A neutralisation reaction is a reaction between an acid and a base in which the acidity or alkalinity of a substance is destroyed, ie, either removing H+ ions by reaction with a carbonate or metal base, or removing OH- ions by reaction with an acid. To get a perfectly neutral solution of pH7, the H+ and OH- ions must be in exactly equal amounts.
The ionic equation for neutralisation is:
H+ + OH- H2O
Heat of neutralisation or enthalpy change of neutralisation is the amount of heat energy given out when one mole of hydrogen ions is neutralised by one mole of hydroxide ions.
POSSIBLE INPUT VARIABLES:
* Concentration of acid
* Concentration of alkali
* strength (pH) of acid
* strength (pH) of alkali
* volume of acid
* volume of alkali
* basicity of acid (the no of H+ ions that can be replaced to form a salt)
I know that the concentration of the reagents affects the heat of neutralisation, however this is not as advanced an experiment, and would not give me much to investigate/analyse and so I shall not choose this variable.
The basicity of the acid affects the heat of neutralisation too, however I shall not investigate this variable because I am told that sometimes the results for this can be unreliable. Also, as there are not many acids available to me, I may not be able to get a wide enough range of measurements to make my investigation worthwhile.
The volume of the reagents also effects the heat of neutralisation, however I shall not choose this variable as again the investigation would not be so advanced, or give me as much to investigate/ analyse as my chosen variable, nor do I find it as interesting.
My chosen investigation is that into how the strength (pH) of the reagents affects heat of neutralisation.
From past experiments I know that neutralisation is an exothermic reaction, i.e. the energy released in bond formation is greater than the energy taken in and used up in breaking bonds. Because of this the DH will be negative. Therefore I predict that when reacting an acid with an alkali in a neutralisation reaction, energy will be given to the surroundings as heat and so there will be a rise in temperature. From book research I know that if the acid and alkali are fully ionised (there H+ and OH- ions are fully dissociated), the heat of neutralisation is always – 57 KJ. Therefore this should be the heat of neutralisation attained from my experiments with fully ionised reagents, however it may well be slightly less as some heat is bound to be lost to the surroundings. When a weak acid or base is involved however, I predict that the heat produced will be less. This is because some energy must be supplied to ionise the acid or alkali fully first (before the ions can react and neutralisation can occur), and so the overall heat given out over the heat taken in will be reduced.
* From this I predict that the highest DH will be reached with the two fully ionised reagents (-57KJ)
* The next highest will be reached by a fully ionised reagent with a partially ionised reagent.
* The lowest DH will be reached by two weak (only partially ionised) reagents.
Of the two weak acids (ethanoic and propanoic ) I predict that the propanoic will create a lower heat of neutralisation, as it has a longer chain length and is less soluble in water, and from research I believe that the longer the chain length, the less ionised the acid.
TO MAKE MY EXPERIMENT A FAIR TEST I SHALL:
* Make sure that all other input variables are kept constant, ie:
1. I shall use the same concentration for all the reagents- 2M solutions of everything.
2. I shall use equal volumes of acid and alkali – 30cm3
3. I shall use only monoprotic/ monobasic acids.
* Make sure that the temperature has reached its maximum before I record it- I shall wait for it to stay constant for some while or start to drop again, and then I shall see the true maximum temperature.
* I shall stir all the experiments equally.
* Repeat all experiments a further two times to give a chance for anomalous or inaccurate results to show up.
* When stirring the solution, I must be careful not to put my hands around the polystyrene cup, as this would cause the temperature to raise falsely from my body heat. Instead I shall hold it at the top.
TO MAKE MY EXPERIMENT SAFE I SHALL:
* Wear an overall securely tied back throughout the entire investigation.
* Wear goggles throughout the entire investigation
* Wash my hands carefully and thoroughly after the experiments (working with acids and alkalis) -sodium hydroxide particularly is corrosive, and hydrochloric acid and the carboxylic acids are irritants.
* Ensure that adequate ventilation is in place- some chemicals such as ammonia solution have very potent smells
* Handle all glassware with care- broken glass is a hazard, and also it may be hot after the experiments.
