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Investigating The Rate Of Reaction Between Sodium Thiosulphate And Hydrochloric Acid Essay, Research Paper

Investigating the rate of reaction between Sodium

Thiosulphate and Hydrochloric Acid Diagram

Aim : We did 4 experiments to find out how

the rate of reaction changes with differing concentrations of Sodium

Thiosulphate, Hydrochloric Acid and water. As an inert and stable liquid, water

was used to alter concentration of Sodium Thiosulphate without changing the end

amount of solution. All the atoms in a water molecule have a full outer shell,

so they would not react with the other chemicals.Equipment ???? Beakers, ??????????????????????? Measuring cylinders Clamp stand and Clamps ??????????????????????? Black paper tube ??????????????????????? Light probe and Blue box ??????????????????????? Datalogger ??????????????????????? Lamp ??????????????????????? Total of 30cm3 H 0,

50cm3 Na S O , 12cm3 HClMethod : We wanted to change the concentration

of Sodium Thiosulphate and Hydrochloric Acid, but without changing the overall

quantities. To do this, the Sodium Thiosulphate and water were mixed at

different ratios, with always a constant amount of acid. The table below shows

the 4 different experiments, and what each solution composed of. The ?Graph?

column relates to the graphs taken from the Datalogger for that experiment,

which are included at the back of this piece. Sodium

Thio-sulphate (cm3) Water

(cm3) Hydrochloric

Acid (cm3) Graph

number 20 0 3 DPPAS_02 15 5 3 DPPAS_03 10 10 3 DPPAS_04 5 15 3 DPPAS_05 ? ??????????? We did not do the experiment in

which 0cm3 Sodium Thiosulphate and 20cm3 water were used, as there would have

been no reaction. In total, there were always 20cm3 of water and Sodium

Thiosulphate, with 3cm3 Hydrochloric Acid, giving a total solution of 23cm3 ??????????? The black tube was put around and

below the beaker to help prevent any unwanted light from entering the light

probe, as this would have impaired our results. The reason we used a Datalogger and light

probe instead of the old ?cross? method, is threefold. First, human error. The

cross would not just disappear ? it would fade. There would be no specific

point at which the cross would disappear, and the results of your experiment

would be based entirely on a person?s eyesight. Second, this method will only

tell you (albeit inaccurately) when the cross disappeared, i.e. how long it

took for the reaction to get to a certain point of cloudiness. It would not

tell you the varying rates of the reaction. You would not be able to tell if

the reaction speeded up, slowed down, went steady all the way e.t.c. Lastly,

what would you do if the reaction never got as far as making the cross

disappear? Or what if the reaction took a number of hours to get that far? This

traditional method is about as accurate as taking the temperature from a beaker

of water with your finger. To do the experiment, we set up the

apparatus as explained above. We put the various amounts of chemicals into the

beaker, and used the Datalogger and blue box to record the first 3 minutes of

the experiment, and then used the computer to draw up a graph. The blue box was

set to SLOW

and 10k lux.Prediction : I predict that the rate of reaction

will increase (and get more cloudy, more quickly) when the solution of Sodium

Thiosulphate and Hydrochloric Acid are strongest, and there is no water. The

reason for this is that it will be easier for the Sodium Thiosulphate to react

with the Hydrochloric Acid, as they are the only two chemicals in the beaker,

and there is not water to hinder the rate of the reaction. There will also be more Sodium Thiosulphate to react with

the Hydrochloric Acid, regardless of how much water there is.Results : The graphs from the Datalogger are

included in this project. The filename of each graph correspond to the

filenames (DPPAS_xx) listed in the table above. The table below shows the

amount of time that each graph took to level out, i.e. how long the experiment

took to finish. To work out the rate of reaction over the whole reaction (up to

the point where the reaction levelled), I divided the light depreciation (k

lux) by the time taken (minutes) to give a rate of k lux/min. Light

depreciation is k lux at start minus k lux at end of reaction. Here are the

results: Graph Reaction

Finished in k lux at start k lux at end k lux dep./s DPPAS_02 70s 8.9 2.1 .097 DPPAS_03 80s 8.9 2.7 .078 DPPAS_04 100s 8.4 3.2 .052 DPPAS_05 N/A 8.3 N/A N/A ??????????? Analysis

/ Conclusion : Our

results show that, as predicted, the more concentrated solutions reacted more

quickly than the weaker ones. As the concentration got weaker, the reaction was

slower. I would expect the same pattern of you swapped Sodium Thiosulphate and

Hydrochloric Acid for two other chemicals, which are not affected by water, but

will react with each other. There were a few anomalies at the beginning of two of

our graphs, but the end results were all in proportion. The experiment shown in

DPPAS_02 twice as concentrated as the experiment in DPPAS_04. In theory, this

means that the light depreciation was twice as much. 0.052 times two is 0.104,

which is very close to the result we got of 0.097. This is assuming that the

result of 0.052 was correct. Giving leeway for inaccuracies, this was as good a

correlation as we could have expected ? only 0.008 k lux inaccurate. Also, as

you can see from the k lux/sec depreciation rate, I was not able to write up

figures for the last experiment. This was because, in the time we recorded the

experiment for, the chemicals never stopped reacting. As with most of the

graphs we never, for some reason, recorded a full 3m (all our graphs finished

at between 2m45s and 2m55s). This didn?t really matter, as all the other graphs

levelled out within the time recorded. But we stopped recording DPPAS_05 before

the reaction stopped, and as a result were not able to work out a finishing

time, and thus an overall rate of reaction. For this reason, I did not think it

fair to put in the results we got from the partial experiment into the main

table above. We can still however work out a rate for the amount of the

experiment that we did. Start light of 8.3, minus finish light of 4.6 is 3.7,

divided by time, which is 165, equals 0.022 k lux dep./s. Assuming this is the rate of reaction throughout the experiment, we

could times this number by four (as DPPAS_05 was four times weaker than

DPPAS_02), and end up with 0.088, which is only 0.009 k lux dep./s out. So by

correlation these three rates of reaction against concentration, we can see

that the results are related, and that the experiment worked well.? Evaluation : In all, our experiment worked quite

well. We did come across a few problems with accuracy though. One of the most

significant one was that the Dataloggers are not 100% accurate, and do not give

very detailed readings So when you take results into account, you have to give

some leeway for the inaccuracy of the Datalogger. ??????????? On DPPAS_05, we had a huge anomaly,

of nearly 4 k lux. This appeared about 20s into our experiment (according to

our graph). We have no idea how or why this happened. My best guess is that

somebody knocked the experiment and/or the Datalogger whilst it was recording. ??????????? I would do many things to improve

the experiment if I did it again in the future. Firstly, I would leave the

experiment for a long time (at least an hour) so that all the experiments would

finish. Second, I would do the experiment in a darkroom, and make the tube

extend upwards as well so that it covered the lamp, and all the light coming

out of the lamp would go through the beaker. If I could, I would also use a

more expensive and better quality Datalogger that could record the light more

precisely and accurately, but this would prove impractical for just one

experiment. ??????????? One thing I would be hard pushed to

overcome would be the light that passes down through the glass sides of the

beaker, via TIR. This is however not that important, as any light entering the

light sensor via this route would remain constant.

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