Experiment-11: Analysis of Vinegar

Acid- Base Titration

Mass percent calculations

Molarity calculation

Acid and Base terminologies
In 1884 Svante August Arrhenius defined acids and bases in terms of the species they
produce upon dissolution in water. An Arrhenius acid is a hydrogen-containing
+
compound that produces hydrogen ion (H ) upon dissolution in water. For example,
HCl (g)
+
Hydrgen chloride
H2O (l)

water
H
+
(aq)
+
hydrogen ion
Cl
(aq)
chloride
An Arrhenius base is a hydroxide-containing compound that produces hydroxide ion
--
(OH ) upon dissolution in water. For example,
NaOH (s)
+
Sodium hydoxide
H2O (l)

water
Na
+
(aq)
sodium ion
+
OH
(aq)
hydroxide ion
In 1923, Bronsted and Lowery broadened the definitions of acids and bases that apply to
both aqueous and nonaqueous solutions. A Bronsted-Lowery acid is a substance that can
+
donate a proton (H ) to another substance. A Bronsted-Lowery base is a substance that
+
can accept a proton (H ) to another substance.
Some common acids are: HCl -Hydrochloric acid (stomach acid), HNO3 - Nitric acid
H2SO4 - Sulfuric acid, H2CO3 -Carbonic acid, HC2H3O2 - Acetic acid (CH3COOH)
Some common bases are: NaOH- Sodium hydroxide, KOH- Potassium hydroxide,
NH4OH -Ammonium hydroxide (Ammonia plus water)
In 1909, Sorensen proposed pH scale to identify and compare acid or base strength in
dilute aqueous solutions. Expressed mathematically, pH = - log [H3O+]. The pH scale
ranges from 0 to 14. The pH value of a solution sample can be experimentally determined
in a chemistry laboratory using a pH meter. Acids have a pH value less than 7.00 and
bases have pH value more that 7.00. The pH value of 7.00 indicates a neutral solution
such as pure water. The lower the pH value, greater the acidity. The higher the pH value,
greater the basicity.
Concentration (strength) of an acid or base may be expressed in terms of Molarity or
mass percent. Molarity is defined as the number of moles of solute per liter of solution.
M = molarity = moles/liter = (# moles of solute/ volume of solution in liters)
Where # moles of solute = grams of solute/ molar mass of solute; and 1000 mL = 1 L.
Percent by mass is defined as the mass of solute divided by the mass of solution
multiplied by 100.
% Mass = (mass of solute/ mass of solution) x 100
Where mass of solution = (mass of solute + mass of solvent)
An acid reacts with a base to produce a slat and water. This popular reaction is known is
the Neutralization reaction. For example,
HCl (aq)
Acid
+
NaOH (aq)
base

NaCl (aq)
salt
+
H2O (l)
water
Vinegar is essentially a very dilute aqueous solution of acetic acid, CH3COOH. There
are many vinegars on the market: name brands, generic brands, flavored vinegars and
cider vinegar. But they all contain acetic acid as the active ingredient, which has several
household uses such as pickling, dressing, and seasoning. As per the federal guidelines,
vinegar must contain at least 4% of acetic acid. It means that 4.0 g of acetic acid must be
present in 100 g of vinegar solution.
The process by which vinegar is made is
theoretically quite simple (liquid ingredients are fermented until they turn acidic, and
then the resulting liquid is filtered and diluted) but the steps actually taken are many and
complex.
One way to determine the amount of acetic acid present in vinegar is through titration.
Titration is a technique used to determine the concentration of a solution and hence the
amount of solute in the solution. A solution whose concentration is known is termed as a
standard solution. In a titration, a solution of known concentration (called the titrant) is
reacted with a specific volume of the solution of unknown concentration (the analyate).
The titrant is added in a controlled and measured manner to the analyate until each
reactant is completely consumed; that is, there is no unreacted titrant or analyate in the
reaction vessel. This point is known as the equivalence point and at this point the moles
of titrant and moles of analyate are stoichiometrically equal, according to the balanced
equation. The equivalence point can be seen by a color change in an indicator, a
molecule that changes color depending on the reaction conditions. The concentration of
the analyate, expressed as molarity or moles per liter, can thus be found knowing the
concentration and volume of the titrant, its stoichiometric relationship to the analayte and
the volume of the analyate titrated. Many types of solutions can be titrated, but the most
common type of titration is an acid-base titration. Determining the concentration of acetic
acid in vinegar is an example of this type of titration. Here the acetic acid is the analyate
whose concentration is to be determined. A base, such as NaOH, is the titrant. At the
equivalence point, the moles of base equal the moles of acid according to the reaction
stoichiometry:
CH3COOH (aq) +
Acetic Acid
At the equivalence point:
NaOH (aq)
base

