CH3CO2H CH3CO2 + H
This is a system in equilibrium – the forward and reverse reactions happen at the same rate. The molecules are always moving so this is a dynamic equilibrium. When the acetic acid is added top the NaOH during neutralisation, the H are removed (the H ions form the acid are being ‘counteracted’ by the OH ions released to ‘cancel each other out’) and the acid molecule therefore disassociates in order to replace the lost H ions. In this case, the CH3CO2H disassociates to produce more H ions and the reaction speeds up. A new equilibrium is set up. So although acetic acid is weak, it will eventually use the same amount of alkali, just in a longer time (for reasons mentioned above). (Other factors, which affect the way in which the equilibrium will move, are temperature and pressure. All reactions are endothermic at one end and exothermic at the other. If we raise the temperature, the endothermic reaction will increase to use up the extra heat. If we decrease the temperature the exothermic reaction will increase to give out more heat. Many reactions also have a greater volume on one side, either of products or reactants. If we raise the pressure if will encourage the reaction which produces less volume. If we decrease the pressure it will encourage the reaction that produces more volume).
I hypothesise that the colour change will happen very slowly with both acids but will then change immediately to pink i.e. not changing to all of the different colours on the PH range of colour changes. This is because when you have a strong acid and a weak base the colour changes very slowly until eventually you get a very large drop in PH. The PH will not change gradually according to the PH range of colour changes but will change rapidly from PH10 to PH3, meaning that the universal indicator will turn pink. When you have a weak acid and weak base (in this case NaOH and acetic acid), the PH changes from PH10 to PH3 i.e. universal indicator will change form blue to pink and not the other colours in between. You will not be able to see when the solution has actually neutralised. When acid is added to an alkali, a graph of PH changes looks like the one below. Depending on whether you have a strong or weak acid/alkali will determine the exact shape of the graph. They do however all follow a similar pattern. Ideally an indicator should change colour at the point marked x – the middle of the line i.e. in between an acid/alkali – neutral. This is not the case however, as has been explained above. (A weak acid and a weak base will not produce a straight line; there will be a gradual change. A strong acid and base will have a vertical line extending from PH10 and PH3. Strong acid and weak base will have a rapid PH change from 7 to 3 and a strong base and weak acid will produce a vertical line form PH10 to PH3).
Preliminary Results:
Colour Change:
Preliminary Conclusion/Evaluation: From my preliminary I have found out that the temperature change was the same for the same volume and molar of both acids added to the alkali. This indicates that the energy change would be the same for the same amount of moles, regardless of the acid or alkali. This would make sense because the same number of bonds are formed with the acid as with the alkali and therefore the same amount of heat energy would be necessary to break theses bonds. I am going to see whether this is true from the results in my investigation and also I am going to time how long each reaction takes to see if it takes longer for weak acids to neutralise than for strong acids. This is what I stated in my hypothesis. I can account for the fact that a lot more HCl needed to be added to neutralise the NaOH than CH3CO2H, by suggesting that perhaps the acid was in some way contaminated and therefore was not 100% HCl.
Heat is produced when the bonds in the product form. The temperature will not rise steadily because, as mentioned above, this is because at the beginning of the experiment there are lots of molecules to combine and no reactant has formed as yet. The temperature will increase greatly due to the large amounts of collisions. Towards the end of the experiment however there are fewer molecules (and more reactant) leaving less to react and therefore give off heat. So although heat is still being given off it is not in as large a quantity as it was towards the beginning of the experiment – the temperature will not increase so dramatically. I think that the product will have less energy than the reactant because it is an exothermic reaction. This means the energy released is more than the energy put in. The energy is transferred from the reactants to the air. The heat of reaction will therefore be negative because Heat of reaction = energy of products – energy of reactants. Seen as though the temperatures both before and after the experiment were the same for both acids, the amount of heat produced was also the same – 3 C. Heat of neutralisation is the heat absorbed when 1 mole of hydrogen ions neutralise 1 mole of hydroxide ions or in other words, how much energy is released for every mole of water made (water because H + OH → H2O). I know from my background knowledge that the heat energy released per mole of water for
HCL + NaOH → NaCl + H2O is –57.9KJ/mole. The minus simply means heat is released, as this is an exothermic reaction. (As heat = energy of products – energy of reactants and the energy in the products is less than that in the reactants because this is an exothermic reaction so energy is given off making less energy at the end). If we times this by the number of moles our product has we can therefore work out an expected energy rise. This is because rise in energy = heat energy x moles.
