Chemistry lab report about TITRATION OF THE ACETIC. Instructions will be attached.

Experiment 6
Titration
by Me and ?
Get the following for today:
• Buret stand
• Buret clamp
• Buret
• 2 Erlenmeyer flasks, 125 mL
• 25 mL vinegar in 50 mL beaker
• 65 mL NaOH in 250 mL beaker
• 10 mL graduated cylinder
• Large waste beaker
• Distilled water bottle
Molarity, M = (moles solute)/ (VL solution)
•
•
•
•
Clean buret with a little bit a NaOH
Repeat two more times
Fill the buret with NaOH
Make sure the meniscus is above zero
• Add 10.00 mL of vinegar to the first
Erlenmeyer flask
• Wash out the graduated cylinder with
distilled water and add to the first
Erlenmeyer flask
• Add two drops of phenolphthalein to
the first Erlenmeyer flask
• Make up a second Erlenmeyer flask the
same way
• Titration
– Determining the amount of a substance by a reaction with
another substance with known concentration
– Detection of the chemical reaction by some kind of signal,
e.g., color change of an indicator
• Purpose of the Experiment
– Determine the amount of acetic acid in vinegar by titration
with sodium hydroxide
CH3COOH + NaOH CH3COONa + H2O
H+ + OH- H2O
Neutralization Reaction
– Indicator is phenolphthalein
• Changes from colorless (acidic) to pink (basic)
• 20 drops ~ 1 mL, a clear solution is acidic and a dark color is basic
• Add & stir 1 mL at a time from the buret to the Erlenmeyer flask until you
have approximately 5 mL transferred ~ What happened?
• Add & stir 0.5 mL at a time from the buret to the Erlenmeyer flask until you
have approximately 5 mL more transferred ~ What happened?
• Stir each time you add a little NaOH to the vinegar and a color change occurs
~ why are you adding less and less NaOH to get a color change?
• Add one more drop and stir ~ what did the solution stay a light pink color?
• Analysis by Titration
– When acetic acid initially in solution equals sodium
hydroxide added, you have reached the equivalence point
(neutral solution)
– When more sodium hydroxide than acetic acid is present,
the solution turns basic
• Indicator will become pink
• When the solution becomes just pink with one drop of
NaOH, you have reached the “endpoint” of the
titration
• Calculations for CH3COOH (the acid) in Vinegar:
• Write down the volume from # 6, V = 16.2 mL NaOH
• Write down the molarity from # 7, M = 0.52 M
• Find the number of moles of NaOH for #8, n(NaOH) = MV
= (0.52 M)(16.2 mL)(L/(1000 mL) = 0.008424 mol ~ 0.0084 mol
• Find the number of moles of CH3COOH for #9, n(acid) =
0.008424 mol NaOH (1 mol acid)/ (1 mol NaOH) = 0.008424 mol
~ 0.0084 mol CH3COOH
• Find the number of grams of CH3COOH for #10, m(acid) =
(0.008424 mol)(60.0 g/mol) = 0.50544 g ~ 0.51 g CH3COOH
• Mass of vinegar for #11, m(vinegar) = 10.00 g (1.005 g/mL)
= 10.05 g vinegar
• Percent by mass of CH3COOH in the vinegar sample for #12,
(0.50544 g CH3COOH)(100%)/(10.05 g vinegar) = 5.029% by
mass ~ 5.0% by mass
CHEM 110L-002
General Engineering Chemistry I
Fall, 2019
(DETERMINATION OF EMPIRICAL FORMULA FOR MAGNESIUM OXIDE)
Date: 10/10/2019
Introduction
According to Dalton’s Law of Multiple Proportions, elements combine in specific whole numbered ratios to
form specific compounds. Chemical formulas can be written as a smallest whole number ratio of combined
atoms. When written in this form, the chemical formula is known as an Empirical Formula. Empirical means
“experimentally determined” in this case. Empirical formula is often different from the molecular formula.
