Indirect thermometric Titration Essay Example for Free

Indirect thermometric Titration Essay* School Name Al Mashrek International School* School Code 2108* Subject alchemy* Topic Indirect A thermometric Titration.* Assessment information Collection, Data Processing Presenting, Conclusion Evaluation.* Candidate Name Bassam Al-Nawaiseh* Date 20/5/2007* getThe 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 potpourri for each reaction, enthalpy of neutralisation reaction.* Data CollectionTable 1 the temperature change for the HCl beginning and CH3CO2H solution afterwards adding 5 cm portions of 1M NaOH on each acid. * Data Processing PresentingGraph 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 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 closely 34 C instead of being 32 from the normal graph.* Amount of NaOH = c x v = 2 x 0.05 = 0.1 gram molecule NaOH* Amount of waken power 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 s trong acid which completely ionizes and dissociates. On the other hand, CH3COOH is a weak acid which partially ionizes in water.* part 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 ( 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 lost to the surrounding during the reaction. Water temperature decreased as a c onduct 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 tear 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 ceaseless all over the reaction, which caused a partial gain of heat.* Improvements1. The polystyrene cup should be covered with a lid, which will increase its insulation and will decrease the join 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 both 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 much accurate.