CH3COOH And Naoh Lab Report

Introduction
Strong acids and strong acids both dissociate completely in water forming ions. However, strong acids donate a proton to form H3O+ along with a conjugate base and strong bases accept a proton to form OH- along with a conjugate acid. The chemical behavior of acids and bases are opposite. When they are together, their ions cancel out and form a neutral solution. In this experiment, HCl and NaOH will react to form NaOH and H2O with these two steps:

The overall reaction is:

Both Na+ and Cl- ions combine to form NaCl. H3O+ and OH- ions combine to form H2O. H2O is part of the overall reaction because it is always the product in acid-base neutralization reaction. Based on this reaction, by adding a known base, the amount of acid can

The mass of vinegar used during the experiment was 4.108 grams. It was determined that there were .003129 moles of CH3COOH in the vinegar sample. Using this information and the molar mass of CH3COOH, which was 60.05 g/mol, the mass of acetic acid in the vinegar was calculated:

4.Vinegar is a 5% aqueous solution of acetic acid. Since the mass of acetic acid within the vinegar was calculated as .18789 g in step 3, the percent of CH3COOH was calculated using the following equation:

To calculate the percent error, the experimental value of 4.5% acetic acid in vinegar was subtracted by the theoretical value of 5% and divided by 5% to yield a percent error of 8.54%. The following is a copy of the calculations done using decimals:

5.The equivalence point of the titration curve measured in step 1 was 25.25 mL of NaOH. Half of this value is 12.63 mL. By interpolating the graph, the pH at this volume was 4.80, which is equivalent to the pKa of acetic acid. According to the tabulated data, the pKa was 4.90 at 15 mL of NaOH. At this point, the change in pH with respect to volume was minimal since these values were far from the equivalence point, which occurred experimentally at 27.41 mL. This can also be seen on the graph as the plateau before the inflection point occured. To calculate the Ka of the acid, the following formula is

The calculated value was 1.6 x 10^-5.

Conclusions
The resulting Ka of the acetic acid from this experiment’s calculations was consistent with the experimental results.
The experimental percent of CH3COOH was calculated at 1.6 x 10^-5, while the actual value was 1.8 x 10^-5. The calculated value is much lower because the pH read from the graph at half of the equivalence was higher than the actual value. To have gotten a 0% error between the experimental and actual value for CH3COOH, the pH would have been measured at about 4.75, which is slightly more acidic than 4.80. The percent error was calculated to determine how accurate the Ka of acetic acid was:

Since the calculations yielded a 20% error, this shows that experimental error occurred during the experiment. Factors that could have affected the results included improper reading of the meniscus for volume of NaOH, not allowing the NaOH to fully drip into the buret after removing the funnel, adding too much acetic acid after the indicator flashed pink to get an inaccurate equivalence point, and not allowing the solution in the beaker to mix thoroughly to get an accurate reading from the pH