Don King O. Evangelista1, Geneizzel B. Gotuato1, Jackielyn B. Ocbina1, Jann Andre H. Padua1
Group 2, III-Bachelor of Science in Chemistry for Teachers, Philippine Normal University
Performed June 21, 2012
Professor Allen A. Espinosa, CHEM5 Professor
The experiment aims to determine the accurateness of the
data presented by vinegar-manufacturing factories in the
Philippines as their acetic acid content in their labels using the
direct titration method, specifically the analyte Datu Puti. Three
samples of 25.0 mL aliquot volumes were titrated using the KHPstandardized NaOH sample which is found to be of 0.05962 M
using the indicator phenolphthalein with observance to its
equivalence point. Data shows that the percent acetic acid of
Datu Puti whether by weight or by volume is less than the
alleged 8% acetic acid content according to its label, but with
consideration to the computed relative error of 35.73%. The
experimenters commend the titration process but recommend
to lessen errors such as systematic and instrument errors.
I. Introduction
Acid-base titrations are used routinely in virtually all fields of chemistry and in
related areas, such as biology, pharmacy, medicine and geology. In addition to
inorganic compounds, thousands of organic compounds exhibit sufficient acidity and
basicity that they can be determined by titration. Excellent understanding of acid-base
properties, together with the relative ease, speed and low-cost of performing titrations,
are major factors contributing to the popularity of the acid-base titration.1 In this
experiment, an organic compound, household vinegar which is mainly used in cooking
and bleaching, is subjected to the titration process, specifically direct titration, for
analyses of its concentration and percent acetic acid as compared to its reported
percent acetic acid written on its labels. Direct titration is a kind of titration where the
reactants involved are primarily the titrant and the analyte aA + tT οƒ  pP. 2
This experiment aims to (1) Test the reliability of the reported vinegar content of
acetic acid in the label by titration process, (2) Experience how analysts go over the
of quantitative
through the experiment and (3) Manifest
Determination of Total Acidity of Vinegar | GROUP 2 – III-BSCT
appreciation of Chemistry by maintaining accuracy and precision in the experiment
without sacrificing time and resources.
Generally, titrations are performed to determine the amount of analyte present
in a sample. To do this, the volume of titrant needed to reach the equivalence point
must be measured. An “equivalence point” by definition means “occurs when all the
moles of H+ ions present in the original volume of acid solution have reacted with an
equivalent number of moles of OH- ions added from the buret.4” These equivalence
phenolphthalein, primarily because of the inherent simplicity of this technique, but also
because of availability of many excellent indicators. These acid-base indicators are
weak organic acids or bases whose conjugate forms different colors. The indicator acts
as a second acid or base in the solution being titrated and must be weaker than the
analyte acid or base so that it reacts last with the titrant. The amount of indicator
added must be kept small compared to that of the analyte present so that it does not
consume an appreciable amount of titrant in the process of “indicating” or changing
from one conjugate form to the other. For this reason, indicators must be very intensely
colored so that only a few drops of a dilute solution, for this experiment,
phenolphthalein with a color of pink, are needed to produce a color that is observed
by the eye. 1
Before titration, the standardization of the molarity of NaOH must be made using
potassium hydrogen phthalate, often abbreviated KHP with a molecular formula of
KHC8H4O4. This compound has a high equivalent weight and is not hygroscopic. This
compound, being a weak acid, the pH at the equivalence point of its titration with a
strong base is in the alkaline region makes it as a good primary standard on
standardizing bases.