* Handle the thermometer with particular care as it contains mercury, which is poisonous.
THE EQUIPMENT AND CHEMICALS I SHALL HAVE AVAILABLE AND USE ARE:
* One 500 ml beaker, for housing the polystyrene cup
* Two 50cm3 measuring cylinders (one for the acid, one for the alkali)
* One polystyrene cup with lid
* Two 100ml beakers for storing the chemicals in use
* One thermometer
* 2 pipettes (one for the acid, one for the alkali)
* Labels for the beakers, measuring cylinders and pipettes- this will ensure that chemicals do not get mixed up.
* 3 Acids- Hydrochloric (strong), ethanoic and propanoic (weak).
* 2 alkalis- Sodium hydroxide (strong) and Ammonia solution (weak)
OVERVIEW OF METHOD
* I shall perform this experiment by reacting different strengths of reagents in neutralisation reactions. I am using 1 strong acid and two weaker ones, 1 strong alkali and 1 weaker one. I am going to look at the various energy changes when reacting a strong alkali with a strong acid, a strong alkali with two different weak acids, a weak alkali with a strong acid and a weak alkali with two different weak acids. (this means their pHs will all be different).
* At the end of my experiment I should be able to tell which of the two weaker acids is the weakest; (whichever creates the lowest heat of neutralisation when reacting with a fully ionised alkali).
* I shall measure out the acid and alkali to be used and react them in a polystyrene cup within a beaker. I know that polystyrene is an insulator, in preliminary work it helped to get more accurate heats of neutralisation by not letting too much heat escape. However having said that I expect that some heat will be lost to the surroundings, and so the heats of neutralisation I attain will probably be lower than those found in a data book.
* I also know from preliminary work that the various different acids and alkalis of different strengths take different times to reach there maximum temperatures- therefore I will not need to put any time restraints or timings on the experiments, as this would cause inaccurate results.
* I know that the maximum temperature change is the heat of neutralisation for the reaction because, as I said before, heat of neutralisation is the amount of heat energy given out when one mole of hydrogen ions is neutralised by one mole of hydroxide ions. This “amount of heat energy” will be the maximum change in temperature that occurs during the neutralisation reaction.
* Also when measuring chemicals I should use a pipette for accuracy- labelled for acid + alkali to avoid contamination.
* My outcome variable will be the maximum temperature reached.
* I have decided to do each experiment 3 times to make sure that any anomalous or inaccurate results will have a chance to show up, as I will have 3 sets of data to compare. Also I think I have chosen a reasonable range of experiments to conduct (6 different ones in total) which I hope will give me a real pattern of results.
Plan Of Results Table -table 1
( ethanoic acid)
NaOH (sodium hydroxide)
Initial temp 0C
Final temp 0C
Temp change 0C
NH4OH (ammonia solution)
Initial temp 0 C
Final temp 0 C
Temp change 0 C
* In previous experiments I have used a calculation to find the heat of combustion, and I can amend this calculation to find the heat of neutralisation.
The calculation that must be used is DH= MC , where M= mass, C= specific heat capacity, = change in temperature.
* I also know form previous experiments that when using dilute solutions, one can take the specific heat capacity and density to be the same as water. The specific heat capacity for water is 4.2, and as the mass will equal the volume (true for water) this will be 60cm3 – (the combined acid and alkali volumes).
* so MC = 60 * 4.2 * temp change Joules
* To get my answer in KJ I must divide by 1000.
* This will give me the total KJ of energy for my experiment, and I wish to know the KJmol-1 .To find this I must divide by the number of moles for one of the reagents.
* To find the no. of moles of alkali:
volume (dm3 ) * concentration = no. of moles.
=0.03 * 2 = 0.06
so KJmol-1 = – (MC / 1000 / 0.06). (this is negative as the reaction is exothermic)
1. Set up apparatus as shown below (labelled appropriately):
2. Fill the two labelled beakers with the appropriate substances (working through in order of results table)- so firstly with HCl and NaOH.
3. Measure very carefully 30cm3 NaOH into the polystyrene cup using the measuring cylinder and pipette, both labelled “alkali”.
