Table 2 trial 1 trial 2 average vinegar volume ml 001

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TABLE 2 Trial 1 Trial 2 Average Vinegar Volume (mL) +/- 0.01 mL 5.00 mL 5.00 mL 5.00 mL NaOH Volume (mL) +/- 0.05 mL 41.94 mL 41.26 mL 41.60 mL In experiment 1, Equation 1 and the results of table 1 are used to determine the molarity of NaOH. In experiment 2, the molarity found for NaOH in trial 1 and equation 1 are used to find the molarity of vinegar. The molarity of NaOH is approximately 0.1276 M/L and the molarity of H C 2 H 3 O 2 is approximately 1.062 M/L. Discussions: Since the actual molarity of vinegar and NaOH were not given, I cannot determine how my results correspond with the “real-life” values. However, I can discuss what could have gone wrong in the experiment that could, perhaps, have skewed my results. 1) If there was bubble in the tip of the buret, it would have given me an inaccurate volume for NaOH. Also, if I did not get the HCl and vinegar close enough to the line on the pipette, it would not have been exactly 25.00 mL or 5.00 mL of solution,
which could effect my molarity results. In this experiment, I learned how to determine molarity when using acids and bases. Also, I learned how to use a buret and pipet considering I have never worked with either pieces of equipment. In the future, I will be able to successfully do a titration and other experiments regarding acids and bases.

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