The purpose of this lab was to find the percent ionization of vinegar. To find the percent ionization, the volume of NaOH or sodium hydroxide. Using titration to see how much of NaOH made the liquid into a slight pink color, the the volume was discovered to be 27.1 on the first trial and 26.3 on the second trial. The pH of vinegar was given to be 2.4. With all this information in mind, the MV = MV equation could be used to find the molarity of acid.
Trial 1: (X M) (7.3) = (0.25 M) (27.1) --> 0.94 M
Trial 2: (X M) (8) = (0.25 M) (26.3) --> 0.82 M
Average Concentration of the vinegar = 0.88 M
[H3O+] = 3.98e-3
The percent ionization was 0.45% found by dividing the hydrogen ions in vinegar water by the CH3COO- times 100 to find the percent.
The percent ionization was so small because the acid is a weak one.