4. Measure and record the initial temperature of the solution
5. Using the other measuring cylinder and pipette, both labelled “acid”, measure 30cm 3 of HCl very accurately in the measuring cylinder.
6. Pour the acid into the cup to join the NaOH, and immediately push down the lid and stir gently. Make sure you only hold the cup at the very top.
7. Wait until the maximum temperature has been reached, and when it shows no signs of rising any further, record it.
8. Wash out the cup, thermometer and measuring cylinders (just in case they have been contaminated) and repeat the experiment a further 2 times.
9. Use the same method for conducting all the other experiments (using the different combinations of acids and alkalis), ensuring that all beakers and equipment is washed in between changing the chemicals they contain.
10. Using the calculation method explained in the method overview, calculate the heat of neutralisation to one decimal place for all sets of data, and then work out an average for each pair of chemicals reacting.
Due to unavailability of chemicals I was unable to use 2M solutions as stated in the plan. Instead I used 1M solutions for all chemicals. This should not affect my investigation as long as it is taken into account when calculating the heat of neutralisations, and the calculations are altered accordingly
( ethanoic acid)
NaOH (sodium hydroxide)
Initial temp 0C
Final temp 0C
Temp change 0C
NH4OH (ammonia solution)
Initial temp 0 C
Final temp 0 C
Temp change 0 C
From my results I can now calculate the heat of neutralisation for each experiment. NB. the no. moles of alkali used is now 0.03 as the solution is 1M.
For HCl and NaOH: Heat of neutralisation= -( MC / 1000 / no of moles of alkali)
= 1. – (60 * 4.2 * 6.8 /1000/ 0.03)
2. – (60 * 4.2 * 6.7 /1000/0.03)
3. – (60 * 4.2 * 6.6 /1000/ 0.03)
= 1. -57.1 KJmol-1
2. -56.3 KJmol-1
3. -55.4 KJmol-1
To find an average of these figures: (57.1 + 56.3 + 55.4) / 3 = – 56.3 KJmol-1
For Ethanoic acid and NaOH, heat of neutralisation= – (MC /1000/no of moles of alkali.)
= 1. – (60 * 4.2 * 6.4 /1000 /0.03)
2. – (60 * 4.2 * 6.3 /1000 / 0.03)
3. – (60 * 4.2 * 6.4 / 1000 /0.03)
= 1. -53.8 KJmol-1
2. -52.9 KJmol-1
3. -53.8 KJmol-1
To find an average of these figures: -(53.8 + 52.9 + 53.8) / 3
= -53.5 KJmol-1
For propanoic acid and NaOH, heat of neutralisation= – (MC /1000 / 0.03)
=1. – (60 * 4.2 * 5.6 /1000 / 0.03)
2. – (60 * 4.2 * 5.5 / 1000/ 0.03)
3. – (60 * 4.2 * 5.5 / 1000/ 0.03)
=1. -47.0 KJmol-1
2. -46.2 KJmol-1
3. -46.2 KJmol-1
To find an average of these figures: – (47.0 + 46.2 + 46.2) /3
= -46.5 KJmol-1
For HCl and ammonia solution, heat of neutaralisation= -(MC /1000/ 0.03)
=1. – ( 60 * 4.2 * 6.5 /1000 / 0.03)
2. – (60 * 4.2 * 6.4 / 1000 / 0.03)
3. – (60 * 4.2 * 6.4 / 1000 / 0.03)
= 1. – 54.6 KJmol-1
2. -53.8 KJmol-1
3. -53.8 KJmol-1
To find the average of these calculations: – (54.6 + 53.8 + 53.8) / 3
= -54.1 KJmol-1
For ethanoic acid and ammonia solution, heat of neutralisation = – (MC /1000 / 0.03)
=1. – (60 * 4.2 *6.2 /1000 /0.03)
2. – (60 * 4.2 * 6.0 /1000 /0.03)
3. – (60 *4.2 *5.9 / 1000 / 0.03)
= 1. -52.1 KJmol-1
2. -50.4 KJmol-1
3. -49.6 KJmol-1
To find the average of these figures: – (52.1 + 50.4 + 49.6) / 3
= -50.7 KJmol-1
Finally for propanoic acid and ammonia solution, heat of neutralisation = – ( MC /1000/0.03)
=1 – (60 * 4.2 *4.9 /1000 /0.03)
2. – (60 * 4.2 * 5.1/1000 /0.03)
3. – (60 * 4.2 * 5.3 /1000 / 0.03)
= 1. – 41.2 KJmol-1
2. -42.8 KJmol-1
3. -44.5 KJmol-1
To find an average of these figures: – (41.2 + 42.8 + 44.5) /3
= – 42.8 KJmol-1
Table 3- showing average heats of neutralisation for each pair of chemicals, to extend original findings
From these results we can see that they do indeed follow the pattern that I outlined in my predicition, and as it is supported it is likely to be correct. The temperatures in all my reactions went up and so all the reactions were exothermic, as I predicted. The Highest heat of neutralisation was reached by the HCl and NaOH (two fully ionised reagents) which was -56.3 27 KJmol-1 only slightly below that of the data book,
– 57 KJmol-1 , which is what I predicted would happen.