Empirical formula is the simplest whole number ratio of elements that combine to form a compound, while
molecular formula represents the specific proportions present in a compound. For example, the Empirical
formula for benzene is CH, while the molecular formula is C6H6.
The purpose of this experiment was to experimentally determine the Empirical Formula for an oxide of
magnesium (MgOx). Magnesium metal burned in air combines with the oxygen in the air with a primary
product of magnesium oxide.
2 Mg (s) + 1 O2 (g) → 2 MgO (s)
A side product is also produced by combination of magnesium and nitrogen in the air. This side product is
magnesium nitride.
3 Mg (s) + 1 N2 (g) → 1 Mg3N2 (s)
Addition of drops of distilled water converts the magnesium nitride to magnesium hydroxide. Magnesium
hydroxide is converted by further heating to magnesium oxide by loss of water.
1 Mg(OH)2 (s) → 1 MgO (s) + 1 H2O
2
The mass of magnesium is subtracted from the final weight of magnesium oxide after heating to yield the mass
of oxygen consumed. By using the molecular weights of magnesium and oxygen, moles of magnesium and
moles of oxygen are calculated. The ratio of these moles is the experimentally-determined ratio of magnesium
to oxygen in the product, MgOx.. This ratio, rounded off and expressed as whole numbers, provides the
experimentally-determined Empirical formula for magnesium oxide, MgOx. In this lab report, the purpose,
background, and chemical reactions have been covered in the Introduction . The materials used in the
experiment are covered in the Materials section. The stepwise procedure is covered in great detail in the
Procedure Section. In the Data and Observations sections, photos of the actual experiment in progress are
presented with descriptions. A photo of the Raw Data recorded in the Student Laboratory Notebook is provided.
These Raw Data are then used in the Calculations section to show how mass of magnesium, mass of magnesium
oxide, and mass of oxygen are calculated. By use of atomic weights, detailed calculations are shown for
determining moles of Mg and O. Calculation of a mole ratio for Mg : O is then shown and explanation is made
as to how Empirical Formula of MgOx is determined from this mole ratio. An application of Empirical Formula
determination is discussed in Application. In the Conclusion section, the technique and calculations are again
detailed. Discussion is made of a laboratory error in the experiment and the affect on the final Empirical
Formula result is carefully explained. Finally, there is discussion of several things that were learned by me from
conducting this experiment. A Reference page is attached.
3
Apparatus Setup
Porcelain Crucible and Lid
Materials
Crucible with lid, Clay triangle, Wire gauze, Iron ring, Ring stand, Bunsen burner, Dropper, Crucible tongs,
Magnesium metal (about 0.3 grams)
Procedure
1. Wash the crucible and lid well with water.
2. Dry crucible and lid over the Bunsen burner with a cool flame for a few minutes.
3. Cool crucible to room temperature.
4. Weight crucible and lid together and record the mass.
5. Add approximately 0.3 g of magnesium metal to the crucible.
6. Weigh crucible, lid, and magnesium metal and record the mass.
7. Cover crucible with lid.
4
8. Heat over low flame for 2 or 3 minutes.
9. Increase the air flow to the Bunsen burner to produce a hot flame.
10. Keep crucible covered with lid.
11. Heat crucible over hot flame while the bottom glows red hot for 10 minutes.
12. After 5 minutes, carefully life crucible lid to see if magnesium has begun to react.
13. If white smoke or bright white glow is seen, close the lid, adjust the flame and heat only a couple more
minutes.
14. If no white smoke or bright white glow is seen, tip the crucible lid open slightly and heat while red hot
for 5 more minutes.
15. A grayish-white powder should be seen at this point.
16. Remove flame and cool crucible for 2 or 3 minutes.
17. Carefully add 5 drops of distilled water. This converts unwanted magnesium nitride to magnesium
hydroxide, which is later heated to eliminate. Ammonia odor may be detected.
