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GSU CHEM 1211K - Chem 1211K Lab Report

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Chemistry 1211K Lab ReportAlexis Hammond1211K- Fall 2019Friday 2:00-5:00L-4806 December 2019Unknown Organic Acid #P5152105-LF-191Introduction Chemistry is a means to understanding what everything is made of and how it came to be, down to the most subatomic detail. Essentially, chemistry is the study of everything in the surrounding of a person. Chemistry can be found in the reactions caused by food ingestion or ventilation response to exercise. It is how one can see certain light frequencies and not others. It is a way to explain the physical and chemical properties of life. When viewing an object, a person can’t help but wonder what it is comprised of. Using chemistry, we are able to find the identity of many compounds using multiple different strategies.For this lab, each chemist has the objective to unveil which unknown organic acid has been given. In this lab report, the identification of an unknown organic acid was determined by comparing the data taken from a few experiments. The experiments done to help the search were the melting point experiment, solubility experiment, and a variety of titrations to establish the equivalent weight and pKa. In the quest to find the name of the acid, the amateur chemists will learn proper technique and procedure, as well as apply chemistry concepts to actual chemical compounds.2Procedures For melting point, the chemist will start by taking 2-3 mm amount into a capillary tube. To find an approximate melting point range, set the Stuart melting point apparatus to 200˚C. Place the tube into the apparatus so that the unknown can be seen from the eye viewer. Press the start button and observe the unknown melting. Next, finding the melting point measurement, choose astandard closest to the melting point range. Place a tube filled with the standard and another with the unknown into heating apparatus. Set plateau at 15˚C lower than melting point range. Record when the sample and standard melt. The melting point range is from when the unknown first started to melt until it finished melting. Turn apparatus off and allow to cool. Next, solubility can be determined by weighing out 0.1 grams of the unknown into a test tube.If the compound goes into solution at room temperature, the solvent is composed of 100% DI water. If compound does not become a solution at room temperature, try adding heat with a waterbath. If the test tube’s contents dissolve, allow the test tube to cool. Should the solvent stay dissolved, it can be determined as 100% DI water, but if crystals form, an ethanol/DI water mixture should be used. Repeating the previous steps with 25% ethanol mixture and 50% ethanolmixture. If this doesn’t work, the chemist should try a 75% ethanol mixture. Now, for titrations, an NaOH solution must be made first. The chemist must weigh out 3.6 grams of NaOH into 900 mL of DI water to attain a solution with a molarity of 0.100 M and store it in a closed bottle. With KHP titrations, cleanse the buret with small amounts of NaOH and fill up to the 0.00 mLline. Once done, weigh out 0.3-0.5 grams of KHP. Dissolve the KHP in 15-20 mL of DI water. Use a hot plate to dissolve if needed. Add a single drop of phenolphthalein indicator in solution. Start to titrate by adding NaOH until the solution turns a very light pink. Record the initial and 3final volumes of NaOH added to the flask. The experiment should have three properly done trials. As for the HCl titrations, cleanse the buret with the NaOH solution and fill to the 0.00 mL line. Record initial volume to two decimal places. With an automatic pipettor, gather 10.00 mL ofHCl in an Erlenmeyer flask. Add about 20 mL of DI water to flask. Add one drop of phenolphthalein indicator into the flask. Now, start to titrate the solution. Once, a very faint pink color is reached, the chemist can add one more drop to the solution. It should lose the color. If one more drop is added, the solution will have reached its end point. Record final volume and repeat steps for two more trials. Titrating the unknown organic acid, weigh out 0.3-0.5 grams of unknown and dissolve with previously discovered solvent. Add a single drop of indicator. Once buret is cleaned, fill with NaOH. Record initial volume and start titrating. Record final volume and use the weight of the acid to find the equivalent weight of unknown acid. To do computer search, input the estimated melting point and equivalent weight with a deviation of about 10 to get a good selection of results. Finding pKa, weigh out 2 millimoles of the unknown and dissolve in previously selected solvent. Add a Teflon coated magnetic stirring bar to the beaker and place on the magnetic stirrer.Stirring velocity should be enough to cause a smooth but quick current. Carefully place the electrode of a calibrated pH meter in the beaker. Clean and fill an buret with NaOH to the 0.00 mL line. Record the initial pH and start titrating using 2 mL increments until there is a jump in pH value. Allow time for the solution to mix between each round. Then, start titrating by 1 mL until the pH is at 5. Now, add drops to solution until a pH of 9 is reached. Lastly, add 1-2 mL each time to a pH of 12 to 13 is reached. Graph NaOH volume vs. pH using Excel.4ResultsTable 1. Results of Melting Point ExperimentPlateau used in determination of melting point 94°CMelting Point for Unknown Acid107°CB.) Experimental ValueA.) Printed ValueMelting Point Range for the Standard98-10099-100Table 2. Results of Solubility ExperimentTrial% EthanolDissolved with Heat1 0%2 25%3 50%4 75%Solubility SolventDissolved at RoomTemperatureNoNoNoNoYes50% Ethanol, 50% DI WaterNoNoYesTable 3. Results from KHP TitrationsTrial Mass of KHP Moles of KHP Vol. NaOH Molarity of NaOH1 0.4025 g 0.001971 moles 20.00 mL2 0.4495 g 0.002201 moles 22.01 mL3 0.4627 g 0.002260 moles 22.98 mLAverage Molarity0.09854 M0.1000 M0.09859 M0.099 M ± 0.0011%% Deviation50.4025 ×1mol KHP204.22 g×0.00197091 mol KHP0.02 L NaOH=0.099 MTable 4. Results from HCl TitrationsTrial Vol. HCl Moles of HCl Vol. NaOH Molarity of HCl1 10.00 mL 0.001962 moles 19.82 mL2 10.00 mL 0.001969 moles 19.89 mL3 10.00 mL 0.001960 moles 19.80 mLAverage Molarity0.1962 M0.1969 M0.1960 M0.196 M ± 0.001% Deviation0.5%1mol HCl1 mol NaOH /¿¿(0.01989 L NaOH) (0.099 M NaOH)¿¿Table 5. Results from Unknown Organic Acid TitrationsTrialMass ofUnknownVol. NaOH Equivalent Weight1 0.3069 g2 0.3028 g3 0.3079 gAverage Equivalent


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