Also I think that ethanoic acid and ammonia solution are likely to be of similar strength as the results obtained when these two chemicals both reacted with fully ionised counterparts (which act as controls ) were very similar.
I think that the weakest acid is the propanoic acid as this reached the lowest heat of neutralisation when reacting with the fully ionised acid (again acting as a control). After checking in a data book, I saw this was correct. I think this is because it has a longer chain length than ethanoic acid, and is less soluble, and therefore has fewer dissociated hydrogen atoms, resulting in more energy being taken in by the reactant to fully ionsise, and so a lower heat of neutralisation.
My results do clearly show that the weaker the reagents involved the lower the heat of neutralisation. I think this is because it is the H+ and OH- ions that react. Because of this weaker reagents must use up energy to fully ionise themselves before taking part in the neutralisation reaction. As breaking bonds uses up energy (taken in from the surroundings) this means that overall less energy will be given out to the environment in the neutralisation reaction. Although all the reactions were overall exothermic, some are more than others, ie. the stronger reagents will react more readily, giving off more energy overall and being more exothermic.
I think overall my experiment was very successful as I achieved the results I predicted I would gain, and I do not seem to have any anomalous results. My experiments were fair and safe as I followed all the guidelines I set in my plan, and I believe it was a relaible method
After consulting a data book I realised that, as I had predicted, my results were not 100% accurate as my equipment was not terribly advanced, and so some heat was lost to the surroundings. However this occured by equal proportions in each experiment, and so although my results were not 100% reliable, they were very accurate relatively speaking. However there are some improvements I could make for if I ever conduct this experiment again:
* Make sure I get very precise volumes as this is needed for accurate results. Perhaps I could be even more careful, using the pipette more skilfully
* Perhaps I could repeat my experiments one more time to make sure that my readings are accurate as I will have more data to average, and compare.
* I could use more advanced equipment to avoid heat loss such as a calorimiter. This is the most important area for improvement, as this is what caused my results to be slightly below those in the data book, resulting in slight inaccuracies.
* Next time I could revert to my original method and use 2M chemicals, although my experiments did seem to be unaltered by the adjustment to 1M chemicals.
Also there are a few ways in which I could extend my investigation:
* Investigate some of the other variables listed in my plan, and see the effects they have on heat of neutralisation.
* Take my current investigation further by investigating a wider range of weaker carboxylic acids, and seeing whether the pattern of lower heats of neutralisation with longer chain lengths continues. This would be very interesting and seems to me a very worthwhile extension
* Do more research and book work to find out exactly why the carboxylic acids do get weaker with longer chain lengths.
Even though my results may not have been exactly those in the data book, they were only slightly below, and all relative to one another (ie. they were lower by the same amounts proportionally). This indicates that it was the heat lost to the surroundings causing this inaccuracy, which was the same in each case, as the same equipment was being used in each case. At least this indicates that my experiment was a fair test. Altogether I think my results are relevant and noticeably >90% accurate/reliable.