Ensure all equipment is thoroughly cleaned before using them in the next experiment- If the equipment is not cleaned properly, any impurities left in cylinders etc., this could affect the results. If different alkalis and acids are mixed together, then the correct effects are not being measured.
Ensure the same polystyrene cup is used – Different polystyrene cups may have different levels of insulation. This would mean that some experiments would experience less heat loss than others, causing comparisons to be unfair and incorrect.
Predictions
All neutralisation reactions have the same reaction, i.e. ACID + ALKALI ➔ WATER + SALT. This reaction can also be expressed as H + + OH - ➔ H2O. This information makes it evident that all neutralisation reactions should react in similar fashions.
I predict that during the reaction, the temperature will increase until the point of neutralisation (at which point the temperature is at its highest), and after this point, the temperature will start to decrease. This is because neutralisation is an exothermic reaction. The heat enthalpy of neutralisation is –57kJmol-1, and therefore this should be the value that is calculated at the end of each experiment. However, I predict that this will not be case, as, during the experiment, heat will be lost to the surrounding area, and the cup etc., hence the heat energy given out will be less. It will decrease because cooler acid is being added to the solution cooling it down.
All reactions will give out 57kJ/mol of energy. However not all the reactions will neutralise at the same point. I predict that Sulphuric Acid should neutralize at 25cm3 instead of 50cm3. This is because sulphuric acid is a dibasic acid, this means that it has 2 hydrogen ions attached to as part of its molecule as opposed to one. So half the amount of the acid can react with the alkali (the H+ ions are the ones that react) therefore the acid and alkali will neutralize at 25 cm3.
The line of the graphs will go up steadily until the point of neutralization when it will decrease steadily. The line will curve upwards due to heat loss, if heat loss did not occur during the experiment then the line would go straight up without curving.
Safety
In order to ensure the safety of myself and my classmates when performing these experiments I:
Wore safety goggles at all times to prevent and acid or alkali being splashed into my eye
Kept the beakers containing the acid and alkali in the middle of the desk to prevent spillage and in case spillage did occur it would not spill onto the floor.
I used an electric thermometer instead of a mercury one because as well as being more accurate there is a much lower chance of it smashing which if a mercury thermometer smashed toxic mercury would be released. This would not occur with an electronic thermometer.
When handling the chemicals used I took great care and caution
If a spill of a chemical occurred I would clean it up straight away with a wet paper towel.
Good ventilation was required in the laboratory, especially with the ammonia which is very potent, to do this we opened both doors to the lab.
Observation
I performed each experiment twice and then averaged out these results, this will authenticate my results and identify any anomalies if any have occurred.
Preliminary Experiment – HCl and NaOH
I decided to perform this experiment exactly as I plan to perform my main experiments. By doing this I should be able to identify any problems that may occur with my method. The heat of neutralization temperature should be 57kJ/mol however it will not be this due to heat loss. If the result I gain from my preliminary experiment is drastically different from this I know that there is something wrong with my method or calculation, which I shall have to rectify before my final experiments.
HCL and NaOH
Step 1 calculating the heat of neutralization
Amount of acid needed to neutralize alkali + the amount of alkali = 50 + 50 =100
100 x 4.18 x 12.6 (beginning temp – max temp reached)
=5266.8 J
Moles = 2.0 x 50/100 = 0.1
= 5266.8 J/0.1
=52668/1000
=52.668kJ/mol
Rounded up = 53kJ/mol
The results I have gained follow my predictions quite accurately. The heat of neutralization I have calculated is 53kJ/mol, which is quite close to the actual heat of neutralization. The graph shows a steady curve until the point of neutralization, this means heat has been lost. This supports my prediction that heat will be lost as if it were not it would be a straight line. The acid, alkali solution neutralized at 50cc as predicted for a monobasic combination. There were no anomalous results and the heat of neutralization I calculated was quite accurate if you consider heat loss, therefore I do not believe that I have to change anything when I perform my final experiment.
Final Experiments.
These experiments I carried out more carefully than my preliminary experiment. The acid/alkali combinations I have chosen will help me to support my prediction.