18. Heat crucible over low flame for 2 or 3 minutes.
19. Adjust flame to heat crucible glowing red hot for 10 minutes.
20. Shut off flame and cool crucible to room temperature.
21. Weigh crucible.
22. Heat crucible for 5 minutes.
23. Cool crucible.
24. Repeat steps 21 through 23 until a constant mass is obtained.
25. Record the mass when a constant mass is obtained.
5
Data and Observations
Photos of experiment in-progress:
Shows magnesium metal before heating.
Experiment materials
Shows magnesium metal during heating.
Heating to a glowing red to react Mg metal
6
Spreading out Mg metal for reaction
Removing crucible.
Spreading out Mg metal for reaction
Addition of water to convert Mg3N2
7
Heating crucible with lid tipped to absorb O2
Picture # . Data from Student Lab Notebook
8
Table 1. Raw Data
Raw Data
A.
Mass of crucible + lid, empty
22.03 g
B.
Mass of crucible + lid + magnesium metal, before
22.80 g
heating
C.
Mass of crucible + lid + magnesium oxide, after
22.93 g
heating
Raw data as transcribed from the Student Laboratory Notebook (see previous page)
Calculations
Table 2. Calculation of mass of magnesium
Mass of magnesium
= B. – A.
Mass of crucible + lid + magnesium metal,
22.80 g
before heating
Mass of crucible + lid, empty
22.03 g
Mass of magnesium = B. – A.
0.77 g
9
Subtraction of the empty crucible + lid mass from the mass of the reactant, magnesium, before heating (and
before reaction with Oxygen from the air) produces the mass of the reactant, magnesium.
Table 3. Calculation of mass of magnesium oxide
Mass of magnesium oxide
= C. – A.
Mass of crucible + lid + magnesium oxide,
22.93 g
after heating
Mass of crucible + lid, empty
22.03 g
Mass of magnesium oxide = C. – A.
0.90 g
Subtraction of the empty crucible + lid mass from the mass of the product after heating produces the mass of the
product, magnesium oxide.
Table 4. Calculation of mass of oxygen
Mass of oxygen
= Mass of magnesium oxide – mass of
magnesium
Mass of magnesium oxide = C. – A.
0.90 g
Mass of magnesium = B. – A.
0.77 g
Mass of oxygen
0.13 g
10
Since MgOx is made up of only Mg and O, when the mass of Mg is subtracted from the mass of MgOx, the
result is the mass of O which was consumed by the reaction of Mg with Oxygen from the air.
Table 5. Calculation of moles of magnesium and moles of oxygen
Moles of magnesium:
0.77 g Mg * (1 mol Mg / 24.3050 g Mg)
0.03168 moles of Mg
Moles of oxygen:
0.13 g O
0.008125 moles of O
* (1 mol O / 15.9994 g O)
Dividing masses of both Mg and O by their respective atomic weights converts masses to moles.
The smaller number of moles is the moles of O. By dividing both moles of Mg and moles of O by the
smaller number of moles, a ratio of moles of Mg to moles of O is obtained. Rounding this off to the closest
whole number ration provides the coefficients for the Empirical Formula of MgOx.
Table 6. Calculation of Mole ratio for Mg and O
0.03168 moles of Mg / 0.008125 moles of O
3.8990
0.008125 moles of Mg / 0.008125 moles of O
1.0000
Dividing both moles of Mg and moles of O by the smaller amount of moles of Mg and O produces a smallest
whole number ratio of the moles of Mg to O.
Therefore, a ratio of approximately 4:1 for moles of Mg : moles of O was determined in this experiment. The
experiment determined the Empirical Formula for MgOx is Mg4O.
11
What’s up
In the field of polymer engineering, compounds with a molecular weight of hundreds of thousands of units
exist. These include plastics, rubber, and textile materials. Polymers are made up of hundreds or thousands of
repeating units of the same empirical formula to make up an extremely large compound that often resembles a
long chain. These compounds can have the same empirical formula but greatly different molecular weights. By
knowing the molecular weight and the empirical formula, the molecular weight can be divided by the weight of
one unit, called a monomer. From this calculation, the number of units in the long chain can be determined. The
relationship between molecular weight and Empirical Formula for a polymer can greatly affect the physical
properties of these materials. Depending on the chemistry of the monomer units, there is potential for
interaction of monomer units between different polymer molecules, which is known as crosslinking. Longer
chains, shorter chains, and crosslinking may produce stronger materials and different thermal properties. The
weight density of the polymeric material may also depend on the number of monomer units in the polymer.