Acetic acid, the acid found in vinegar, is in definition a weak acid. Therefore, the
titration to be made is a weak acid titration to a strong base, which is NaOH. The
reaction of this titration is:
CH3COOH + OH- οƒ  CH3CO2- + H2O1
In this reaction, we can divide it into four. Before the addition of the titrant, the
main constituent is only the acetic acid, which the experimenters will treat as a weak
acid. Before the equivalence point, acetic acid is also present plus the presence of
Determination of Total Acidity of Vinegar | GROUP 2 – III-BSCT
CH3COONa that is now a buffer. At the equivalence point, CH3COONa is only present
which a weak base is now. After the equivalence point, CH3COONa is present plus
NaOH, which is now a strong base. 1
Another concept is that vinegar has a high acid concentration, thus it is good to
dilute the sample by incorporating aliquots to it. By diluting it, the quantity of the acetic
acid in the bottle of vinegar can be determined by introducing a “dilution factor.” This
factor will relate the amount of acetic acid in the diluted vinegar solution per volume of
portions of the aliquot. 3
With these concepts in hand, the percent by weight and percent by volume of
acetic acid in the sample can now be easily determined and calculated.
II. Methodology
Ample amounts of vinegar from the bottle of Datu Puti randomly picked in the
supermarket were undergone in an analysis. This was undergone into a quantitative
analysis using direct titration and phenolphthalein as an indicator against sodium
hydroxide to determine the acetic acid content. The following are the procedures in
the direct titration process:
First, standardize the prepared assumed .1M NaOH against KHP in three trials with
ranges of weights from 0.2 to 0.3 g. These samples were distilled in 50 mL of distilled
water and was titrated with phenolphthalein indicator as endpoint. According to the
data, the NaOH called for 16.07 mL, 18.22 mL and 17.32 mL for the 0.195 g, 0.2171 g and
0.2074 g samples of KHP respectively.
Next, the density of the vinegar, denoted by d = m/v, is measured by the
different groups by the process of weighing and concept of the Archimedes principle
which deducts the mass of the container itself. From these data, the experimenters got
the average mass of the 10 mL vinegar versus the volumes and found out that the
average density of vinegar is 0.98706 g/mL.
Lastly, the standardized NaOH from the first part, which is now measured with
concentration of 0.05962 mol/L (M) was now utilized to direct titrate the three samples
of vinegar aliquot with phenolphthalein indicator. The vinegar aliquot is produced by
diluting a 25-mL sample of pure vinegar with 250-mL of volumetric flask. Then, the
Determination of Total Acidity of Vinegar | GROUP 2 – III-BSCT
titration proceeded. Phenolphthalein endpoint is indicated by a faint pink color,
exceeding this pink color means that the titration is above its endpoint. According to
the data tables, each 25.0 mL aliquot of acetic acid needed volumes of 37.4 mL, 35.9
mL and 46.9 mL of NaOH for the three trials respectively.
The photo that follows shows the reactions and the succession of the titration
Figure 1: The Titration Process (from: Chemistry, The Molecular Nature of Matter and Change 4th Edition by Silberberg)
After gathering all the data from the three parts, these data is plotted in data
tables and are subject for statistical treatments, specifically the usage of the mean,
which indicated the central measurement and the deviation, standard deviation,
variance and coefficient of variance which are measures of variability.
The schematic diagram on the following page explains and summarizes all
procedures done in the experiment.
Determination of Total Acidity of Vinegar | GROUP 2 – III-BSCT
Standardization of the Prepared 0.1 M NaOH Solution
KHP Weighing with ranges
from 0.2 to 0.3 g
Dissolving it in 50 mL distilled
Titrating it using the
prepared NaOH
Titrate until end point of
phenolphthalein is reached
Measurement of Density of Vinegar Sample
Measure mass of dry 50 mL beaker, 50
mL beaker and 10 mL vinegar poured
via pipeting.
Get the value of the mass of vinegar
using these data.
Get density of vinegar using the
computed mass on Step 2 and the
volume of vinegar (10 mL).
Determination of Acetic Acid Content in Vinegar
Make the vinegar aliquout by diluting
25 mL acetic acid in 250-mL volumetric
flask by adding sufficient water. Add 2-3
drops phenolphthalein afterwards.
Titrate these aliquot portions.
Titrate until end point of
phenolphthaleinis reached. Note the
final volume of NaOH in the base buret.
Statistical Treatment of Data
Determination of Total Acidity of Vinegar | GROUP 2 – III-BSCT
III. Data and Results with Sample Calculations
The following data tables were computed first using a scientific calculator then
rechecked using MS Excel.
