BCHCL138 Chemistry Lab IV

EXPERIMENT 10
TO STUDY THE
KINETICS OF
ACID-CATALYSED
HYDROLYSIS OF AN ESTER
BY TITRIMETRY
Structure
10.1 Introduction 10.4 Procedure
Expected Learning Outcomes 10.5 Observations
10.2 Principle 10.6 Calculations
10.3 Requirements 10.7 Results

10.1 INTRODUCTION
In the previous experiment you have learnt about and studied the kinetics of
persulphate-iodide ion reaction by initial rate method. In this experiment you
would study the kinetics of acid-catalysed hydrolysis of an ester. The progress
of the reaction will be followed titrimetrically. In fact, there are two parts of the
experiment. In the first part you would study the acid-hydrolysis of methyl
acetate and in the second part the same reaction will be used to compare the
strengths of two mineral acids.

In the next experiment you would learn about and perform the kinetics of base
catalysed hydrolysis of an ester i.e., saponification reaction.

Expected Learning Outcomes

After studying about and performing this experiment, you should be able to

 explain the principle of studying the kinetics of acid-hydrolysis of methyl

acetate by titrimetry;
 explain the term, ‘pseudo first order reaction’;
 perform the acid-hydrolysis of methyl acetate and follow the progress of
162 reaction by titrating the reaction mixture against standard NaOH solution;
Experiment 10 To Study The Kinetics of Acid Catalysed Hydrolysis of Ester by Titrimetry
 process the data obtained to calculate the rate constant for the acid-
hydrolysis of methyl acetate;

 graphically determine the rate constant for the reaction; and

 compare the strengths of two mineral acids by studying the rate of hydrolysis
of methyl acetate by them.

10.2 PRINCIPLE
You would recall from your school chemistry that the esters can be hydrolysed
by mineral acids or bases. The hydrolysis of an ester by itself is quite slow and
reversible. However, when catalysed by acid and alkalies the reaction
proceeds in the forward direction. In this experiment, you would study the
hydrolysis of methyl acetate in presence of hydrochloric acid. The reaction can
be given as

H
CH 3 COOH3 ( aq)  H 2 O(l) 
 CH 3 COOH( aq)  CH 3 OH( aq) …(10.1)

The rate of the reaction is expected to depend on the concentrations of methyl

acetate, water and the acid. We can write the generalised rate expression as

Rate = k [CH3COOCH3]a [H2O]b [H3O+]c …(10.2)

However, since the reaction is done in aqueous medium, the concentration of

water is much larger than that of methyl acetate and the acid. Therefore, when
the reaction progresses the concentration of water does not change
significantly. That is, the concentration of water remains constant. Further,
since H3O+ ions act as a catalyst and as you know the concentration of the
catalyst does not change during the reaction i.e., the concentration of
H3O+ions also remain constant. Under such conditions the Eq. 10.2 gets
modified to

Rate = k* [CH3COOCH3]a …(10.3)

Where, k* = k [H2O]b[H3O+]c

This implies that the rate of the reaction depends only on the concentration of
methyl acetate. The value of a, the order of the reaction with respect to methyl
acetate is found to be equal to 1. That is it is a first order reaction and the rate
equation becomes

Rate = k* [CH3COOCH3] …(10.4)

So, we can say that the reaction shows first order kinetics under given set of
conditions. Such reactions, where the order of reaction is 1 due to the
condition that the concentrations of other components remain constant, are
called pseudo first order reactions.

Using the integrated rate equation for a first order reaction which you have
learnt in Unit 13 of BCHCT-137 course we can write,

2.303 c0
k log ...(10.5)
t ( c 0  x)
163
BCHCL138 Chemistry Lab IV
Where,c0 is the initial concentration of methyl acetate and (c0-x) is the
concentration of the methyl acetate at a time, t. Thus, to get the rate constant,
we need to know the value of c0 and (c0-x) as a function of time.