Other physical properties, such as melting point, viscosity, and opaque nature may be manipulated with control
over the number of monomer units in the polymer molecule.
12
Conclusion
The goal of the experiment was to present a technique for experimentally demonstrating that chemical
compounds form in specific whole number ratios of atoms to form specific compounds. By reacting metallic
magnesium metal contained in a crucible with the oxygen in air at a very high temperature with a Bunsen
burner, magnesium oxide was formed. By knowing the mass of the empty crucible and lid, the mass of the
magnesium metal added before heating, and the mass of magnesium oxide formed, calculations were performed
to determine whole number ratios of magnesium and oxygen that reacted in this experiment. Mass of
magnesium was converted to moles of magnesium using the atomic weight of magnesium. Mass of oxygen was
determined by subtracting the weight of magnesium before heating from the weight of magnesium oxide
formed after heating. Magnesium oxide is composed only of elemental Mg and O, thus allowing for the gain in
mass after heating to be attributed totally to gain in mass of oxygen. The smaller of the moles of magnesium
and oxygen (which was found to be moles of oxygen) was used as a divisor for both the moles of magnesium
and moles of oxygen. What resulted was a ratio for moles of magnesium to moles of oxygen that made up the
compound formed, magnesium oxide.
A ratio of 3.8990 to 1.0000 (approximately 4:1) for moles of Mg : moles of O was determined in this
experiment. Therefore, the experiment determined the Empirical Formula for MgOx to be Mg4O. Magnesium is
an alkaline earth metal and is in group II in the Periodic Table of Elements. Group II elements form ionic bonds
by losing 2 electrons, and therefore, have a most stable oxidation state of Mg2+. Since oxygen is in Group VI in
the Periodic Table of Elements, its most stable oxidation state occurs when it gains 2 electrons to form O2-.
Given these two most stable oxidation states, magnesium oxide will form a solid in which 1 atom of Mg2+
combines with one atom of O2- to form magnesium oxide with a molecular formula of MgO.
In reviewing the raw data, it is seen that the mass of oxygen consumed by reacting metallic magnesium is used
to calculate moles of oxygen. In the experimental data for this experiment, moles of oxygen consumed in the
13
reaction were found to be nearly four times less than the moles of magnesium reacted. The weight of oxygen
consumed has led directly to an Empirical Formula result which is far from the expected value of a 1:1 ratio for
moles of magnesium : moles of oxygen.
During the experiment, experimental errors were noted that led to this apparent low amount of oxygen
consumed in the reaction. After heating, and before the final constant mass of the crucible + lid + magnesium
oxide could be recorded, we observed that a considerable amount of the magnesium oxide was inadvertently
lost from the crucible and could not be recovered. This led to a much lower mass of crucible + lid + magnesium
(as recorded in C. in Raw Data of this report). This led to an erroneously low mass and moles of oxygen to be
found in the calculations. When moles of oxygen was found to be the lower of the moles of magnesium and
moles of oxygen, then moles of oxygen was used to divide both moles of magnesium and moles of oxygen. The
result of dividing moles of magnesium by an erroneously low moles of oxygen was that the mole ratio of
approximately 4:1 for moles of magnesium : moles of oxygen was obtained. Thus, the experimentallydetermined Empirical Formula of magnesium oxide was greatly in error, but is reported here as determined as
Mg4O. By reviewing most common oxidation states for Mg and O, as pointed out above, a molecular formula of
MgO with a 1:1 mole ratio is seen to be expected.
From this experiment, I learned that empirical formula in terms of moles of reactants (Mg and O) can be
calculated from weighing the starting material (Mg) for the reaction before reaction and heating and from
weighing the products (MgOx) of the reaction after heating. The masses of the elements and compounds
involved (mass of O) can be determined by subtracting final vs. original masses (mass of MgOx – mass of Mg).