Table 1: Data for the Standardization of prepared 0.1 M NaOH
Mass of KHP (g)
mmol of KHP (204.23 g/mol)
mmol of NaOH
Final volume of NaOH (mL)
Initial volume of NaOH (mL)
Delivered volume of NaOH (mL)
Molarity of NaOH (mmol/mL or M)
Average Molarity (mmol/mL or M)
Relative Deviation
3.35 x 10-4
Standard Deviation
Coefficient of Variance
1.7611 %
Supporting Calculations for Table 1 (using Data in Trial 1 with reference to Table 1)
1 π‘šπ‘œπ‘™π‘’ 𝐾𝐻𝑃
οƒΌ mmol of KHP = (0.195 𝑔 𝐾𝐻𝑃) (204.23 𝑔 𝐾𝐻𝑃) (
1000 π‘šπ‘šπ‘œπ‘™ 𝐾𝐻𝑃
1 π‘šπ‘œπ‘™π‘’ 𝐾𝐻𝑃
= 0.9548 π‘šπ‘šπ‘œπ‘™ 𝐾𝐻𝑃
οƒΌ Balanced equation of NaOH and KHP
NaOH + KHC8H4O4 οƒ  KNaC8H4O4 + H2O
Mole ratio: 1 mmol of NaOH = 1 mmol of KHP, thus, mmol of NaOH corresponds also to
each mmol of KHP, therefore, mmol of NaOH = mmol of KHP.
0.9548 π‘šπ‘šπ‘œπ‘™ π‘π‘Žπ‘‚π»
16.02 π‘šπΏ π‘π‘Žπ‘‚π»
οƒΌ Molarity of NaOH = (
= 0.0596 𝑀 π‘π‘Žπ‘‚π»
0.0596 𝑀+0.0589 𝑀+0.06037 𝑀
οƒΌ Average Molarity of NaOH = (
= 0.05962 𝑀 π‘π‘Žπ‘‚π»
οƒΌ Deviation (|d|) = M – Mave
 d1 = 0.0596 – 0.05962 = 0.00002
 d2 = 0.0589 – 0.05962 = 0.00072
Determination of Total Acidity of Vinegar | GROUP 2 – III-BSCT
 d3 = 0.06073 – 0.05962 = 0.00075
οƒΌ Relative deviation (dr) = di/M
 dr1 = 0.00002 / 0.05962 = 3.35 x 10-4
 dr2 = 0.00072 / 0.05962 = 0.01207
 dr3 = 0.00075 / 0.05962 = 0.01257
Σ𝑑 2
οƒΌ Standard Deviation (s) = √ 𝑛−1
οƒΌ Coefficient of Variance (CV) = 𝑀 × 100% =
= 1.05 × 10−3 π‘œπ‘Ÿ 0.00105
× 100% = 1.7611%
Table 2: Determination of Average Density and Mass of Vinegar
Volume Added
Volume of Vinegar (mL)
Mass of 50-mL beaker (g)
Mass of beaker + 10 mL vinegar (g)
Mass of 10 mL vinegar (g)
Average mass of 10 mL vinegar (g)
Average density (g/mL)
0.98706 g/mL
Volume of Vinegar taken from bottle (mL)
Mass of vinegar (g)
24.6765 g
Supporting Calculations for Table 2
Mass of 10 mL vinegar (g) = (Mass of beaker + 10 mL vinegar) – Mass of beaker
= 48.0853 – 38.0696
= 10.0157 g
Average Mass of 10 mL vinegar = (Sum of Three masses) / 3
= (10.0157 + 9.6447 + 9.9515)/3
=9.8706 g/mL
Average Density = Average Mass / Volume
= 9.8706 g / 10.00 mL
= 0.098706 g/mL
Determination of Total Acidity of Vinegar | GROUP 2 – III-BSCT
Mass of vinegar from Vinegar Taken from Bottle:
Density of vinegar = mass of vinegar / volume of vinegar