Now the question comes up is that how do we measure the concentration of

the reactant (methyl acetate) as a function of time. A look at the Eq. 10.1
shows that as the reaction proceeds, each molecule of methyl acetate gives a
molecule of acetic acid and methanol. Also, we know that to begin with, the
reaction mixture contains methyl acetate, water and hydrochloric acid (which
acts as a catalyst and its concentration does not change). This means that as
the reaction progresses the total acid content (HCl + acetic acid) of the
reaction mixture will increase as a function of time. So, if we measure the total
acid content of the reaction mixture as a function of time, we can follow the
progress of the reaction.

In a typical method to study the kinetics of this reaction we mix known

Every time, in the amounts of the aqueous solution of methyl acetate and hydrochloric acid to
given mixture there start the reaction. We then withdraw equal and known volumes of the reaction
is a definite amount mixture at definite time intervals and add it into vessels containing ice-cold
of hydrochloric acid water. It is then titrated against standardised alkali using phenolphthalein as
and also the acetic the indicator. The end point is the appearance of pink colour which does not
acid formed as a disappear within ten seconds. This provides the acid content of the reaction
consequence of the mixture at the time of withdrawal of the reaction mixture. The ice-cold water is
hydrolysis of the added to significantly decrease (you know that the rate of reaction depends on
ester. Thus, the titre temperature) the rate of the reaction so that it does not proceed much during
value corresponds the titration. The titre values (volume of NaOH solution required to neutralise
to the volume of
the acids in the withdrawn reaction mixture) at various time intervals can be
NaOH required to
used to determine the rate constant as explained below.
neutralise both the
acids. Suppose that V0 ,Vt and V  respectively are the titre values initially (t = 0), at
a time t, and at the end of the reaction (infinite time reading). You can see that

V0 will be the volume of NaOH required for the neutralisation of hydrochloric

acid present in a definite volume of reaction mixture, when no acetic acid is
formed i.e., in the beginning of the reaction.

V  will be proportional to the amount of hydrochloric acid, and acetic acid

present in a definite volume of reaction mixture after the completion of the
reaction i.e., when all the methyl acetate has been hydrolysed.

Vt will be proportional to the amount of hydrochloric acid and acetic acid

present in a definite volume of reaction mixture at time, t, after the beginning
of the reaction.

Further, V  V0 would be proportional to the amount of acetic acid formed

when the hydrolysis is complete, or it is proportional to the initial concentrate
of methyl acetate (c0) i.e.,
V  V0   c 0 …(10.6)

Similarly, Vt  V0 would be proportional to the acetic acid formed at a time t or

we can say that it is proportional to the concentration of methyl acetate
164 hydrolysed (x) at time, t.
Experiment 10 To Study The Kinetics of Acid Catalysed Hydrolysis of Ester by Titrimetry
Hence the concentration of ester remaining unhydrolyzed at a time t i.e., (c0 -x)
would be proportional to

[(V  V0 )  (Vt  V0 )] = V  Vt i.e.,  c0  x   V  Vt  …(10.7)

Substituting Eq. 10.6 and 10.7 in Eq. 10.5, we can write

2.303 V  V0 
k log10 …(10.8)
t V  Vt 
Thus, the values of k can be determined by substituting the values of V  V0
and V  Vt at different times in the Eq. 10.8.

Alternatively, we can simplify and rearrange Eq. 10.8 to get

kt
log10 (V  Vt )  log(V  V0 )  …(10.9)
2.303

This is an equation of a straight line. This means that if we plot a graph

between log10( V  Vt ) and time t we should get a straight line. The slope of
the line would be equal to –k/2.303 from which the value of k can be obtained

k = – 2.303  slope …(10.10)

Thus, we can find the value of rate constant by graphical method also.
Therefore, there are two methods to determine the rate constant for the acid
hydrolysis of methyl acetate studied by titrimetry.

