CHEM 102 Lab 7 Acids and Bases

Chem Lab 102-102 02/22/2023 Acids and Bases

I. Photo of Experimental procedure II. Signed Data Page III. I n t r o d u c t i o n The purpose of this experiment is to determine the differences, relationships between acids and bases that were gotten from the experiment. We will use titration, also pH indicator and burette to observe and find the equivalence point.

Volume of NaOH pH 0 2.3 3.1 2.7 6 3.1 9 3.3 12 3.5 15 3.7 18 3.8 21 4 24 4.1 27 4.4 30 4.6 33.1 5 33.4 5.1 33.7 5.2 34 5.3 34.3 5.4 34.6 5.5 34.9 5.6 35.2 5.8 35.5 5.9 35.9 6.3 36.2 7 36.3 8.4 36.4 8.6 36.5 8.7 36.8 9.1 39.8 10.7 41.8 11.3 44.8 11.6 47.4 11.8 50 12

The equivalence point is a (36.3, 8.4) with the values of Volume of NaOH = 36.3mL, and pH = 8.4. The moles of NaOH and CH 3 COOH = 0.0363L * 0.0970 = 0.0035211 moles The mass of CH 3 COOH in vinegar: 0.0035211 moles CH 3 COOH * 60.052grams/mole CH 3 COOH = 0.2114grams Percent of CH 3 COOH in vinegar: % CH 3 COOH = (mass of CH 3 COOH/ mass of vinegar ) * 100% = (0.2114/4.05)*100% = 5.22% Percent error: Percent error = ((experimental value – actual value)/actual value)*100% [(5.22-5.00)/5.00]*100% = 4.4% K a of the acid K a =10 -pH at half volume Half volume = 36.3/2= 18.15 At 18.15 pH ~ 3.8 10 -3.8 = 1.58*10 -4 V. Conclusion The main purpose of this experiment was to identify the differences between acids and bases. The equivalence point was found through titration curve was measured to be a pH of 8.4 and Volume of NaOh of 36.3mL The percentage of CH 3 COOH in vinegar is around 5.22% which is close to the theoretical value of 5.00%. The percent error was 4.40%, which shows that there was a little uncertainty in results. The K a experimental result is 1.58*10 -4 which is close to the theoretical result = 1.84*10 -4 . From this experiment, I have learned to do accurate titration method, how to calculate K a constant, pH of the solutions, and differentiate the acids and base. There can appear some mistakes while adding volumes during titration which