
Chapter 13 – Volumetric analysis
(acid – base titrations)


Volumetric analysis is a well established and
versatile form of quantitative chemical
analysis. The purpose of this type of analysis
is to use an accurately known volume and
concentration of one solution to find the
accurate concentration of a second. The
experimental procedure which allows us to do
this is called a titration. This procedure is
described in detail below.
Open the tap and run out some of the
liquid until the tap connection is full of
acid and no air remains (air bubbles
would lead to an inaccurate result as
they will probably dislodge during the
titration).
Remove the funnel (stops dripping
while you read the meniscus).
Release the liquid until the bottom of
the meniscus is on the 0ml.
Preparing the pipette
Titration procedure
Filling the burette
0
10
20
30
Burette
40



50






Clean the burette, rinse it and dry the
outside.
Rinse it with the solution it is going to
contain (acid).
Fill the burette to above the 0ml
mark.
Check for air bubbles and invert to
remove any, if required.


Wash and rinse well.
Rinse with the solution it is to contain.
Suck up solution with a pipette filler,
above the grad mark. dry outside.
Release the solution until the bottom
of the meniscus is on the grad line .
Tip off any hanging drop (this should
not be counted).
Allow to drain under gravity (do not
blow).
When drained touch the tip off the
side, any drops which should be
inlcuded will drain in. Leave the rest.
Preparing the conical flask
0
10
20
Conical flask
30
40
50




Rinse several times with deionised
water.
Dry outside.
Add base solution as described above,
from the pipette.
Rinse down walls of the flask with
deionised water (you know exact
volume added of base)
Titration procedure
 Add indicator to the flask, 2 or 3
drops are enough because all
indicators are weak acids or bases.
 Rinse down the sides with water.
 Run the solution into the flask from
the burette, slowly.
 Rinse the sides of the flask regularly.
 Swirl the flask constantly, to ensure
thorough mixing of reagents.
 As the end point nears, add the
solution drop by drop.
 When the end-point is reached the
indicator will change colour suddenly.
 At this point the acid will have exactly
neutralised the base.
 Now read the meniscus of the
burette, from the bottom, at eye
level.
 Use a filter paper, if necessary, to
make the meniscus more readable.
 Record your result.
 Repeat the titration several times.
 Get the average value.
 Only include the values that agree
within 0.2ml of each other.
To prepare a standard solution of sodium
carbonate.
Weigh the sample
 Weigh 1.30 g of sodium carbonate on an
electronic balance, as accurately as you
can. Use a clock glass.
 Two places of decimals would be best.
Transfer to beaker
 Use a spatula to transfer the sample
to a beaker of warm water (100ml).
 Rinse the clock glass.
 Rinse the remaining grains into the
beaker with deionised water.
 Rinse the spatula into the beaker also.
 All traces must be transferred.
Pour the washings into the volumetric
flask.
 Pour the washings into the volumetric
flask, using a funnel and a glass rod.
 Wash the rod as well.
 Rinse the beaker several times with
deionised water.
 Pour these washings into the
volumetric flask.
 Top up the volumetric flask with
deionised water, until just below the
graduation mark.



Top up to the graduation mark with a
dropper.
Read the bottom of the meniscus at
eye level.
Invert and mix to ensure proper
mixing of the contents.
Calculations
Number of moles of sodium carbonate
1.3

 0.012 moles
106
0.012 moles in 250 cm 3
0.012  4  0.048 moles in 1 litre
 0.048 Molar
To use this standard sodium carbonate
solution to find the concentration of
(standardise) a given hydrochloric acid
solution.
Procedure
 Place 20 ml of 0.0.48 Molar sodium
carbonate into a conical flask using a
pipette.
 Add two drops of methyl red
indicator. This will give a yellow colour
to the solution. Note; the number of
drops of indicator should be kept to a
minimum as most indicators are either
weak acids or bases and will therefore
take part in the neutralization
process.
 Place the hydrochloric acid in the
burette and adjust the level to zero,
taking all of the usual precautions.
 Titrate in the usual manner.
 Record the volume of acid required to
neutralise the sodium carbonate. The
point of neutralisation is reached when
the indicator just turns red (pink).

Repeat the titration several times
until two titration values agree to
within 0.2 ml of each other.
Equation for the titration
2HCl  Na2CO3  2NaCl  H2O  CO2
Results
 Volume of the acid = 19.2 ml
 Factor for the acid = 2 (the number in
front of HCl in the balanced equation)
 Molarity of the acid = ?
 Volume of the base = 20 ml
 Factor for the base = 1 (the number in
front of sodium carbonate in the
balanced equation)
 Molarity of the base = 0.048 M
Calculations
Va  Ma Vb  Mb

