25cm of the 1.0 mol dm Nitric acid added to 1.0 mol dm Sodium Hydroxide
25cm of the 0.5 mol dm Sulphuric acid added to 1.0 mol dm Sodium Hydroxide
From the graph I can write down the temperature rise and the volume of alkali needed to exactly neutralise the acid concerned. Below are the results,
For Hydrochloric Acid
Temperature rise = 26.50 c
Volume of sodium hydroxide needed to neutralise the HCL = 25.0 cm
For Nitric Acid
Temperature rise = 27.50 c Volume of sodium hydroxide needed to neutralise the HNO = 25.50 cm
For Sulphuric Acid
Temperature rise = 27.0 c
Volume of sodium hydroxide needed to neutralise the H SO = 25.0 cm
Analysis Experiment 1 :- 1.0 dm of HCL added to 1.0 dm Sodium Hydroxide.
Ionic Equation for neutralisation :- H+(aq) + OH-(aq) H O(l)
Equation :- HCL(aq) + NaOH(aq) NaCl(aq) + H O(l)
Mole Ratio:- 1:1
Heat Transfer = Mass x specific heat capacity x temperature change
Heat = m x c x _T
Specific heat capacity means the amount of energy required to raise the temperature of 1g substance by 1k. Water is 4.2jg k.
_T is temperature change which is the subtracting of the temperature before the experiment to the maximum temperature, this = temperature change.
M means mass, for this experiment total volume = mass. Total volume can be achieved by adding the volume required to neutralise the acid and the actual volume which is 25 cm
M = (25.5 + 25) = 50.5, mass
C = 4.2
T = (26.5 - 20) = 6.5c= 50.5 x 4.2 x 6.5 = -1378.65 heat given off (joules, j)
For 1 mole
Moles of HCL = volume/1000 x concentration
= 25/1000 x 1.0 = 0.025 moles
1 x 1 = 1 mole = 0.025 x 1 = 0.025 moles
Energy change per mole of HCL = -1378.65/0.025 = -55146/1000 = -55KJ
Experiment 2 :- 1.0 mol dm Nitric Acid added to 1.0 dm Sodium Hydroxide
Equation :- HNO (aq) + NaOH(aq) H O(l)
Mole Ratio :- 1:1
Heat Transfer = m x c x _T
M = (25.5 + 25) = 50.5
C = 4.2
T = (27.5 - 21.5) = 6
= 50.5 x 4.2 x 6 = -1277.6 J
For 1 Mole.
Moles of HNO = volume/1000 x concentration
= 25/1000 x 1.0 = 0.025 moles
1 x 1 = 1mole = 0.025 x 1 = 0.025 moles
Energy change per mole of HNO = -1277.6/0.025 = -51104/1000 = -51KJ
Experiment 3:- 0.5 dm , H SO added to 1.0 mol dm NaOH
Equation:- 0.5H SO (aq) + NaOH(aq) H 0(l)
Mole Ratio :- 0.5:1
Heat Transferred:- m x c x _T
M = (25 + 25) = 50
C = 4.2
T = (27 - 20.5) = 6.5
= 50 x 4.2 x 6.5 = -1365 J
For 1 Mole
Moles of H SO = volume/1000 x concentration
= 25/1000 x 0.5 = 0.0125 for 0.5 Moles
For 1 Mole
Ratio = 0.5 x 2 = 1mole
0.0125 x 2 = 0.025 moles = 1mole
-1365/0.025 = -54600/1000 = -55KJ, correct to 1 mole
From the results a trend or a pattern can be seen. The temperature of the mixture for all three acids rises and then falls again in the end.
Evaluation
The experimental errors, which could have affected my results, were inaccurate readings of the equipment such as the thermometer, this would have affected my final result. Each time I stirred the mixture, for each volume I may of stirred it differently for all the others, this would mean that more or less heat would of escaped, then required, which affect my thermometer readings. After doing the experiment with each acid, I cleaned the burette, pipette and the polystyrene cup, if I did not clean this properly then the equipment would have been contaminated when using the next acid, this would have affected my overall result as well. As this is an exothermic reaction heat is generated or absorbed in solution, therefore the extent to which heat is kept inside the cup, will affect my results, this is not helped by using a cup with no lid on as heat can easily escape.
As you can see the final neutralisation results per mole were, -55kj mol for HCL, -51kj mol for HNO and -55kj mol for H SO. Two of the results were the same while the other one was slightly different, this may be affected by the sources of errors listed above. I also believe the differences may also be due to the chemical behaviours for each acid.
The experiment generally went well, but problems aroused when heat was lost from the cup, ss there was no lid on, so the results are not accurate. Also, it was difficult, to keep the air bubbles out, of the burette and pipette.
The main improvements to the methods would be to use a cup with a lid on top, with a small hole, to keep the thermometer, this would achieve a more accurate result. To extend this experiment, other alkalis could be introduced.
This experiment has produced reasonably reliable results using the simple equipment. The conclusions which can be drawn from this experiment are that when a acid ie HCL, and an alkali ie NaOH, neutralise each other the reaction is relatively exothermic. This can be more clearly seen, by looking at the graphs. For example, for HCL, once the acid and alkali have neutralised at point 25cm, the temperature is 26.5 c, the starting temperature was 20.0 c. So there has been a rise of 6.5 c, until the neutralising point, so the reaction is exothermic. Once, the heat energy given to the mixture is used up, the temperature decreases, end temperature for HCL is 23 c. This example is a reflection of all three experiments.