These masses can be used along with atomic weights of the reactants to determine moles of reactants involved.
From the moles of the two species participating in the reaction (Mg and O), a mole ratio can be found. This
mole ratio is the same ratio of moles which exist in the final product of the reaction, and is directly applied to
produce an Empirical Formula for the compound formed (MgOx).
14
I also learned to be very careful to avoid spills of any materials where accurate masses are going to be required.
This experience demonstrated how a simple and inadvertent error in technique can greatly affect the results
when determining Empirical Formula.
In this experiment, I learned technique of working with crucibles and how heating to extremely high
temperature causes an oxidation reaction with O2 in the air. I learned safety precautions to undertake when
working with hot crucibles and laboratory apparatus at very hot temperatures. I learned that careful heating and
cooling must be undertaken to ensure that the crucible heats at an appropriate temperature and to avoid cracking
the crucible by heating excessively at levels of heat that are too high. I learned safety in working with a very
reactive metal like magnesium, which is oxidized and catalyzed further by the extreme heat.
15
References
Chemcollective.org. (no date). Stoichiometry Tutorials: Determining the Empirical Formula of a Compound from Its
Molecular Formula.
Retrieved from: http://www.chemcollective.org/activities/tutorials/stoich/ef_molecular
16
Experiment 6
Titration
by Me and ?
Get the following for today:
• Buret stand
• Buret clamp
• Buret
• 2 Erlenmeyer flasks, 125 mL
• 25 mL vinegar in 50 mL beaker
• 65 mL NaOH in 250 mL beaker
• 10 mL graduated cylinder
• Large waste beaker
• Distilled water bottle
Molarity, M = (moles solute)/ (VL solution)
•
•
•
•
Clean buret with a little bit a NaOH
Repeat two more times
Fill the buret with NaOH
Make sure the meniscus is above zero
• Add 10.00 mL of vinegar to the first
Erlenmeyer flask
• Wash out the graduated cylinder with
distilled water and add to the first
Erlenmeyer flask
• Add two drops of phenolphthalein to
the first Erlenmeyer flask
• Make up a second Erlenmeyer flask the
same way
• Titration
– Determining the amount of a substance by a reaction with
another substance with known concentration
– Detection of the chemical reaction by some kind of signal,
e.g., color change of an indicator
• Purpose of the Experiment
– Determine the amount of acetic acid in vinegar by titration
with sodium hydroxide
CH3COOH + NaOH CH3COONa + H2O
H+ + OH- H2O
Neutralization Reaction
– Indicator is phenolphthalein
• Changes from colorless (acidic) to pink (basic)
• 20 drops ~ 1 mL, a clear solution is acidic and a dark color is basic
• Add & stir 1 mL at a time from the buret to the Erlenmeyer flask until you
have approximately 5 mL transferred ~ What happened?
• Add & stir 0.5 mL at a time from the buret to the Erlenmeyer flask until you
have approximately 5 mL more transferred ~ What happened?
• Stir each time you add a little NaOH to the vinegar and a color change occurs
~ why are you adding less and less NaOH to get a color change?
• Add one more drop and stir ~ what did the solution stay a light pink color?
• Analysis by Titration
– When acetic acid initially in solution equals sodium
hydroxide added, you have reached the equivalence point
(neutral solution)
– When more sodium hydroxide than acetic acid is present,
the solution turns basic
• Indicator will become pink
• When the solution becomes just pink with one drop of
NaOH, you have reached the “endpoint” of the
titration
• Calculations for CH3COOH (the acid) in Vinegar:
• Write down the volume from # 6, V = 16.2 mL NaOH
• Write down the molarity from # 7, M = 0.52 M
• Find the number of moles of NaOH for #8, n(NaOH) = MV
= (0.52 M)(16.2 mL)(L/(1000 mL) = 0.008424 mol ~ 0.0084 mol
• Find the number of moles of CH3COOH for #9, n(acid) =
0.008424 mol NaOH (1 mol acid)/ (1 mol NaOH) = 0.008424 mol
~ 0.0084 mol CH3COOH
• Find the number of grams of CH3COOH for #10, m(acid) =
(0.008424 mol)(60.0 g/mol) = 0.50544 g ~ 0.51 g CH3COOH
• Mass of vinegar for #11, m(vinegar) = 10.00 g (1.005 g/mL)
= 10.05 g vinegar
• Percent by mass of CH3COOH in the vinegar sample for #12,
(0.50544 g CH3COOH)(100%)/(10.05 g vinegar) = 5.029% by
mass ~ 5.0% by mass
Exp. No.