E.g. If we use 25ml of HCl and 25ml of NaOH, both of 1M then:
Concentration x Volume = moles
= 1 x (25/1000) = 25 x 10 ³ for both HCl and NaOH
because they are both the same concentration and volume.
As HCL + NaOH → NaCl + H2O we can see that if 25 x 10 ³ of acid is used, then
25 x 10 ³ of H2O is also produced. The predicted energy rise is therefore 25 x 10 ³ x
-57.9 (heat energy x moles) = 1.4475 KJ/mole. I therefore predict that the energy rise for the reaction with HCl and NaOH, both of 1M, will be -1.4475KJ/mole.
I will now predict the energy rises for all of the neutralisation reactions I intend to do:
-
HNO3 + NaOH → NaNO3 + H2O. 25ml of both acid and alkali will be used, both of 1M. Concentration x Volume = moles so 1 x (25/100) = moles
= 25 x 10 ³. This is the same for the H2O as 1 mole of HNO3 makes 1 mole of
H2O. Predicted energy change is therefore 25 x 10 ³ x –57.6 =
- 1.44KJ/mole.
-
CH3CO2H + NaOH → CH3C02Na + H2O. 25ml of both acid and alkali will be used, both of 1M. Concentration x Volume = moles so 1 x (25/100) = moles = 25 x 10 ³. This is the same for the H2O as 1 mole of HNO3 makes mole of H2O. Predicted energy change is therefore 25 x 10 ³ x –56.1 =
-1.4025KJ/mole.
These predictions show that the stronger the acid used in neutralisation, the more energy is released per mole of water. I will be able to compare these predicted energy changes with my actual results after my investigation. When I have my results – the temperature change from beginning to the end of the experiment - I can use the equation Energy change = mass of solution x temperature change x specific heat capacity of water to work out the actual energy change. In neutralisation we assume that the density of the acid/alkali is the same as that of water, so volume of acid = mass of acid. We know that the specific heat capacity of H2O is always 4.2 so the equation will end up as 50 (25ml of acid + 25ml of alkali which is equivalent to 50cm³) x temperature change x 4.2. As the mass and specific heat capacity of water will stay constant and the temperature change will increase by the same amount regardless of the strength of acid (it will just take longer), the energy change each time will be the same. This theory is called the Standard Enthalpy (or heat) of neutralisation. As can be seen above, the energy changes for each of the acids rounds up to the same amount of energy per mole of water - -1.4, thus supporting the Standard Enthalpy Theory.
As I now have a predicted energy change for each equation I can now predict, using this result and the equation ∆H = volume x temperature change x specific heat capacity (sph), the temperature change of the different strength acids upon neutralising the NaOH. I can do this by rearranging the above formula to: ∆H/(volume x S.H.C) = temperature change.
-
For HCl +NaOH: -1.4475/(50 x 4.2) = 7 C
-
For HNO3 + NaOH: -1.44/(50 x 4.2) = 7 C
-
For CH3CO2H + NaOH: -1.4025/(50 x 4.2) = 7 C
All of the predicted energy changes are the same suggesting that, regardless of the strength of the acid, the eventual temperature rise will always be the same. This supports my results from my preliminary experiment. Although the temperature changes in my preliminary were slightly less – 3 C as opposed to 7 C – I can account for this by saying that a small amount of heat could have escaped as I used a beaker instead of a polystyrene cup i.e. I did not use an insulator. I will use a polystyrene cup in my actual investigation so hopefully the temperature change will be as predicted. The only difference with the strong and weak acids should be that the time taken for the reaction to take place is much less for the stronger acid i.e. the stronger the acid, the quicker the reaction. I will use data logging equipment to show this as the temperature will be accurately recorded every 5 seconds and I will be able to see when the experiment has finished from the straight line visible on the graph drawn by the data logging equipment.
