Titration Lab

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In the titration experiment, the endpoint was recorded in the experimental data to be at 21.30 mL of NaOH and at a pH of 10.44. However, when all of the data from the table was graphed, the observed endpoint was too high up and on a part of the upper concave down portion of the graph. To ensure that the proper equivalence point was used, a new point had to be extrapolated that was roughly the point at which the graph went from concave up to concave down. This point was at 21.28 mL of NaOH added and pH of 9.20. Dividing both of these points by two, the half equivalence point was found to be at a pH of 5.30 and 10.64 mL of NaOH added. The pH is equal to the pKa here, so the pKa was found to be 5.30. Using data from the equivalence point extrapolated from the graph, the molar mass of the unknown was calculated to by 148 grams per mole. Lastly, because there was only one region of …show more content…

Error analysis calculations showed a 1.4 % error. A possible source of error is extrapolating a value for the freezing point that was too high. The Y-axis of the graph was done in intervals of two degrees Celsius, so precisely finding the value from the graph has the possibility to be estimated incorrectly. Extrapolating too high of a value would cause ΔT to be too low. This would then cause molality to be too low, moles of solute to be too low and the molar mass to be falsely high. After doing an IR spectrum for unknown 149, the only information that could be determined about the structure was that it contained a hydroxide group, a carbonyl and a C-C aromatic bond somewhere in the structure. There was a lot of trouble getting a readable IR spectrum from unknown 149 but after numerous pellets made, it was later found that the printed spectrum would suffice for comparison to trans-Cinnamic and o-Chlorobenzoic acid

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