Therefore: Mass of vinegar = (density of vinegar)(volume of vinegar)
Since: Density of vinegar does not change (Intensive property)
Mass of Vinegar = (0.098706 g/mL)(25 mL)
= 24.6765 g
Table 3: Data on Determination of Datu Puti’s Acetic Acid Content
Average Molarity of NaOH (Table 1) (mmol/mL)
Final volume of NaOH (mL)
Initial volume of NaOH (mL)
Delivered volume of NaOH, Vf – Vi (mL)
Volume of aliquot (mL)
mmol NaOH
mmol CH3COOH in aliquot
Volume of Aliquot (mL)
Volume of Diluted Vinegar in Volumetric flask (mL)
Dilution Factor
mmol CH3COOH in volumetric flask (mL)
Mass of CH3COOH (60.06 g/mol) (g)
Mass of 25.0 mL vinegar (from table 2)
% CH3COOH (w/w)
Average % of CH3COOH
Molarity of Vinegar (Acetic acid sol’n, mmol/mL)
% CH3COOH (w/v)
Average % CH3COOH (w/v)
Relative Deviation
Standard Deviation
Determination of Total Acidity of Vinegar | GROUP 2 – III-BSCT
Coefficient of Variance
Accepted % acid value (from vinegar label)
Absolute error (E)
Relative error (Er ,%)
| -2.6428|
Supporting Calculations for Table 3 using Trial 1
0.05962 π‘šπ‘šπ‘œπ‘™ π‘π‘Žπ‘‚π»
π‘šπΏ π‘π‘Žπ‘‚π»
οƒΌ mmol of NaOH = (37.4 π‘šπΏ π‘π‘Žπ‘‚π») (
= 2.2297 π‘šπ‘šπ‘œπ‘™ π‘π‘Žπ‘‚π»
οƒΌ Balanced equation of NaOH and KHP
CH3COOH + NaOH οƒ  H2O + NaC2N3O3
Mole ratio: 1 mmol of NaOH = 1 mmol of acetic acid, thus, mmol of NaOH corresponds
also to each mmol of acetic acid, therefore, mmol of NaOH = mmol of acetic acid.
οƒΌ mmol of acetic acid = 2.2297 mmol NaOH
οƒΌ mmol
250 π‘šπΏ π‘£π‘–π‘›π‘’π‘”π‘Žπ‘Ÿ
25 π‘šπΏ π‘Žπ‘™π‘–π‘žπ‘’π‘œπ‘‘
(2.2297 π‘šπ‘šπ‘œπ‘™ π‘Žπ‘π‘’π‘‘π‘–π‘ π‘Žπ‘π‘–π‘‘) (
22.297 π‘šπ‘šπ‘œπ‘™ 𝐴𝑐𝑒𝐴𝑐𝑖𝑑
οƒΌ mass
60.06 π‘šπ‘šπ‘œπ‘™ π‘Žπ‘π‘’π‘‘.π‘Žπ‘π‘–π‘‘
1 π‘šπ‘šπ‘œπ‘™ π‘Žπ‘π‘’π‘‘.π‘Žπ‘π‘–π‘‘
1000 π‘šπ‘”
(22.297 π‘šπ‘šπ‘œπ‘™ π‘Žπ‘π‘’π‘‘π‘–π‘ π‘Žπ‘π‘–π‘‘) (
= 1.3393 𝑔 𝐴𝑐𝑒𝐴𝑐𝑖𝑑
οƒΌ Molarity = 1.3393 g / 25 mL = 0.053572 M
οƒΌ % acetic acid (w/v) = (1.3393 g/ 25 mL) x 100 % = 5.35%
οƒΌ % acetic acid (w/w) = (1.3393 g/(25 mL x 0.98706 g/mL)) x 100% = 5.4272%
οƒΌ Absolute error for (w/v) = 5.3572 % - 8% = |-2.6428|
οƒΌ Relative error for (w/v) = (2.6428/8) x 100% = 35.73%
IV. Discussions and Guide Questions
According to the data tables shown in the third part of the report, these three
points are noted.