Experiment two-HCL and KOH
STRONG – STRONG COMBINATION
HCL and KOH
Heat of neutralization
100 x 4.18 x 12.7 = 5308.6 J
Moles = 2.0 x 50/1000 = 0.1
5308.6 J/0.1= 53086/1000 =53kJ/mol
Experiment 3-H2SO4 and NAOH
STRONG DIBASIC ACID WITH STRONG MONOBASIC ALKALI
Heat of Neutralization
80 x 4.18 x 15.9 = 5316.96J
Moles = 2.0 x 50/1000 = 0.1
5316.96J/0.1=53169.6/1000
=53kJ/mol
Experiment four-HNO3 and NAOH
STRONG MONOBASIC ACID WITH A STRONG DIBASIC ALKALI
HNO3 and NaOH
Heat of neutralization
100 x 4.18 x 12.6 = 5266.8J
Moles = 2.0 x 50/1000 = 0.1
5266.8J/0.1 = 52668/1000
=52.6kJ/mol
Rounded up = 53kJ/mol
Experiment five-CH3OOH and NAOH
WEAK ACID WITH A STRONG ALKALI
CH3OOH and NaOH
Heat of neutralization
100 x 4.18 x 12.2 = 5099.6J
Moles = 2.0 x 50/1000 = 0.1
5099.6J/0.1 = 50996/1000
51kJ/mol
Experiment six-HCL and NH3
STRONG ACID WITH A WEAK ALKALI
HCL and NH3
Heat of neutralization
100 x 4.18 x 12.3 = 5141.4J
Moles = 2.0 x 50/1000 = 0.1
5141/0.1 = 51414/1000 = 51kJ/mol
Analysis
These results and graphs show that overall my prediction was supported. All graphs show that the temperature steadily increase up until the point of neutralisation, before decreasing relatively sharply. The graphs also support my prediction that heat loss will occur during the experiment; hence the heat of neutralisation will be less than the theoretical value, as all the graph lines curve, as opposed to them being straight lines.
As I predicted all points of neutralization are where they should lie, at 50cm3 apart from sulphuric acid, which I found was neutralized at 30cm3.this, contradicts my prediction as I had predicted it would neutralize at 30cm3.The reason for this could be that incorrect amounts of acid were added to the solution, e.g. 8ccs instead of 10ccs added at a particular point.
All values gained were very close to the theoretical value of 57kJ/mol. All of the values I achieved were in a range of 6 of the actual value and were all below the theoretical value as predicted. This discrepancy between the values can be accounted for by heat loss.
The two results that were most off the theoretical value were the ones that involved a weak acid or alkali. Both of the values I gained were 51kJ/mol, this is two off the value I gained for all the other experiments.
In a strong acid there are more H+ ions, which are given more readily than in a weak acid i.e. not all the acid molecules of an acid form ions when in solution. This means that there are less H+ ions to react. Now, the theory of exothermic reactions states that when bonds are formed, energy is given out. It is this energy that we measure during this experiment. However, if there are less H+ ions to react, then fewer bonds are formed, hence less energy will be given out.
Additionally, if the bonds of the weak acid are broken, then much more energy will be taken in to break them than in a strong acid. This would cause some of the energy that is given out during bond-formation to be cancelled out. Therefore, even though the overall reaction is exothermic, less energy is being given out, as more of it is being used up to break the bonds of the weak acid.
. A weak alkali means that only some molecules form ions in solution. This means there are fewer ions to form bonds and give out energy, and therefore less energy still is given out.
Overall the results I obtained I graphs I have drawn all support the predictions I had previously made.
Evaluation
The experiments I performed I believe were carried out fairly and accurately. However they could be improved in the following ways:
When measuring out the acid to be added to the concentration more care is needed to ensure the volumes of the acid are accurate, as the acid needed to be added quickly to prevent further cooling of the solution, incorrect amounts were sometimes measured out which then gave an incorrect point of neutralization, (e.g. H2SO4 and NaOH).
A better insulator could be used to contain the mixture as the heat of neutralization figures that I calculated were lower than the theoretical value due to heat loss. In order to lower the amount of heat loss an enclosed insulator could be used to prevent the heat escaping. The only problem that this could pose would be that the acid would not be able to be added quickly enough to the solution.
I would like to perform another experiment combining a weak- weak combination, which would go further to help me prove my predictions were correct. I was not able to do this combination because there was not enough time.
An example of a neutralization reaction in every day life is the neutralization of a wasp sting. The venom that a wasp injects into the body is alkaline and can therefore be neutralized by an acidic substance e.g. vinegar a mildly acidic solution.
In this coursework I was guided by the following:
Letts revision textbook
Chemistry for GCSE textbook
www.chem.uidaho.edu/~honors/neutral.html