NaCH3COO (aq)
+
Sodium acetate, a salt
H2O (l)
water
moles of acid = moles of base or
(molarity of acid x volume of acid) = (molarity of base x volume of base)
where molarity of acid = unknown, to be found at the end of experiment
volume of acid = known, you will use 5.0 mL
molarity of base = known, 0.1 M NaOH standard solution provided
volume of base = known, final minus initial buret reading.
Mass percent of acetic acid can easily be calculated from the molarity.
Indicator : 3 drops Phenolphthalein in each trail. End point of titration: when colorless
solution in Erlenmeyer flask changes to lightest permanent pink.
Procedure:
1. Transfer approximately 20 mL Vinegar into a clean and dry small size beaker and
take it your bench. Do not take the big bottle to your desk.
2. Using a small graduated cylinder, transfer exact 5.0 mL vinegar into a clean
Erlenmeyer flask. If inside of this Erlenmeyer flask is wet, it will not affect your
results. Add 25 mL distilled water into this flask. Add 3 drops of phenolphthalein.
So there will be three items in this Erlenmeyer flask (5.0 mL vinegar + 25 mL
distilled water 3 drops of phenolphthalein). Swirl the flask for 1 minute to mix the
contents thoroughly.
3. Transfer approximately 100 mL 0.1 M NaOH into a 250 mL beaker and take it to
your bench. You will use this NaOH to fill-up the buret. Do not take the big bottle
of NaOH to your desk.
4. Review the figure. Wash and rinse a buret with distilled water. Then rinse this
buret twice with 5 mL portions of the 0.1 M NaOH, and drain into the designated
waste container. Fill the buret almost to the top with 0.1 M NaOH. Clamp this
buret to the iron stand. Open the stopcock/control knob slowly to drain out NaOH
until the lowest level of meniscus is at 0.00 mL mark. Make sure there are no air
bubbles. 0.00 mL is your initial reading.
5. Place the Erlenmeyer flask from step-2 under the buret and start titration as
demonstrated by your teacher. Remember to keep swirling the flask while adding
NaOH in a drop wise and slow manner. One may place a white piece of paper
under the flask to observe the end point (colorless to permanent lightest pink).
Once you reach the end point add no more NaOH. Read the buret. This is your
final buret reading. The difference between final and initial reading is the volume
of 0.1 M NaOH used in this trail#1 (__________mL)
6. Discard the contents of Erlenmeyer flask into the designated waste container.
7. Using a small funnel carefully refill the buret to 0.00 with 0.1 M NaOH solution.
8. Repeat step-2. This is your second trial.
9. Repeat step-5. Volume of 0.1 M NaOH used in this trail#2 (__________mL)
10. One by one discard the contents of Erlenmeyer flask, buret and left over solutions
into the designated waste container.
Laboratory Report#11: Analysis of Vinegar
Last Name_____________________________, first name________________
Date of Experiment___________ Instructor’s Initials_______________
Data and Calculations
1. Volume of Vinegar used in each trail
= 5.00 mL
2. Molarity of NaOH used in each trail
= 0.1 M
3. Volume of NaOH used in trail#1 (Final – Initial buret reading) = ________ mL
4. Volume of NaOH used in trail#2 (Final – Initial buret reading) = ________ mL
5. Average Volume of NaOH used = (#3 + #4) / 2 = _________ mL
6. Calculate Molarity of Acetic acid in vinegar = _________ M
(molarity of acetic acid x volume of acid) = (molarity of base x volume of base)
volume of acid = step-1, molarity of base = step-2, volume of base = step-5
7. Calculate the molecular weight of Acetic acid using a Periodic table=
_____________ g/mol
8. Grams of acetic acid in one liter of Vinegar = Molarity x Molecular weight
= step-6 x step-7 = _________________ g
9. Grams of acetic acid in 100 mL of Vinegar = step-8/10 = _________________ g
10. Calculate % of acetic acid in vinegar:
(Assuming density of vinegar is 1.00 g/ml so100 mL of Vinegar = 100 g Vinegar)
% of acetic acid in vinegar = (step-9 g/100 g ) x 100 = ______________ %
{numerically #9 and #10 are same}
Show all your work on the backside! Use proper units and significant figures!!
1. A 10.0 mL vinegar sample was completely neutralized by 22.5 mL 0.2 M NaOH
solution. Calculate molarity and % of Acetic acid in vinegar.