Question

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

7. Assuming that the density of vinegar is the same as the density of water, find the mass of vinegar for each titration, and calculate the mass percent of acetic acid forthat titration. (You looked up the density of water back in the measurement lab.)

8. Average your three mass percents of acetic acid to get your final answer for the

mass percent of acetic acid.

9. Find the volumes of base solution for each of your unknown acid titrations.