(1) The expected molarity of the standard base NaOH falls short than expected.
According to Table 1 which shows the standardization of prepared 0.1 M NaOH
against KHP, the average molarity of the solution is roughly 0.05962 M, almost half than
expected. Some factors that can be accounted for is the reaction of solid sodium
hydroxide with atmospheric carbon dioxide. This is the most troublesome aspect of
NaOH, as it reacts with atmospheric carbon dioxide in a reaction CO2 + 2OH- οƒ  CO32- +
Determination of Total Acidity of Vinegar | GROUP 2 – III-BSCT
H2O. Both the solid reagent and its solutions can react this way; consequently, an
already standardized sodium hydroxide solution must be protected from atmospheric
CO2, thus before preparation, distilled water is boiled because often distilled water is
saturated with carbon dioxide and after preparation it is immediately capped.
Although the concentration of NaOH is reduced when it comes to contact with carbon
dioxide, its reactive capacity towards acids may not change. Unfortunately, the most
common application of sodium hydroxide titrants are for the titration of weak acids,
which require an indicator with a basic transition range, for this experiment, is
phenolphthalein. In this case, each carbonate ion will be titrated to bicarbonate,
reacting with only one hydronium ion for the analyte in the reaction: CO 32- + H3O+ οƒ 
HCO3- + H2O. As a result, the effective concentration of the base solution is decreased
and a determinate error will be incurred as we can see in the results.
(2) The concentration of acetic acid in the vinegar is considerably low as compared to
its reported concentration.
Seen on Table 3, we can infer that more or less, the amount of acetic acid in the
vinegar itself is 71% lower than expected. The experimenters got an average of 5.7386
than of the accepted value of 8%. A factor that can be considered is that they may
have other acids that can interfere with the results of the experiment, for example, the
carbonic acid formed with carbon dioxide.
(3) The data is not accurate but precise.
According to the statistical treatment done to the data tables, the data is
precise because the deviations between the data is small and is supported by the
coefficient of variance, however, it is not accurate because there is a large error in the
data as manifested by the results of the absolute and relative errors computed.
The following are other points and the answers to the guide questions for
discussion in reference to the guide questions drawn from the laboratory manual.
[1] There is a need to boil distilled water since that distilled water is saturated with
carbon dioxide. The reaction is discussed above on the first point.
[2] The reaction of phenolphthalein with NaOH is as follows:
HC2H3OH (aq) + NaOH (aq) οƒ  H2O (l) + NaC2H3 (aq) (TITRATION RXN)
+ H2O
οƒ  H3O- + Indic-
Determination of Total Acidity of Vinegar | GROUP 2 – III-BSCT
Figure 2: Shift of electrons in phenolphthalein that indicates its color. (Retrieved from: Fundamentals of Analytical Chemistry
5th Edition)
[3] The faint pink color of phenolphthalein indicates that it reached its endpoint,
therefore, the number of moles of H+ is equal to the number of OH – ions in the solution.
This color must be taken because excess of this will lead to an error (darker pink color),
as the equilibrium is disturbed. The factors that affect the endpoint sharpness in this
titration is the pH range of the selected indicator and the distinction of the person of the
color itself.
[4] The standard reagent is a strong base rather than a weak base because a weak
base will result in partial dissociation of the ions rather than if the base is strong, even if
the acid is weak, the dissociation becomes complete and a 1:1 stoichiometric ratio is
[5] As said in the preceding page, the data is not accurate but precise.
[6] The errors that are committed in this experiment are the following:
6.1 Determinate Errors
6.1.a Method Error. As said in the first point of the discussion part, the
molarity of the standardized NaOH is lower than to the expected. Therefore, the titration
period became longer, and can trigger contamination within the NaOH with its longer
Determination of Total Acidity of Vinegar | GROUP 2 – III-BSCT
contact time with atmospheric conditions. The experimenters accept the fact that the
distilled water is not totally boiled and that NaOH was not secured immediately, that
most probably be the cause of the error. Another point is that three days is the gap
between the preparation of the NaOH stock and the standardization with KHP.
