Indirect thermometric Titration Essay Example for Free

confirmatory thermometric Titration Essay* initiate Name Al Mashrek International School* School Code 2108* Subject Chemistry* Topic Indirect A thermometric Titration.* Assessment Data Collection, Data Processing Presenting, shutdown Evaluation.* Candidate Name Bassam Al-Nawaiseh* troth 20/5/2007* AimThe aim of this experiment is to determine the concentrations of two acids. The two acids are Hydrochloric acid, HCl, and Ethanoic acid, CH3CO2H. This allow for be done by thermometric titration, by calculating the total heat change for each reaction, enthalpy of neutralization.* Data CollectionTable 1 the temperature change for the HCl closure and CH3CO2H reply subsequently adding 5 cm portions of 1M NaOH on each acid. * Data Processing PresentingGraph 1 represents the temperature change in the resolvent when titrated with HCl afterward extrapolation.Graph 2 Represents the temperature change of the solution titrated against Ethanoic Acid after extrapolation.* From grap h 1, it is shown that after extrapolating the final temperature of the solution are 38 C instead of being 34 C from the normal graph.* From graph 2, it is shown that after extrapolating the graph, the final temperature of the solution is about 34 C instead of being 32 from the normal graph.* sum up of NaOH = c x v = 2 x 0.05 = 0.1 seawall NaOH* Amount of Heat vigor 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 aci d which solely 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%* buret (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 incertitude = 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 ( 1.3)* ?H Hydration for CH3COOH is -45.6 KJ/mol ( 0.95)* Percentage Error1. 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 %* Errors1. Some heat was incapacitated to the surrounding during the reaction. Water temperature reduced as a result from the heat loss, which caused a decrease in the final temperature.2. The polystyrene instill 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 unending all over the reaction, which caused a partial gain of heat.* Improvements1. The polystyrene cup should be covered with a lid, which provide 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 fix embedding it in the solution, while using a g lass 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.