Investigating the reaction Between Sodium Thiosulphate (Na2S2 O3) and Hydrochloric acid (HCl)
I am investigating the rate of reaction between Hydrochloric acid
(HCl) and Sodium Thiosulphate (Na2S2O3), when Na2S2O3 is mixed with
The rate of a reaction can be speeded up by increasing the
temperature; at a higher temperature the particles move faster and
collide more often, as a result of this the reaction speeds up.
Increasing the concentration of reactants in water will also speed up
the reaction, as there are more reactive particles in the same volume
and therefore more chance of a reaction taking place, which speeds up
the reaction. This is shown in the diagrams below.
Sulphuric acid (H2SO4)
High Conc. Of Na2CO3
Text Box: Low Conc. Of Na2CO3
These diagrams are from a previous experiment with Sulphuric acid and
Breaking up solids into smaller pieces increases the rate of reaction,
as this increases the surface area of the reactant, therefore there is
a larger area for reaction to take place.
However this is not relevant for my investigation as I am only using
liquids, not solids. Also using a catalyst will increase the rate of
reaction, but this is also not relevant for my investigation, as I am
not using any catalysts.
I predict that as the concentration of Sodium Thiosulphate
is increased, the rate of reaction will also increase.
I think this is because as the Na2S2O3 concentration is increased it
will react more quickly with the Hydrochloric acid
(HCl) as it is more
concentrated so there are more Na2S2O3 particles and therefore more
chance of a Na2S2O3 particle to collide with a HCl particle creating a
reaction and this results in the solution becoming cloudy quicker as
the sulphur is precipitated, as the time taken for the sulphur to go
cloudy (precipitate) decreases, the rate of reaction increases.
The equation for this reaction is:
I will change the concentration of Na2S2O3 with H2O, in order for my
findings to be valid and correct it must be a fair experiment. I will
make sure it is fair by keeping the solution of H2O and Na2S2O3 at
50ml, but the ratio of H2O to Na2S2O3 will change, this will effect
the number of particles in the solution, but not the area of the
solution, making it a fair test. To make it a fair test I will keep
the temperature the same, this is because it would be harder the
control as the main variable, I will also keep the concentration of
HCl the same as, I am looking at how the Na2S2O3 reacts with HCl and
changing the HCl will no effect the number of reactants, but will
effect the time taken to react.
I am doing 6 variations on the concentration to provide a wide range
of readings. The solution of 50ml3 will consist of a range from 5ml3
to 50ml3 of Na2S2O3; the rest of the solution will be H2O. This range
was determined by preliminary tests, where I found the experiments
won't take to long to complete, but will slow enough to easily record
A concentration of Na2S2O3 and H2O to make a solution of 50 cm3
-varying from 5ml3 of Na2S2O3/45ml3 of H2O to 50ml3 of Na2S2O3- is
measured using separate burettes, and then poured in to a conical
flask. 5 cm3 of HCl is measured using a syringe and poured in to a
test tube. Both solutions are poured in to a new conical flask, which
is placed on top of a piece of paper marked with a cross. The
stopwatch will now be started. When the mixture has turned
sufficiently cloudy so that the cross can no longer be seen the
stopwatch will be stopped and the time will be recorded. The
experiment is repeated with solutions increasing concentrations of
Na2S2O3. The whole procedure is then repeated for accuracy and to
provide an average of 2 sets of results, if there are any anomalous
results, the tests will be repeated to get the correct results.
To conduct my experiment safely I will follow normal laboratory rules,
R The wearing of safety goggles to protect my eyes from chemical
R Standing up to conduct the experiment, therefore reducing the risk
of tripping and spilling chemicals.
R Taking care when handling all chemicals, I will not touch my eyes or
mouth until I have thoroughly washed my hands.
R Taking care when using glassware to prevent injury.
Na2S2 O3(ml3 )
Time 1 (sec)
Time 2 (sec)
This is shown on the graph on page 5.
Results in bold are anomalous.
Analysis and Conclusion (N.B. All times used in this section are the
I found out that my prediction was correct; when the concentration of
the Na2S2O3 has gone up, the rate of reaction has speeded up. This is
shown on the graph on page 5. The graph starts off at 5ml3 of Na2S2O3,
where is takes 282 sec, it then drops sharply down to 84 sec at 15ml3
and then drops sharply again down to 57 sec, where it then decreases
slowly down to 26 sec. If I were to carry on this investigation to a
higher concentration I do not think it would decrease much further as
it seems to have nearly reached its saturation. I reached this
conclusion after seeing that the rate of reaction levels out after
There is good negative correlation with all the results on the graph,
as the graph shows the time taken for the cross to disappear, not the
rate of reaction.
Time 2 of my results for 15ml3 of Na2S2O3 was anomalous, however this
is probably due to human error, because of the interpretation as to
when the experiment ended.
I conclude that the more concentrated a reactant is, the quicker the
rate of reaction time will be. I have come to this conclusion because
of several reasons. Firstly, my results give conclusive evidence that
as the amount of Na2S2O3 in the H2O solution decreases there are less
reactive particles to collide with and therefore less reactions taking
place, so the reaction rate is slower. In a more concentrated
solution, there are more reactive particles to collide so the reaction
time is quicker.
My experiment went according to plan but there were flaws in it. For
example, I only recorded two sets of results and then worked out an
average. I should have done more tests to gain a more reliable average
and this would have solved my anomalous results. However overall my
results were accurate.
I think there is also a human error factor involved when you are
measuring liquids and looking for an end point in the reaction.
Although the reaction I chose had a fairly definite end point it was
still hard to tell whether the whole cross had disappeared or not. I
could do more results to find a better way of working this out.