CHEMISTRY FOURTH LABORATORY REPORT
ACID-BASE TITRATION
Written by:
Amelia Quinta Jasmine (CHE)
Hanifa Karmelia (FT)
Syadza Luthfiyya (FT)
Date of Experiment: March 18th, 2016
Date of Submission: April 1st, 2016
Department of Chemical Engineering
Department of Food Technology
Faculty of Life Science
International University of Liaison Indonesia
1.1 Purpose
In this experiment, the titration of 2 solutions such as CH3COOH and NaOH was
performed to determine the concentration of an acid solution.
1.2 Abstract
Basic acid-base titration is generally used to obtain the molarity of a solution given the
molarity of other solution that involves neutralization between acid and base. This
experiment was done to determine the concentration of the acid solutions. Identifying
concentration of an acid solution was given attention in the experiment. The whole
experiment dealt with the acid and base solutions. Throughout the experiment, the
concentration of 5 ml of NaOH was dissolved with deionized water. The solution was
moved inside the burette. Then 5 ml of CH3COOH was dissolved with deionized water
inside the beaker glass, then 5 drops of Phenolphtalin was dropped gently using pipette.
It was mixed using the stirrer and placed below the burette. Then NaOH was poured
into the beaker glass. The solution started to change color. Finally, after the experiment,
it was known that the average of the second try and final try solution was 45.40 mL.
1.3 Tools and Materials
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Beaker glass
Burette
Pipette
Erlenmeyer flask
Magnetic stirrer
Stirrer
0.05 mL of NaOH
5 mL of CH3COOH (vinegar)
Deionized water
5 drops of Phenoftalein
1.4 Procedure
Before the experiment was started, the neutralization reaction of CH3COOH + NaOH
had to be known first. The molarity of NaOH was given by the instructor (0.05 M). It
can be done by using this calculation:
1 CH3COOH + 1NaOH  1CH3COOH + 1H2O
[ Z= mol/l]
5 ml
[0.05 M]
vol = 45.40 mL
Then, n of NaOH and Z of CH3COOH had to be determined. It can be done by using
this calculation:
n NaOH
=Mxv
= 0.05 M x 45.40 mL
= 2.27 mol
2.27 𝑚𝑜𝑙
Z CH3COOH =
5 𝑥 10−3 𝐿
= 454 mol/L
0.454 mol/l =
0.454 𝑚𝑜𝑙 𝐶𝐻3𝐶𝑂𝑂𝐻
100 𝑚𝑙
= 27.254
𝑔𝑟
𝑚𝑙
Then, the Mr and gr of CH3COOH were calculated. We had to look up to the periodic
table to see for the mass for each elements. The calculation can be done as follows;
Mr CH3COOH
= (2 x 12.001) + (4 x 1.00794) + (2 x 15.994)
= 60.03256 gr/mol
Gr CH3COOH
= 60.03256 gr/mol x 0.454 mol
= 27.25478224 gr
For the experiment, firstly, deionized water was put into burette to clean up the burette
twice. It was filled in until a half burette. Then, 0.05 mL of NaOH was put into burette
to clean up again the burette. After that, 0.05 mL solution of NaOH was put into the
burette for the titration. 5 mL of vinegar then was taken by using pipette and was put
into a beaker glass. Deionized water later was added into the beaker glass for about 10
mL. Then 5 drops of Phenoftalein was added by using pipette into beaker glass. The
beaker glass was put above the stirrer, and the magnetic stirrer was used to mix the
solution. It was done for about 350 rpm. Then the burette was put above the beaker
glass. After it was stirred for a minute, NaOH was started to drop slowly inside the
burette. Then it took for a while until the color of vinegar solution plus Phenoftalein
changed into soft pink.
1.5 Results and Discussion
2.1 Results
A. First try
Start point
: 0.30
End mark
: 42.80
Volume
: 42.50 ml (failed)
B. Second try
Start mark
: 1.20
End mark
: 47.50
Volume
: 46.30 ml
C. Final try
Start mark
: 1.20
End mark
: 45.70
Volume
: 44.50 ml
Average
=
44.50 𝑚𝑙 + 46.30 𝑚𝑙
2
= 45.40 mL
2.2 Analysis of Data
The concentration of vinegar was calculated for the experiment back then. It was
more like a prediction. It can be done by using this equation:
M1V1 = M2V2
0.5 mol/l . V1 = 0.05 mol/l . 100 mL
0.5 . V1 = 0.05 x 100 mL
5 𝑚𝐿
V1 =
0.5
V1 = 10 mL
Description:
M= Molarity (mol/l)
V= Volume (l, which later then converted to mL)
2.3 Discussion
Titration is the way of analysis of the measurement of the amount of solution
needed to react with substances contained fixed with another solution.
In this experiment, the molarity was determined the molarity of NaOH using
titration process between CH3COOH solution of 10 ml with 0.5 M NaOH solution.
In this titration experiment, 10 ml of CH3COOH solution was titrated with NaOH
to produce the equation as follows;
CH3COOH + NaOH  NaCH3COO + H2O
H2O  H+ + OH•
First titration
CH3COOH 5 ml of 0.3205 M put into a glass flask, 5 drops of Penoftalin was added.
50 mL of NaOH was put into a burette, then Penoftalin was dropped slowly until
the indicator changes color, and NaOH was reduced up to 42.80 ml mark and the
color was dark pink. The first titration was failed due to the our inaccuracy, so it
caused the solution became dark pink colored.
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Second titration
CH3COOH 5 ml of 0.3205 M put into a glass flask, 5 drops of Penoftalin was added.
50 mL of NaOH was put into a burette, then Penoftalin was dropped slowly until
the indicator changes color, and NaOH was reduced up to 47.50 ml mark and the
color was dark pink. The second titration was a success because the solution became
pale pink colored.
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Final titration
CH3COOH 5 ml of 0.3205 M put into a glass flask, 5 drops of Penoftalin was added.
50 mL of NaOH was put into a burette, then Penoftalin was dropped slowly until
the indicator changes color, and NaOH was reduced up to 45.70 ml mark and the
color was dark pink. The final titration was a success as well because the solution
became pale pink colored.
1.6 Conclusion
A titration is a valid form of finding the concentration of a substance within a solution.
In this experiment, calculation of concentration must be calculated precisely. If
experimenters made a single mistake, it would impact to the whole experiment. That is
why cautiousness on observing the titration is a vital thing on this experiment.
1.7 Reference
S. Zumdahl and Susan A. Zumdahl, Chemistry, 9th edition
Laboratory module (Nugraha, Tutun)
Laboratory journals (Jasmine, Karmelia and Luthfiyya)