Experiment/Subject
6
Tilration
Dage/10/19
Name
Lab Partner
Locker/
Desk No.
Course &
Section No.
(84
Initial value Vi=0
Concentration of the NaOH 40 Ч M
Volume of NaOH used during trial Etration 21.0.Me
Titrations
x 42 43
Final buret reading
21.0 45
65.. 87.ML
210.45
Iurticel buret reading
21.0 45
65 ML
Volume of NaOH Soleition 40-363 24
20
22.Me
Concentratish of
HC2 H₂O2 l
Average concentration
Standard deviation
21.0
M
frath 87
* the inittim value for the first it was 21.0 to tuner Pink
7 second 45
calculation
thing 65
Vi Mevz. 0.400M (loom
loomL)
-0.0403632
Д M
= 40.036327mal
clean [oHJ [ H]
0.991 an
Zero
21
45
65
87
Data collection
report sheet 450 U U
Signature
Note: Insert Divider Under Copy Sheet Before Writing
THE HAYDEN-MCNEIL STUDENT LAB NOTEBOOK
EXPERIMENT #6: TITRATION OF THE ACETIC
ACID IN VINEGAR
Purpose:
The purpose of this experiment is to determine the amount of acetic acid in vinegar by titrating it with a standard (known
concentration) solution of NaOH in the presence of the acid-base indicator phenolphthalein.
Special Apparatus and Chemicals:
Standard NaOH Solution
10 mL Volumetric Pipet
50 mL Buret
Dropper bottle with phenolphthalein
5% white vinegar
Buret Clamp and Stand
Discussion:
The concentration of a solution is often determined by volumetric analysis in which a given volume of the solution, measured
with volumetric glassware, is analyzed against a knownsubstance or solution. In our experiment we will start with a known
volume of the vinegar whoseconcentration is to be determined, and amounts of a standard solution of sodium hydroxide of
known concentration will be added incrementally until the reaction (neutralization) is complete. This process of incremental
addition until a complete reaction is achieved is called titration. The standard solution is referred to as the titrant and the
substance being analyzed the analyte. The volume of the titrant needed for neutralization is determined by using a buret. Before
beginning this experiment be sure to read about using a buret beginning on page 31 of this manual.
In this experiment you will be determining the concentration of a vinegar solution (dilute acetic acid) using a standardized
solution of sodium hydroxide. Because this and most other acid-base neutralizations occur with no visible reaction, an acid-
base indicator such as phenolphthalein is used to give a visible sign that neutralization has occurred. Phenolphthalein is
colorless in acidic or neutral solutions, but pink in basic solutions. Thus, by putting phenolphthalein in our sample of vinegar
that is being titrated, the initial solution is clear but becomes pink upon neutralization of the acetic acid and addition of excess
sodium hydroxide. Thestage at which the vinegar has been neutralized and one extra drop of titrant causes the indicator to
change colors is called the end point. In very precise work, the sodium hydroxide itself would be standardized against a
primary standard such as potassium hydrogen phthalate.
Initially you will know the concentration of the standard sodium hydroxide solution and the volume of vinegar that you are
titrating. After carrying out the titration you will also know the volume of NaOH that is needed to neutralize your given volume
of vinegar. Using this information you can easily calculate the concentration of acetic acid in your vinegar. Because the
reaction of acetic acid and sodium hydroxide occurs in a one to one molar relationship,
HC2H302 + NaOH –> NaC2H302 + H20
the following simple relationship holds true when concentration is in terms of molarity:
(Concentration HC2H2O2) = (Concentration NaOH X Volume NaOH/Volume HC2H302)
This simple relationship works because: the concentration in molarity (moles/liter) multiplied by a volume gives the number of
moles for a compound, and the number of moles of HC2H302 must be equal to the number of moles of NaOH at neutralization.