Main Method: As above but this time I will use a wider variety of both weak and strong acids. As well HCl and acetic acid, I will also use HNO3. I will repeat each experiment three times as a measure of accuracy and make sure to again use burette and pipette filler, washed out after each experiment, as an added measure of accuracy. A burette and pipette filer are both more accurate than a measuring cylinder because they are correct to roughly 0.05ml. I will therefore be able to measure the amount of acid and alkali needed, as accurately ads possible. This time I will also use a polystyrene cup to hold the acid and alkali mixture. This is because polystyrene is an insulator and so less heat will be lost making my experiment more accurate. To record the temperature changes I will use data logging equipment to record the temperature every 10 seconds. This way I will not only know what the temperature change was (more accurately then by recording it myself) but I will also be able to see how long the whole experiment took with each individual acid because when temperature remains constant (this will be visible on the graph drawn up by the data logging equipment), this signals the end of the reaction. I can then see whether my prediction is correct that the stronger the acid the faster the reaction will be. Also, data logging is safer to record the temperature incase the temperature gets too high and snaps. I will use Universal Indicator to test the PH, as opposed to litmus paper or methyl orange because it changes colour the closest to PH7 (at PH8), which is the most accurate with regards to a neutralisation reaction because we want the closest possible to PH7.
Results:
For HCl of 1m reacting with 25ml of 1M NaOH:
Temperature Changes:
Colour Changes:
For HNO3 of 1M reacting with 25ml of 1M NaOH:
Temperature Changes:
Colour Change:
Analysis:
See graphs
I can see from my results that the reaction with HCl acid and NAOH took the same amount of time to finish as the experiment with HNO3 and NaOH. I can also tell from my results that there was a greater temperature change with the reaction between HCl acid and NAOH than there was between HNO3 and NaOH. According to my results and calculations based on my background knowledge and my preliminary results, these results are wrong. I used the same molar (1M) of both acids. This would mean that the same number of bonds are formed with the acid as with the alkali and therefore the same amount of heat energy would be necessary to break these bonds, meaning that the temperature change should be the same. My hypothesis that the energy changes would be the same is therefore incorrect.
My second hypothesis was that the strength of the acid would have no effect on the energy change. This is because I am not changing the molarity of the acid and so there are no extra molecules floating about to increase the time taken to break the bonds. My results do show this: the average time taken for the acids to neutralise the NaOH was exactly the same – 32 seconds, which shows that the strength of the acid did not have any effect.
A lot more HNO3 was needed then HCl, to neutralise the NaOH. This should not have been the case: equal amounts should have been sufficient going by the above hypothesis. I can account for this by saying that the HNO3 could have been contaminated. This is because I had to mix up the HNO3 myself. There was no 1M left and so I therefore had to dilute an equal amount of 2M HNO3 and use this. I may not have measured accurately and may have come across some paralex error, which meant that the HNO3 was not actually 1Mbut slightly less (meaning it would take longer). The same could apply to the HCL. I may not have diluted it enough, meaning that it was actually higher then 1M and therefore reacted quicker than it should have if it was just 1M.
Evaluation: My experiment was very inaccurate and therefore did not fit in entirely with what I predicted. I did not follow my method in the sense that I did not use three different strengths of acid. I only used two – HCL and HNO3 and not CH3CO2H. This was due to lack of time but because of this I am now not able to use my results because they do not show me a great enough comparison between strong and weak acids. I did not stir the solution between each titration, which meant that the colour could have changed a lot quicker but as I did not stir it I did not realise this. A lot more acid was used then was necessary in order to make the colour strong enough for me to see. This makes my results inaccurate. If I were to improve my experiment I would have to stir the solution of acid and alkali in between each 5mlk titration.
I diluted the acids myself. This was because there was no 1M acid left and so I had to take the 2M and dilute it accordingly. I may have been inaccurate whilst diluting it and did not take into account paralex error. I could therefore have ended up with either a higher or lower concentration of acid than previously thought. This also makes my experiment and results inaccurate and probably played a big part in why my results were so inaccurate. I also had to dilute my own NaOH and the same errors could have occurred. By diluting my own acid and alkali meant that I could not use a pipette filler or burette for measuring. The purpose of using this equipment was to make my measuring more accurate. I could not use them and therefore my measurements were not as accurate as they could have been.
The prongs of the data logging equipment were not washed between each titration. This means that some acid could have still been present in between each experiment and could have started to neutralise the alkali before the experiment was actually began and data logged.
If I were to do this experiment again I would not only change what is written above but I would use a third acid (as I had planned to do).
