chance sodium hydroxide absorb atmospheric carbon dioxide. Next, operate the
experiment using an indicator such as phenolphthalein. Also, avoid using a pH meter
because it will taking a lot of time. Besides, assure that the tip of the burette is filled
with NaOH so that there are no air bubbles in the tip. Moreover, check that the position
of the eye is directly perpendicular to the meniscus when reading the volume of the
solution to avoid imprecision. Last but not least, it is recommended to carry out three
accurate titrations to enhance the quality of average value of the data.
11.0 References
[1]
Penn State University, “Analyzing the Acid in Vinegar,” Penn State J., vol. 2, no. 3,
2017.
[2]
“Lab 8: Determination of Acetic Acid Content in Vinegar.”
[3]
I. Encyclopaedia Britannica, “Acetic Acid.” p. 1, 2020.
[4]
D. Li and S. Liu, “Seawater Quality Detection,” Water Qual. Monit. Manag., pp. 233–
249, 2019.
[5]
P. A. Jennings, C. A. Mullen, and M. Roy, “Titration and pH Measurement,” Encycl.
Life Sci., no. August 2018, 2010.
[6]
A. Marie, “What Is a Primary Standard in Chemistry?,” ThoughtCo, pp. 1–2, 2020.
[7]
Dr. Walter Scharf and Dr. Charles Malerich, “Determination of Acetic Acid Content of
Vinegar,” Japanese J. Radiol. Technol., vol. 44, no. 9, p. 1364, 1988.
[8]
C. LibreTexts, “Ways of Expressing Concentration,” 2020. [Online]. Available:
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry__The_Central_Science_(Brown_et_al.)/13%3A_Properties_of_Solutions/13.4%3A_W
ays_of_Expressing_Concentration.
[9]
K. Academy, “Acid-base tra on curves,” pp. 1–20, 2021.
[10]
Radiometer Analytical SAS, “Titration Theory and Practice Table of contents,” Notes,
pp. 1–46.