Indirect thermometric Titration Essay
Sorry, but copying text is forbidden on this website!
* School Name: Al Mashrek International School
* School Code: 2108
* Subject: Chemistry
* Topic: Indirect A thermometric Titration.
* Assessment: Data Collection, Data Processing & Presenting, Conclusion & Evaluation.
* Candidate Name: Bassam Al-Nawaiseh
* Date: 20/5/2007
The aim of this experiment is to determine the concentrations of two acids. The two acids are Hydrochloric acid, HCl, and Ethanoic acid, CH3CO2H. This will be done by thermometric titration, by calculating the enthalpy change for each reaction, enthalpy of neutralization.
* Data Collection:
Table 1: the temperature change for the HCl solution and CH3CO2H solution after adding 5 cmï¿½ portions of 1M NaOH on each acid.
* Data Processing & Presenting:
Graph 1: represents the temperature change in the solution when titrated with HCl after extrapolation.
Graph 2: Represents the temperature change of the solution titrated against Ethanoic Acid after extrapolation.
* From graph 1, it is shown that after extrapolating the final temperature of the solution are 38 ï¿½C instead of being 34 ï¿½C from the normal graph.
* From graph 2, it is shown that after extrapolating the graph, the final temperature of the solution is about 34 ï¿½C instead of being 32 from the normal graph.
* Amount of NaOH = c x v = 2 x 0.05 = 0.1 mol NaOH
* Amount of Heat Energy for HCL solution
= m x s x ?T = (100/1000) x 4.18 x (38 – 23) = 6.27 KJ
* Molar Heat Energy for HCL solution
= – 6.27 x (1 / 0.1) = – 62.7 KJ/mol
* Amount of Heat Energy for Ethanoic Acid Solution
= m x s x ?T = (100/1000) x 4.18 x (34 – 23) = 4.56 KJ
* Molar Heat Energy for Ethanoic Acid solution
=- 4.56 x (1 / 0.1) = -45.6 KJ/mol.
(Negative sign was added to both the heat energies because the reaction is exothermic due to the rise in temperature of the solution.)
* Conclusion & Evaluation:
* ?H neutralization for Ethanoic Acid (-45.6 KJ/mol) is lower than that for Hydrochloric Acid (-62.7 KJ/mol). This is because HCL is a strong acid which completely ionizes and dissociates. On the other hand, CH3COOH is a weak acid which partially ionizes in water.
* Percentage Uncertainties is:
* Pipette (Volume of NaOH):
(0.1/50) x100 = 0.20%
* Burette (Volume of HCL):
(0.05/50) x 100 = 0.10%
* Burette (Volume of CH3COOH):
(0.05/50) x100 = 0.10%
* Thermometer (Temperature of HCL):
(0.5/61) x 100 = 0.81 %
* Thermometer (Temperature of CH3COOH):
(0.5/57) x 100 = 0.87 %
* Total Percentage Uncertainty = 0.20+0.10+0.10+0.81+0.87
= 2.08 %
* Absolute Uncertainty for ?H HCL = 62.7 x (2.08/100) = 1.3
* Absolute Uncertainty for ?H CH3COOH = 45.6 x (2.08/100) = 0.94
* ?H Hydration for HCL is -62.7 KJ/mol (ï¿½ 1.3)
* ?H Hydration for CH3COOH is -45.6 KJ/mol (ï¿½ 0.95)
* Percentage Error:
1. Literature value for HCL is -57.6 KJ/mol
= (57.6 – 62.7)/57.6 = 0.0885 x 100 = 8.85 %
2. Literature value for CH3COOH is -36.8 KJ/mol
= (36.8 – 45.6)/45.6 = 0.193 x 100 = 19.3 %
1. Some heat was lost to the surrounding during the reaction. Water temperature decreased as a result from the heat loss, which caused a decrease in the final temperature.
2. The polystyrene cup was not covered with a lid, which also caused heat to be lost to the surrounding.
3. While stirring, the thermometer hit the bottom of the polystyrene cup which caused the thermometer to take the temperature of the cup instead of the water. This affected the readings of temperatures in different intervals which caused an error in drawing the graph.
4. Stirring of the solution was not constant all over the reaction, which caused a partial gain of heat.
1. The polystyrene cup should be covered with a lid, which will increase its insulation and will decrease the amount of heat lost to the surrounding.
2. The thermometer should not hit the bottom of the cup when stirring and friction should be reduced to maximum. This can be done by either holding the thermometer accurately up from the bottom. Or by adjusting it into a clamp embedding it in the solution, while using a glass rod for stirring.
3. Stirring the solution should be constant all over the reaction in order to have accurate readings during all time intervals, which will make the graph and its extrapolating more accurate.