CHEMISTRY II. LAB REPORT
TOPIC: TITRATION
PREPARED BY: MERVE KASIMOĞULLARI 221
08.04.2012
THE GROUP
Merve KASIMOĞULLARI
Zeynep GÜNLER
Nurefşan EFEOĞLU
MERVE KASIMOĞULLARI
08.04.12
PURPOSE
The purpose of the experiment is to train ourselves with glass laboratory equipment, their
safety procedures, the setup and the use a burette, and to make an experiment which is
titration.
HYPOTHESIS
We can determine the concentration of an unknown solution in a solution of known
concentration.
WHAT IS TITRATION?
A titration is a technique where a solution of known concentration is used to determine the
concentration of an unknown solution. Typically, the titrant (the know solution) is added from
a burette to a known quantity of the analyte (the unknown solution) until the reaction is
complete. Knowing the volume of titrant added allows the determination of the concentration
of the unknown. Often, an indicator is used to usually signal the end of the reaction, the
endpoint.
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MERVE KASIMOĞULLARI
08.04.12
EQUIPMENT AND MATERIALS

A conical flask

A pipette

A 50 cm 3 burette

Hydrochloric acid

Sodium carbonate

Methyl orange indicator

A calculator
EXPERIMENT or PROCESS
First, titrate the acid against your standard solution.

Measure 25 cm3 of the sodium carbonate solution into a conical flask, using a pipette.
Add a few drops of methyl orange indicator.

Pour the acid into a 50 cm3 burette. Record the level.

Drip the acid slowly into the conical flask. Keep swirling
the flask. Stop adding acid when a single drop finally
turns the indicator red. Record the new level of acid in the
burette.

Calculate the volume of acid used, as shown on the right.
How much acid was
used?
Final level: 28.8 cm3
Initial level: 1.0 cm3
Volume used: 27.8 cm3
To convert cm3 to dm3:
Divide by 1000
So 27.8 cm3= 0.0278dm3
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MERVE KASIMOĞULLARI
08.04.12
Now calculate the concentration of the acid.
Step 1) Calculate the number of moles of sodium carbonate used.
25
1000 cm3 of 1 M solution contains 1 mole so 25 cm3 contains 1000 ∗ 1 mole or 0.025 mole.
Step 2) Form the equation, find the molar ratio of acid to alkali.
2HCL(aq) + Na2CO3(aq)
2 moles
2NaCL(aq) + H2O(l)+ CO2(g)
1 mole
The ratio is 2 moles of acid to 1 mole alkali.
Step 3) Work out the number of moles of acid neutralized
1 mole of alkali neutralizes 2 moles of acid so 0.025 mole of alkali
neutralizes 2* 0.025 moles of acid. 0.05 moles of acid were neutralized
Step 4) Calculate the concentration of the acid
Concentration =
𝑛𝑢𝑚𝑏𝑒𝑟𝑜𝑓𝑚𝑜𝑙𝑒𝑠
𝑣𝑜𝑙𝑢𝑚𝑒 𝑖𝑛 𝑑𝑚3
0.05
= 0.0278 = 1.8 mol/dm3
So the concentration of the hydrochloric acid is 1.8 M.
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MERVE KASIMOĞULLARI
08.04.12
RESULTS
We obtained 1.8 M of hydrochloric acid by this
experiment.
CONCLUSION
We can learn that how much acid we used in an experiment by the titration. Also, the
experiment confirmed the hypothesis. I learned that what I can do when I don’t know the
concentration of an unknown substance. Titration makes easier when I am in trouble with the
experiment which includes unknown concentrations.
TASTING ACIDS AND BASES USING LITMUS PAPER
Often, to measure the pH, special papers which have been soaked with indicators are used.
These papers change color when they are immersed in acidic or basic liquids. This is the case
of the well-known litmus paper.
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MERVE KASIMOĞULLARI
08.04.12
Litmus paper
Litmus is a substance obtained from certain lichens. It has the property of changing its color
to red with acidic substances and to blue with basic ones. On the packet of the litmus paper,
there is a color scale which indicates the color assumed by the paper as a function of the pH .
Using Litmus paper
Using Litmus paper is simple. First of all, it is
necessary to immerse an end of it in the liquid you
wish to examine and to remove it immediately. The
pH of the liquid is determined by comparing the
color of the paper to the scale of colors which is printed on its packet.
ANOTHER EXAMPLE OF TITRATION
A 25 ml solution of 0.5 M NaOH is titrated until neutralized into a 50 ml sample of HCl.
What was the concentration of the HCl?
SOLUTION
Step 1 - Determine [OH-]
Every mole of NaOH will have one mole of OH-. Therefore [OH-] =
0.5 M.
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MERVE KASIMOĞULLARI
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Step 2 - Determine the number of moles of OH-
Molarity = # of moles/volume
# of moles = Molarity x Volume
# of moles OH- = (0.5 M)(.025 L)
# of moles OH- = 0.0125 mol
Step 3 - Determine the number of moles of H+
When the base neutralizes the acid, the number of moles of H+ = the number of moles of OH-.
Therefore the number of moles of H+ = 0.0125 moles.
Step 4 - Determine concentration of HCl
Every mole of HCl will produce one mole of H+, therefore number of moles of HCl = number
of moles of H+.
Molarity = # of moles/volume
Molarity of HCl = (0.0125 mol)/(0.050 L)
Molarity of HCl = 0.25 M
ANSWER
The concentration of the HCl is 0.25 M.
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MERVE KASIMOĞULLARI
08.04.12
ANOTHER SOLUTION METHOD
The above steps can be reduced to one equation
MacidVacid = MbaseVbase
where
Macid = concentration of the acid
Vacid = volume of the acid
Mbase = concentration of the base
Vbase = volume of the base
This equation works for acid/base reactions where the mole ratio between acid and base is
1:1. If the ratio were different as in Ca(OH)2 and HCl, the ratio would be 1 mole acid to 2
moles base. The equation would now be
MacidVacid = 2MbaseVbase
For the example problem, the ratio is 1:1
MacidVacid = MbaseVbase
Macid(50 ml)= (0.5 M)(25 ml)
Macid = 12.5 MmL/50 ml
Macid = 0.25 M
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MERVE KASIMOĞULLARI
08.04.12
A SHORT EXAMPLE FOR THE TITRATION
Titration reveals that 11.6 mL of 3.0 M sulfuric acid are
required to neutralize the sodium hydroxide in 25.00 mL of
NaOH solution. What is the molarity of the NaOH solution?
Solution:
H2SO4(aq) + 2NaOH(aq)
2O(l)
+ Na2SO4(aq)
= 2.8 M NaOH
REFERENCES:
http://www.funsci.com/fun3_en/acids/acids.htm#contents
http://chemed.chem.purdue.edu/genchem/lab/techniques/titration/what.html
http://chemistry.about.com/od/workedchemistryproblems/a/titrationexampl.htm
http://www.mpcfaculty.net/mark_bishop/titration_example.htm
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