Experiment 20 – Titrations
Name __________________
Lab Section __________________
Experiment 20 – Titrations
Introduction
Neutralization occurs when equal number of moles of acid and base combine to form water. In
general it follows the equation
𝐻𝐴 (𝑎𝑞) + 𝐵𝑂𝐻 (𝑎𝑞) → 𝐴𝐵(𝑎𝑞) + 𝐻𝑂𝐻 (𝑙)
Where, on the reactant side, HA is an acid and BOH is a base; and on the product side, AB is an
aqueous salt solution and liquid water is formed. This water is often written as HOH instead of
the familiar H2O, as it aids in balancing the reaction. This is merely a convention.
In this experiment, the molarity (concentration) of a vinegar solution will be determined using
titration. Titration is the act of neutralizing and acid with a base, or vice versa.
Acid solutions and a base solution are clear and colorless by nature; and the products, water and
an aqueous salt, are also clear and colorless. In order to know when the reaction is done (recall
the evidences of chemical reaction) it is necessary to use an indicator in the reaction. An
indicator is a compound that is not part of the reaction, but signals when the reaction is
complete. This experiment will use phenolphthalein as its indicator. Phenolphthalein is
colorless in an acidic solution but it turns pink in a basic solution. The endpoint, where all the
acid and base have neutralized one another, is a very sharp point. Merely one drop of excess
base is enough to cause the indicator to turn pink and signify the end of the titration.
This experiment will be done in two parts. The first part of the experiment will be to standardize
a base solution against a known acid. Standardization means to determine the concentration
(molarity) of the solution. You will make up an unknown base solution and use a known mass of
potassium hydrogen phthalate, KHC8H4O4 (abbreviated as KHP with a molar mass of 204.23
𝑔
), to titrate the base and determine its molarity. This titration will follow the balanced
𝑚𝑜𝑙
equation:
𝐾𝐻𝑃 (𝑎𝑞) + 𝑁𝑎𝑂𝐻 (𝑎𝑞) → 𝐾𝑁𝑎𝑃(𝑎𝑞) + 𝐻𝑂𝐻 (𝑙)
The second part of the experiment will use the standardized base to titrate an unknown strength
acetic acid, HC2H3O2, solution. This titration will follow the balanced equation:
𝐻𝐶2 𝐻3 𝑂2 (𝑎𝑞) + 𝑁𝑎𝑂𝐻 (𝑎𝑞) → 𝑁𝑎𝐶2 𝐻3 𝑂2 (𝑎𝑞) + 𝐻𝑂𝐻 (𝑙)
It is important to remember that in any titration, indeed in working with pipets and burets in
general, that conditioning of the glassware takes place prior to doing the experiment.
Conditioning is essentially rinsing the glassware; except you rinse it with a portion of the liquid
you will be using (not always water).
Experiment 20 – Titrations
Name __________________
Lab Section __________________
Examples
1) A 1.125 g sample of KHP is dissolved in 50 mL of water and titrated with NaOH. The initial
buret reading is 1.10 mL and the final reading is 27.55 mL. What is the molarity of the NaOH
solution?
Recall the chemical equation:
𝐾𝐻𝑃 (𝑎𝑞) + 𝑁𝑎𝑂𝐻 (𝑎𝑞) → 𝐾𝑁𝑎𝑃(𝑎𝑞) + 𝐻𝑂𝐻 (𝑙)
1 mol KHP
1 mol NaOH
1
1.125 g KHP [204.23 g KHP] [ 1 mol KHP ] [0.02645 L NaOH]
= 0.20826 M NaOH → 0.208 MNaOH
(Note: the volume is the final volume of base minus the initial volume of base, and that is
converted to liters to determine molarity.)
2) A 25.0 mL sample of HC2H3O2 is titrated with 0.310 M NaOH. If it takes 20.50 mL to reach
the endpoint, what is the concentration of HC2H3O2?
𝐻𝐶2 𝐻3 𝑂2 (𝑎𝑞) + 𝑁𝑎𝑂𝐻 (𝑎𝑞) → 𝑁𝑎𝐶2 𝐻3 𝑂2 (𝑎𝑞) + 𝐻𝑂𝐻 (𝑙)
0.310 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻 0.02050 L NaOH
𝐿 𝑁𝑎𝑂𝐻
[
1
][
1 mol 𝐻𝐶2 𝐻3 𝑂2
1 mol NaOH
1
] [0.0250 L 𝐻𝐶
2 𝐻3 𝑂2
]
= 0.2542 M 𝐻𝐶2 𝐻3 𝑂2 → 0.254 M 𝑯𝑪𝟐 𝑯𝟑 𝑶𝟐
𝑔
𝑔
If the density of HC2H3O2 is 1.01 𝑚𝐿 and the molar mass is 60.06 𝑚𝑜𝑙, what is the %HC2H3O2 of
the solution? (remember your definitions of molarity and density)
0.254 𝑚𝑜𝑙 𝐻𝐶2 𝐻3 𝑂2 𝑠𝑜𝑙𝑢𝑡𝑒 60.06 𝑔 𝐻𝐶2 𝐻3 𝑂2 𝑠𝑜𝑙𝑢𝑡𝑒
1000 𝑚𝐿 𝐻𝐶2 𝐻3 𝑂2
[
𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 1 mol 𝐻𝐶
2 𝐻3 𝑂2
1 mL 𝐻𝐶 𝐻3 𝑂2 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
2
] [ 1.01 g 𝐻𝐶
𝑠𝑜𝑙𝑢𝑡𝑒
2 𝐻3 𝑂2
𝑠𝑜𝑙𝑢𝑡𝑒
= 1.51042 % 𝐻𝐶2 𝐻3 𝑂2 → 1.51 % 𝑯𝑪𝟐 𝑯𝟑 𝑶𝟐
] x 100%
Experiment 20 – Titrations
Name __________________
Lab Section __________________
Prelaboratory Questions
1) Please define the following:
Indicator
Titration
Neutralization
Endpoint
Conditioning
Standardization
2) A 50.0 mL sample of phosphoric acid, H3PO4, is titrated with 42.50 mL of 0.250 M NaOH.