Comparing the strengths of two mineral acids

As mentioned above, the rate of acid hydrolysis of methyl acetate depends on

the concentrations of methyl acetate, water and acid (in terms of hydronium or
hydrogen ions). Further, we know that different mineral acids may ionise to
different extents at a given concentration. This means that they would provide
different concentrations of hydronium ions in solution. You may note that in
Eq. 10.3 we defined the rate constant to be

k* = k [H2O]b [H3O+]c

This means that the rate constants would be different for different acids as the
concentration of H3O+ ions will be different in them. We can write

k1* = k [H2O]b [H3O+]1c …(10.11)

k2* = k [H2O]b [H3O+]2c …(10.12)

Where, [H3O+]1 and [H3O+]2 respectively are the concentrations of hydronium

ions provided by a given concentration of acid 1 and 2. Thus, if we determine
the rate constant for acid hydrolysis of methyl acetate using two different
mineral acids under identical conditions of temperature and concentration then
their comparison would provide the relative strengths of the two mineral acids.

k1*

  
c
k H 2 O b H3 O  1

 
H3 O  1
c

H3 O  1  …(10.13)
2 
k 2* k H O b H O  c
3  
2  
c
H3 O  2 H3 O  2  165
BCHCL138 Chemistry Lab IV

10.3 REQUIREMENTS
Apparatus Chemicals

Burettes (50 cm3) 2 Methyl acetate

Conical flasks (100 cm3) 7 Hydrochloric acid

Conical flasks (250 cm3) 2 Sulphuric acid

Pipettes (5 cm3) 2 Phenolphthalein

Water trough 1 Sodium hydroxide

Stop watch 1 Deionised water

Solutions Provided

The following solutions would be provided by your counsellor.

1. 0.5 M HCl: prepared by transferring 45 cm3 of concentrated hydrochloric

acid to a 1 dm3 volumetric flask containing approximately 500 cm3 of
deionised water. The addition of acid is made in a large number of small
lots and the solution is shaken after the addition of each lot of hydrochloric
acid. The volume is made up to the mark with deionised water.

2. 0.5 M H2SO4: prepared by transferring 25 cm3 of concentrated sulphuric

acid to a 1 dm3 volumetric flask containing approximately 500 cm3 of
deionised water. The addition of acid is made in a large number of small
lots and the solution is shaken after the addition of each lot of sulphuric
acid. The volume is made up to the mark with deionised water.

3. 0.10M NaOH: prepared by dissolving8 g of sodium hydroxide pellets in 2

dm3 of deinoised water taken in a 5 dm3 jar.

4. Phenolphthalein indicator: prepared by dissolving 1 g of

phenolphthalein in 100 cm3 of ethanol and adding 100 cm3 of water to it
with constant stirring. If there are any solid particles, the solution is
filtered.

10.4 PROCEDURE
There are two parts of the experiment

a) To study the acid-catalysed hydrolysis of methyl acetate

b) To compare the strengths of two mineral acids (HCl and H2SO4) using
acid-catalysed hydrolysis of methyl acetate

Let us learn about the procedure for these two parts.

a) To study the acid catalysed hydrolysis of methyl acetate

1. Take 50 cm3 of given 0.5 M hydrochloric acid solution in a 250 cm3

166 conical flask by using a burette.
Experiment 10 To Study The Kinetics of Acid Catalysed Hydrolysis of Ester by Titrimetry
3
2. Pipette out 5 cm of methyl acetate in a test tube and keep the flask
and the test tube in a water trough for about five minutes.

3. Take six conical flasks of 100 cm3 each and add about 20 cm3 of ice-
cold water to each one of them.

4. Transfer methyl acetate to the conical flask containing hydrochloric

acid and start the stopwatch when nearly half the volume of ester has
been added. Shake the mixture in the conical flask.