CAUTION: This particular relationship only works for reactions with a one-to-one stoichiometry; otherwise, you
should work out the concentration of unknown using a complete dimensional analysis procedure.
For example: If 34.50 mL of 0.2486 M NaOH is needed to titrate 10.00 mL of a HCl solution, what is the concentration of the
HCI?
HCl + NaOH ——> NaCl + H2O
Concentration HCl = 0.2486M NaOH x 34.50ml NaOH/10.00 mL HCl = 0.8577M HCI
The stoichiometry of NaOH to HCl is 1:1; therefore we can use our simple relationship, and the concentration of the HCI
solution is therefore 0.8577 M.
Concept:
You will determine the concentration of acetic acid in vinegar by titrating a sample of vinegar with a known solution of sodium
hydroxide. Phenolphthalein will be used as an indicator dye to determine when the reaction is complete. In basic solutions
phenolphthalein is pink, in acidic solutions it is clear. NaOH will be added to a known volume of vinegar containing a few
drops of phenolphthalein. As the neutralization reaction is complete and the next drop of NaOH is added, the solution will go
from acidic to basic and change color from clear to very light pink signaling the end-point of the titration. Knowing the
volumes of vinegar and sodium hydroxide used and the concentration of the standard sodium hydroxide solution, the acetic
acid concentration in the vinegar can be determined.
Procedure:
Clean and prepare a buret according to the directions given in the section on “Measurement of Volume” in the beginning of this
manual. Fill the buret with NaOH. Make sure no air bubbles are left in the tip of the buret. Be sure to write down the
concentration of the NaOH and the initial buret reading. Read the buret to within +0.02 mL; remember to read from the bottom
of the meniscus. CAUTION: NaOH may cause chemical burns. If any NaOH comes in contact with your skin or clothes, rinse
the affected area immediately with large amounts of cold water.
Pipet exactly 10 mL of white distilled vinegar into a clean 125 mL Erlenmeyer flask using a volumetric pipet. Note: Pipetting
should never be done out of the original reagent bottle; in order to avoid contamination, transfer some of the liquid being
pipetted to a small beaker or flask before pipetting. Add 10 mL of distilled water and 2 or 3 drops of phenolphthalein solution
to the vinegar. You are now ready to begin the titration. (Note: It is often helpful to do an initial trial titration to find the
approximate volume of titrant needed. Check with your instructor about doing a trial titration.)
Begin the titration by letting a volume of sodium hydroxide run into the flask equal to the volume that you think is needed for
neutralization minus 1 to 2 mL. You should now be within 20-40 drops of the end point (1 mL equals about 20 drops). Swirl
the flask to mix the solution and then rinse down the walls of the flask with distilled water from a squirt bottle. Now add the
sodium hydroxide drop by drop until the end point is reached. The sodium hydroxide should be added while maintaining a
gentle swirling motion of the flask. The solution should be a very light shade of pink when the titration is finished. A piece of
plain white paper placed under the flask is often helpful in determining when the end point is reached. Record the final buret
reading. If you are unsure whether you have reached the end point, record the reading and then add one more drop. If you were
at the end point, the extra drop will give the solution a very definite pink color.
DISPOSAL: All solutions from this experiment may be safely disposed of down the sink with running water.
Refill the buret with NaOH and rinse out the flask with distilled water. Repeat the titration three more times or as directed by
your laboratory instructor. After you have collected a good set of data and with due regard for significant figures, calculate the
acetic acid concentration in the vinegar for each trial; determine the average concentration of acetic acid using only the
measurements that you are confident in (i.e. those in which you did not overshoot the endpoint); and determine the standard
deviation for your concentration values.