na
nb
19.2  Ma 20  0.048

2
1
 Ma 
20  0.048  2
19.2
Ma  0.1 moles per litre
 0.1 Molar  0.1 M
To make up an approximate solution of
sodium hydroxide and standardise it (find
its exact concentration) by titration with
the standard hydrochloric acid solution
above.
Procedure
 Place 20 ml of the sodium hydroxide in
the conical flask. Note; Always place
the base in the conical flask as they
may react with the ground glass in the
tap of the burette.
 Add two drops of methyl red indicator
and a yellow colour is imparted to the
solution.
 Put the hydrochloric acid (previously
standardized) into the burette.
Adjust to the zero level in the usual
way.
 Titrate in the usual manner.
 When the colour of the solution in the
conical flask changes to a faint trace
of permanent pink the end-point has
been reached.
 Record the volume of acid required to
do this.
 Repeat the titration several times
until two titration values agree to
within 0.2 ml of each other.
Equation for the titration
NaOH  HCl  NaCl  H2O
Results
 Volume of base = 20 ml
 Factor for the base = 1
 Molarity of the base = ?
 Volume of the acid = 19.8 ml
 Factor for the acid = 1
 Molarity of the acid = 0.1 M
Calculation
Va  Ma Vb  Mb

na
nb
19.8  0.1 20  Mb

1
1
 Mb 
19.8  0.1
20
 Mb  0.099 moles per litre
 0.099 M
To determine the percentage of ethanoic
acid in vinegar.
Vinegar is a solution of ethanoic acid
dissolved in water. The purpose of this
titration is to find the percentage of this
acid in the vinegar.
Procedure

Add 50 ml of vinegar to a volumetric
flask using a 25 ml pipette twice.

Make up the solution to the 250 ml
mark with deionised water. This is the
solution which will be used for the
titration.

Note; diluting the solution in this
manner is necessary for two reasons
(i) you will use less reagents this way
and (ii) if an error is made measuring a
dilute solution it will not have great
implications for the final answer.

Add 20 ml of 0.1 M sodium hydroxide
solution to the conical flask using a
pipette.

Add two or three drops of
phenolphthalein indicator, just enough
to impart a pink tinge to the sodium
hydroxide solution.

Put the dilute vinegar solution in the
burette.Titrate in the usual manner.


The end-point is reached when the
pink colour changes to colourless.
Record the volume of acid used from
the burette.Repeat the titration
several times until two titration values
agree to within 0.2 ml of each other.
Equation for titration
CH3COOH  NaOH  CH3COONa  H2O
Results
 Volume of acid used = 13 ml
 Factor for the acid = 1
 Molarity of the acid = ?
 Volume of base used = 20 ml
 Factor for the base = 1
 Molarity = 0.1 M
Calculations
Va  Ma Vb  Mb

na
nb

13  Ma 20  0.1

1
1
20  0.1
13
 0.154 moles / litre
 Ma 
 0.154  5 moles / litre in the original vinegar.
 0.77 moles/ litre of solution
 0.77  60 g/L
 46.2 g/L  4.62 g in 100 cm 3
 4.62% (w/v)
To determine the percentage of water of
crystallization in hydrated sodium
carbonate (washing soda).
Water of crystallization is the water which is
found as part of the structure of a
crystalline substance. It has nothing to do
with being wet. The water molecules referred
to in the term occupy positions in the crystal
lattice of the substance. This water of
crystallization is generally represented in the
chemical equations of such compounds, at the
end of the formula e.g.
Na 2 CO 3 .xH 2 O
The ‘x’ here is a number which represents the
number of molecules of water in the crystal.
The purpose of this experiment is to
determine the percentage of water of
crystallization in a substance by titration.
Procedure
 Weigh out accurately 5 g of hydrated
sodium carbonate on a clock glass.
 Make up the solution to 250 ml in a
volumetric flask. Follow the same
procedure as for ‘making a standard
solution’ previously outlined above.
 Pipette about 25 ml of this solution into
a clean conical flask.
 Add a few drops of methyl red
indicator, enough to impart a faint
yellow colour to the solution in the
conical flask.
 Place 0.2 M HCl in the burette and
adjust the level to zero taking all the
usual precautions.
 Titrate in the usual manner until the
yellow colour is replaced by a permanent
pink tinge. This is the end-point of the
titration.
 Record the volume of acid required to
reach the end-point and repeat several
times until two readings (titres) agree
to within 0.2 ml of each other.
Results






Volume of acid used = 23.5 ml
Factor for the acid = 2
Molarity of the acid = 0.2 M
Volume of base = 25 ml
Factor for the base = 1
Molarity of base = ?
Calculations
Va  Ma Vb  Mb

nA
nb
23.5  0.2 25  Mb

2
1
23.5  0.2 4.7
 Mb 

2  25
50
 0.094 Molar sodium carbonate

0.094
 0.0235 moles in 250 cm 3
4
 0.0235  106 g  2.491g

 6.73g - 2.491g  4.239 g
% water of crystallisation

4.239
6.73
 100  63 %
We can now calculate the value
of 'x' in the formula Na2 CO3 .xH2O
We already know that there are
0.0235 moles of sodium carbonate
present in the crystals.
We also know that there are
4.239 g of water present. This is
4.239
moles
18
 0.2355 moles
equivalent to
moles of Na2 CO3 ; moles H2O
0.0235 : 0.2355
1 : 10
 x  10