I will then repeat this experiment but using the same solutions at different concentrations.
I will make this experiment safe as both the acid and the base solutions are corrosive and should be handled with care. I will wash off any acid or base that gets on my skin or clothing. When I will use a pipette I shall use a rubber bung. I will not stir the solution with the thermo probe; so I will use a separate stirrer. I will also wear safety goggles.
Prediction
When I mix the acid with the alkali, a neutralisation reaction takes place. This reaction is exothermic and gives out heat. This temperature rises. If you add a little bit of acid at a time to a known volume and concentration of alkali, I can plot a graph of temperature rise against volume of acid added. At some point, the temperature no longer rises as all the alkali OH- has neutralised the H+ ions of acid. The temperature then begins to fall as you dilute your mixture with cold acid. The point of inflection tells me the end point.
Using stronger concentrations of acid I predict more heat will be released and so the maximum temperature will be higher.
Both the acid and alkali molecules are dissolved in the water that means that the H+ and the OH- ions are swirling around freely mixed in with many water molecules. In order to react and make the heat I am measuring, the acid H+ have to bump into the alkali OH-.
The stronger the concentration of the acid, the more acid molecules will be in any particular part of the solution. Therefore it is more likely that they will bump into the alkali molecules, cause more reactions to happen and more heat released to measure.
HCl(aq) + NaOH(aq) → NaCl + H2O(l)
Acid H+(aq) + alkaline OH-(aq) → H2O(l)-neutral
The heat of the reaction can be worked out by multiplying the total mass of solution by the specific heat capacity and change in temperature.
Heat of reaction =
mass x specific heat capacity x temperature change
The specific heat capacity of a substance is the amount of energy (in joules) that is needed to raise the temperature of 1kg of the substance by 10C.
I will assume that the heat capacity of the mixture is the same as of water, 4200J/kg0C.
For my equation I need to know the mass but I only know the volume. Assuming the density of our solution is the same as water, 1000kg/m3=1g/cm3.
I shall now work out the number of moles in 25cm3 of 2 moles/dm3 of sodium hydroxide.
Number of moles in a solution
= Molarity x volume of solution (cm3)/1000
=2x25/1000
=0.05 moles
I will need to work out how many moles of water have formed from the solutions. If I haven’t then I have just measured the heat of neutralisation for a given mixture and the heat change would increase if the molarity would increase and the volumes remain the same.
If I can calculate the number of moles of water produced then I found the energy change for producing one mole of water. Then the heat of neutralisation will not change when using more concentrated acids and alkalis. This is called the standard enthalpy (or heat) of neutralisation.
The reaction between a strong acid and a strong base is the same irrespective of the acid or base used. If you look at the ionic equations they only include H+ and OH-.
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
Na+(aq)+OH-(aq)+H+(aq)+Cl-(aq) → Na+(aq)+Cl-(aq)+H2O(l)
From the above equation we can see that the sodium and the chlorine ions are present before and present after.
So the real reaction is:
OH-(aq) + H+(aq) → H2O(l)
If I used different acids or alkalis in the neutralisation reaction, enthalpy of neutralisation would be different as well.
Enthalpy is a measure of energy needed to break bonds and make new bonds. Different types of bonds need different amounts of energy and because different acids have different types of bonds in them they will need to have different amounts of energy to break them. This will affect the overall enthalpy of neutralisation.
Obtaining evidence
0.1 moles/dm3 of HCl and NaOH
2 moles/dm3 of HCl and NaOH
Analysis
When I mixed the acid with the alkali, a neutralisation reaction took place. This reaction was exothermic as heat was given out. The temperature went up. I added a little bit of acid at a time to a known volume and concentration of alkali. I plotted a graph of temperature against volume of acid added. At the neutralisation point, the temperature no longer increased as all the alkali OH- had neutralised the H+ ions of acid and formed neutral water. The temperature then began to fall as I diluted the mixture with cold acid. The point of inflection told me the end point.
In 5 out of the 6 experiments, the maximum temperature was reached when 25ml of acid was added. In the last experiment the maximum temperature was reached when 30ml of acid was added.
In the calculations to follow I will assume that in both experiments 25ml of acid was needed to neutralise 25ml of alkali. In the first experiment it was 25ml of 0.1 moles/dm3 of HCl and NaOH and in the second experiment 25ml of 2 moles/dm3 of HCl and NaOH.
In the above graph it can be seen more clearly in the second experiment that the point of inflection is at 25ml.
In the first experiment by redrawing the above graph with a different scale with the temperature change measured from 18.50C to 200C you would see an obvious point of inflection when 25ml of 0.1 moles/dm3 HCl was added.
Calculation to work out the enthalpy of neutralisation of:
HCl(aq) + NaOH(aq) → NaCl + H2O(l)
Using the results from the second experiment as a larger temperature change was noticed in this experiment.
25ml of 2 moles/dm3 HCl neutralised 25ml of 2 moles/dm3 NaOH with an average maximum temperature rise of 9.70C.
Heat of reaction =
Mass x specific heat capacity x temperature change
We assume that 25ml of solution has a mass of 25g.
25g x 4200J/kg0C x (29.83-20.13)
=0.025kg x 4200J/kg0C x (29.83-20.13)
=0.025kg x 4200J/kg0C x 9.70C
=1018.5J for 25ml of 2 moles/dm3 hydrochloric acid.
Moles of acid used = volume x concentration
Where volume is in litres and concentration is in moles/dm3.