6.1.b Instrument Error. Instrument errors are inevitable since they are only
machines that do their functions. The analytical balance, due to its age, is hard to tare
and has the tendencies to have errors. It is greatly manifested by the fact that vinegar is
expected to have a density greater than 1.00 g/mL which is the density of water since
vinegar has more components than of water. Deriving the relationship of mass
(components) with density, it is expected as there is higher mass, the density also
6.1.c Personal Judgment Error. Measuring using pipet and burette is hard,
since there is called a parallax error or viewing error. This can be eliminated as well.
6.2 Indeterminate Error
6.2a Constant Error. As said in the discussion part, there is a speculation of
other acids present in the vinegar, for example, carbonic acid, lactic acid and such
acids that can be formed during the fermentation process of the vinegar. Lactic acid is
produced during fermentation in line with carbon dioxide. Therefore, this error is
constant error and unavoidable.
V. Conclusions and Recommendations
In conclusion, the experiment “Determination of Total Acidity of Vinegar”
assessed the concentration of acetic acid in a famous brand of vinegar “Datu Puti” by
comparing the reported value of acetic acid versus experimental value. Direct titration
process is employed with NaOH as standard base and phenolphthalein as indicator.
Results showed that the concentration of acetic acid in Datu Puti falls considerably
short rather to the reported value, however, this result has an error with a range of 35%
that is accumulated through many errors, most of them manipulative and can be
The experimenters recommend the following points to be considered if the same
experiment would be repeated. First, take note that distilled water must be boiled to
remove the saturated carbon dioxide from the water that interferes with the data.
Determination of Total Acidity of Vinegar | GROUP 2 – III-BSCT
Second, NaOH must be on its considerable concentration, therefore, must be prepared
in advance and standardize in advance to reduce time consumption. Lastly, it would
be best if titration must be done in different varieties of vinegar to find out if which of
the vinegar brands had the closest value to their reported concentration in their labels
and which of them is the most “acidic.”
VI. References
A. Directly Cited Books
(The superscripts here also refer to the superscripts that were used in the introduction,
thus the parts where superscripts are found are cited and the references were found
Harris, Larry G. (1988). Analytical Chemistry (1st Edition). New Jersey. Prentice-Hill Inc.
pp. 94-109.
Chemistry Faculty. (unpublished). Analytical Chemistry 1 Instructional Materials.
Silberberg, Martin S. (2006). Chemistry: The Molecular Nature of Matter and Change.
New York: Mc-Graw Hill. pp.118-119.
Roque, Adolfo P. (unpublished). Analytical Chemistry Laboratory Manual.
B. Photo Credits
Harvey, David. (2000). Modern Analytical Chemistry. New York: McGraw Hill. pp.
284-287 (Electronic Format)
Skoog, Douglas A., et al (2004). Fundamentals of Analytical Chemistry (8th
Edition). Canada: Brooks/Cole. pp. 368-375 (Electronic Format)
Silberberg, Martin S. (2006). Chemistry: The Molecular Nature of Matter and
Change.(4th Edition) New York: Mc-Graw Hill. pp.118-119. (Electronic Format)
If you are wise and understand God’s ways, prove it by living an honorable life, doing good works with the
humility that comes from wisdom. But if you are bitterly jealous and there is selfish ambition in your heart,
don’t cover up the truth with boasting and lying. For jealousy and selfishness are not God’s kinds of wisdom.
Such things are earthly, unspiritual and demonic. For wherever there is jealousy and selfish ambition, there
you will find disorder and evil of every kind.
But the wisdom from above is first of all pure. It is also peace loving, gentle at all times and willing to yield to
others. It is full of mercy and good deeds. It shows no favoritism and is always sincere. And those who are
peacemakers will plant seeds of peace and reap a harvest of righteousness.
James 3:13-18 (NLT)
To God be the glory!
Determination of Total Acidity of Vinegar | GROUP 2 – III-BSCT