What is the molarity of the acid?
𝑔
3) Assuming the density of the phosphoric acid is 1.07𝑚𝐿, what is the % H3PO4 of the solution?
Experiment 20 – Titrations
Name __________________
Lab Section __________________
Procedure
Standardization of the Base
Prepare your base solution by adding approximately 25 mL of 6 M NaOH to a Florence
Flask. Use a large beaker to measure between 500 and 600 mL of DI water and add this to
the Florence flask to dilute the base. Be sure to keep the base stoppered when not in use.
Place an Erlenmeyer flask on the balance and tare (zero) the balance. Weigh between 0.7
and 1 g of KHP into the Erlenmeyer flask. Dissolve the KHP in approximately 25 mL of DI
water. If crystals adhere to the side of the flask, rinse them down into the solution using a
squirt bottle.
Add 2-3 drops of phenolphthalein to the KHP solution.
Condition a buret with your base solution. Fill the buret until the meniscus reads between 0
and 3 mL. Record this initial volume of base. Place the KHP solution on top of a white
sheet of paper under the buret. This will aid in the detection of the color change of
phenolphthalein.
Titrate the KHP solution with your NaOH solution until the solution turns a faint pink.
Record the final volume of base.
𝑔
Calculate the molarity of your base. Note: the molar mass of KHP is 204.23 𝑚𝑜𝑙.
Repeat these steps until three molarities are all within 0.005 M of one another.
Titration of Acetic Acid
Obtain 50 mL of an unknown strength acetic acid solution in a clean, dry, 50 mL beaker from
your instructor and record its number.
Condition a 10.0 mL pipet with the unknown solution. Transfer 10.0 mL of the solution into
an Erlenmeyer flask. Add 20-25 mL of deionized water to the flask.
Add 2-3 drops of phenolphthalein to the KHP solution.
Condition the buret with your NaOH solution. Fill the buret until the meniscus reads
between 0 and 3 mL. Record this initial volume of base. Place the acetic acid solution on
top of a white sheet of paper under the buret. This will aid in the detection of the color
change of phenolphthalein.
Titrate the acetic acid solution with your NaOH solution until the solution turns a faint pink.
Record the final volume of base.
Repeat these steps two more times for reproducibility.
Experiment 20 – Titrations
Name __________________
Lab Section __________________
Data Table
Standardization of NaOH
Trial 1
Trial 2
Trial 3
Trial 4
Trial 5
Mass of KHP
_________
_________
_________
_________
_________
Final buret reading
_________
_________
_________
_________
_________
Initial buret reading
_________
_________
_________
_________
_________
Volume of NaOH
_________
_________
_________
_________
_________
Calculate the molarity of the NaOH for each trial:
Calculated Molarity
Trial 1
Trial 2
Trial 3
Trial 4
Trial 5
________
________
________
________
________
Average Molarity
_______M
Experiment 20 – Titrations
Name __________________
Lab Section __________________
Data Table
Standardization of Acetic Acid
Trial 1
Trial 2
Trial 3
Volume of Acetic Acid
_________
_________
_________
Final buret reading
_________
_________
_________
Initial buret reading
_________
_________
_________
Volume of NaOH
_________
_________
_________
Calculate the molarity of the acetic acid for each trial, the average molarity, and the average %
𝑔
acetic acid (the density of acetic acid is: 1.01 𝑚𝐿 ):
Calculated Molarity
Average Molarity
________
Trial 1
Trial 2
Trial 3
________
________
________
Average % Acetic Acid ________
Experiment 20 – Titrations
Name __________________
Lab Section __________________
Postlaboratory Questions
1) An extra strength antacid tablet contains 750 mg of active ingredient, CaCO3. If it takes
22.25 mL of HCl to neutralize the tablet, how strong is the acid?
2 𝐻𝐶𝑙 (𝑎𝑞) + 𝐶𝑎𝐶𝑂3 (𝑎𝑞) → 𝐶𝑎𝐶𝑙2 (𝑎𝑞) + 𝐻𝑂𝐻 (𝑙) + 𝐶𝑂2 (𝑔)
2) A beaker containing 25.0 mL of 0.360 M H2SO4 spills on the counter. How much baking
soda, NaHCO3 will be needed to neutralize the acid?
𝐻2 𝑆𝑂4 (𝑎𝑞) + 2 𝑁𝑎𝐻𝐶𝑂3 (𝑎𝑞) → 𝑁𝑎2 𝐶𝑂3 (𝑎𝑞) + 2 𝐻𝑂𝐻 (𝑙)
3) If a 35.75 mL sample of 0.175 M HCl solution is neutralized with 0.125 M Mg(OH)2, how
much Mg(OH)2 will be needed
𝐻𝐶𝑙 (𝑎𝑞) + 𝑀𝑔(𝑂𝐻)2 (𝑎𝑞) → 𝑀𝑔𝐶𝑙2 (𝑎𝑞) + 2 𝐻𝑂𝐻 (𝑙)