Chem 1211K Lab Report Essay

1861 WordsMar 9, 20158 Pages
Chemistry 1211K Lab Report Briana Limage Drawer #D20 Tuesday December 2, 2014 Lab Day and Time: Tuesday 2-5 Unknown # 4224034-DF14 Introduction The purpose of this semester long experiment was to determine an unknown organic acid. An organic acid is an organic compound with acidic properties. A base reacts with acids to form salts. Titrations are used to determine the concentration of unknown substances. The purpose of the KHP experiment was to determine the molarity of NaOH. HCl titrations are mainly to check technique and used to verify the molarity of NaOH solution. The hypothesis is that this acid is C4H3OCOOH. Experimental Procedure for melting point: The identification of the melting point of the organic acid was…show more content…
pH was recorded every time 1.00 mL of NaOH was added to beaker. When the amount of NaOH added to the beaker was about 5.00 mL away from the expected end point, NaOH was added very slowly. Approximately 0.20 mL of NaOH was added until the pH made a jump. The pH was recorded until it reached ~12. This was repeated two more times. The pKa of each trial are determined using the graphs made on excel. Results and Discussion Melting Point: Starting (°C) Final (°C) Rough Melting Point 139 159 Melting Point 131 133 Standard Melting Point 138 140 No correction had to be made to the melting points because the standard melted in the range labeled on the bottle. The melted point observed is the correct melting point. KHP: Table 1 Trial Mass KHP (g) Mol KHP / mol NaOH Initial Volume (mL) Final Volume (mL) Total Volume (mL) 1 0.3113 0.001524 5.62 22.70 17.08 2 0.3017 0.001477 9.98 26.40 16.42 3 0.3178 0.001556 14.39 31.85 17.46 Table 2 Trial Mol NaOH (mol) Corrected Vol. (L) Molarity NaOH (M) Average Molarity (M) Molarity Deviation (M) Average Deviation (M) 1 0.001524 0.01708 0.08923 0.08943 0.0002 0.0003 2 0.001477 0.01642 0.08995 0.08943 0.00052 0.0003 3 0.001556 0.01746 0.08912 0.08943 0.00031 0.0003 Percent deviation= (Average deviation/average molarity) X 100 Percent deviation= (0.0003/0.08943) X 100 = 0.38% Calculations for the molarity of NaOH were: The results showed the molarity of the NaOH solution. This experiment was completed twice and a new average molarity

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