5. Using another pipette immediately pipette out 5 cm3 of the reaction

mixture into a 100 cm3 conical flask containing 20 cm3 of ice-cold
water. Add a drop of phenolphthalein and titrate the resultant solution
with 0.10M sodium hydroxide solution. The end point is the
appearance of light pink colour that does not decolourise in about 10
seconds on shaking.

6. Note and record the burette readings in the Observation table-I.

7. After 10 minutes again pipette out 5 cm3 of the reaction mixture into a
100 cm3 conical flask containing 20 cm3 of ice-cold water and titrate
with 0.10 M NaOH solution. Record the burette readings again in the
Observation Table-I.

8. Repeat step 7 at least 4 times.

9. After taking at least six readings, pipette out 10 cm3 of mixture into a
100 cm3 conical flask and close it with a loose cork. Heat the conical
flask on a copper water bath at 500C for about 30 minutes. Then cool
the conical flask to room temperature and titrate 5 cm3 of this mixture
against 0.10M sodium hydroxide sodium using phenolphthalein as
indicator. Record the burette readings in Observation Table-I

b) To compare the strengths of two mineral acids using acid catalysed

hydrolysis of methyl acetate

In order to compare the strengths of two mineral acids say hydrochloric

acid and sulphuric acid, using acid-catalysed hydrolysis of methyl acetate
we need to follow the procedure given above, for both the mineral acids.
That is,

 First we perform the experiment using say 0.5 M HCl and determine
the value of rate constant.

 The same procedure is followed for the determination of rate constant

using 0.5 M of sulphuric acid.

 The two rate constants are then compared to assess the relative
strengths of the two acids as explained above.

It is important to note that in this part of the experiment we need to

standaradise the two given solutions of the mineral acids and dilute
them appropriately to bring them to same concentration before
starting the experiment. 167
BCHCL138 Chemistry Lab IV

10.5 OBSERVATIONS
Record your observations on the time of withdrawal of reaction mixture and the
corresponding burette readings here
Room temperature =…. oC
Observation Table -I
3
S.No. Time Burette reading Titre value / cm
/ min Initial Final (Final-initial burette reading)
1 0
2 10
3 20
4 30
4 40
6 50
7

10.6 CALCULATIONS
You would recall from Section 10.2 above that we can determine the rate
constant by two methods. These are the integrated rate equation method and
the graphical method. We would follow both of these here. Let’s first take up
integrated rate equation method.

I. Note down the following data from Observation Table-I

The titer value at t = 0 i.e., V 0 =

The titer value at t =  i.e., V  =

The initial concentration of methyl acetate=(C0)= V  V0 =

Copy the titre values from Observation Table-I to the column III of the
following table. Calculate and complete the data in column IV to VI in the
same.
Time Titre (C0-x) Log Rate constant
S.No. / min value, (V  Vt ) ( V   Vt ) 2.303 V  V0 
k log10
Vt t V  Vt 
1 0
2 10
3 20
4 30
4 40
6 50
7
Calculate the average value of the rate constant obtained for S.No. 2-6.

168 Average value of the rate constant =

Experiment 10 To Study The Kinetics of Acid Catalysed Hydrolysis of Ester by Titrimetry
II. Graphical method

Plot a graph between (V  Vt ) (on y-axis) and t (on x-axis) using the data
from column V and column II respectively of the table given above.

Calculate the slope of the line = slope=

Calculate the value of rate constant by using the following expression

k = –2.303  slope

= ………. s-1

Part b) of the experiment

Repeat the above procedure and calculations for 0.5 M H2SO4 and
determine the value of k. let the values of the rate constant be k HCl and
k H 2 SO 4 respectively. Then we can calculate the relative strength of the
acids as
(Strength of HCl) k HCl

(Strength of H 2 SO 4 ) k H 2 SO 4

10.7 RESULTS
k (by integrated rate equation) = ……. s-1

k (by graphical method) = ……s-1

The relative strengths of HCl and H2SO4 are:

169