Molarity of standard NaOH solution: x/1 H 2/3 Unknown + 2/3 NaOH-> 2/3 H2O +Na 2/3 Acetate
| Vinegar: | Titration 1 | Titration 2 | Tirtration 3 |
| Initial V of Acid | .01 mL | 5.56 mL | 11.32 mL |
| Final V of Acid | 5.56 mL | 10.71 mL | 16.02 mL |
| Inital V of Base | .01 mL | 27.12 mL | 3.01 mL |
| Final V of Base | 22.23 mL | 44.71 mL | 24.90 mL |
Unknown Acid code and number of reactive hydrogens: H+3
| Unknown Acid | Titration 1 | Titration 2 (Error) | Titration 3 | Titration 4 |
| Mass of Acid | .1024 g | .1028g | .1021g | .1023g |
| Initial V of Base | 24.90 mL | 37.80 mL | 45.5 mL | 9.24 mL |
| Final V of Base | 37.80 mL | 45.5 mL | 9.24 mL | 17.80 mL |
4. Find the molarity of the vinegar for each of your threevinegar titrations, then find the average molarity of the acid.
5. Find the percent deviations for your three vinegar molaritiesfor the vinegar titrations.
6. From the moles of acetic acid for each vinegar titration,find the mass of acetic acid for that titration