=0.025 x 2 = 0.05 moles
1 mole of acid reacts with 1 mole of alkali.
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
The number of moles of NaOH = 0.05 moles
To work out the heat of reaction/mol.
I know that 1018.5J of heat energy is released for 25ml of 2 moles/dm3 hydrochloric acid which we now know contains 0.05 moles of NaOH.
0.05 moles produces 1018.5J
1 mole produces 1018.5/0.05=1018.5 x 20=20370J
=20.37kJ
So the heat of neutralisation per mol
=20370J/mol
=20.37kJ/mol
Looking in the book “Chemistry Counts” by Graham Hill on page 190 he states that when hydrochloric acid and sodium hydroxide react the heat loss is 57.9kJ/mol of each reactant. That is worked out using the best scientific equipment. My result of 20.37kJ/mol is less than half this value but is reasonable given the limits of this experiment.
Using the results from the first experiment.
25ml of 0.1 moles/dm3 HCl neutralised 25ml of 0.1 moles/dm3 NaOH with an average maximum temperature rise of 0.60C.
Heat of reaction =
Mass x specific heat capacity x temperature change
We assume that 25ml of solution has a mass of 25g.
25g x 4200J/kg0C x (21.2-20.6)
=0.025kg x 4200J/kg0C x (21.2-20.6)
=0.025kg x 4200J/kg0C x 0.60C
=63J for 25ml of 0.1 moles/dm3 hydrochloric acid.
Moles of acid used = volume x concentration
Where volume is in litres and concentration is in moles/dm3.
=0.025 x 0.1 = 0.0025 moles
1 mole of acid reacts with 1 mole of alkali.
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
The number of moles of NaOH = 0.0025 moles
To work out the heat of reaction/mol.
I know that 63J of heat energy is released for 25ml of 0.1 moles/dm3 hydrochloric acid which we now know contains 0.0025 moles of NaOH.
0.0025 moles produces 63J
1 mole produces 63/0.0025=63 x 400=25200J
=25.2kJ
So the heat of neutralisation per mol
=25200J/mol
=25.2kJ/mol
Evaluation
In this experiment the reaction took place in a polystyrene cup, as it’s a good thermal insulator however it had no lid. A calorimeter would have been a better insulator.
The improvements I would make by using a calorimeter are that it has a lid, is better insulator and has a mechanical stirrer.
The results of the 2 experiments using the average temperature change both give results just under half of the known value of 57.9kJ/mol.
The experiments were done on more than one day. Looking at first experiment when the temperature started at 18.60C and went up to 19.60C there was a rise in temperature of 10C.
Using only the results of this one experiment I get a much closer result to 57.9kJ/mol.
25ml of 0.1 moles/dm3 HCl neutralised 25ml of 0.1 moles/dm3 NaOH with a maximum temperature rise of 10C.
Heat of reaction =
Mass x specific heat capacity x temperature change
We assume that 25ml of solution has a mass of 25g.
25g x 4200J/kg0C x (19.6-18.6)
=0.025kg x 4200J/kg0C x (19.6-18.6)
=0.025kg x 4200J/kg0C x 10C
=105J for 25ml of 0.1 moles/dm3 hydrochloric acid.
Moles of acid used = volume x concentration
Where volume is in litres and concentration is in moles/dm3.
=0.025 x 0.1 = 0.0025 moles
1 mole of acid reacts with 1 mole of alkali.
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
The number of moles of NaOH = 0.0025 moles
To work out the heat of reaction/mol.
I know that 105J of heat energy is released for 25ml of 0.1 moles/dm3 hydrochloric acid which we now know contains 0.0025 moles of NaOH.
0.0025 moles produces 105J
1 mole produces 105/0.0025=105 x 400=42000J
=42kJ
So the heat of neutralisation per mol
=42000J/mol
=42kJ/mol
The other 2 experiments using 0.1 moles/dm3 were done on a much warmer day so I didn’t notice such a big temperature rise.
When adding the acid in all experiments I let in 5ml of HCl and then measured the temperature. Maybe adding 2ml at a time and measuring the temperature rise would have given me a more accurate experiment.
When adding the 5ml of HCl each time this should have been done at equal time intervals and quickly because if a large amount of time was left before more acid was added the temperature of the reactants would cool down.
To make the experiment better I would have allowed the acid to run in to the alkali at a constant rate and get the computer to plot a graph of time against temperature change. As I know the time it takes for all the acid to run into the alkali I could then turn that graph into volume of acid against temperature change. This would be more accurate.
This experiment could be repeated using different concentrations of the same acids or using other acids and alkali at different concentrations.
When using different acids and alkalis it is important to workout how many moles of water are formed. If you don’t then you can just measure the heat of neutralisation for a given mixture and the heat change would increase if the morality increased and the volumes remained the same.
If you do know the number of moles of water produced you can find the energy change for producing 1 mole of water. In this case the heat of neutralisation will not change if you use more concentrated acids and alkalis. This is then called the Standard Enthalpy (or Heat) Of Neutralisation.
The reaction between a strong acid and a strong base is the same irrespective of the acid or base used. If you look at the ionic equations they only include H+ and OH-.
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
Na+(aq)+OH-(aq)+H+(aq)+Cl-(aq) → Na+(aq)+Cl-(aq)+H2O(l)
From the above equation we can see that the sodium and the chlorine ions are present before and present after.
So the real reaction is:
OH-(aq) + H+(aq) → H2O(l)