Question

1. Volume of Vinegar used in each trial = 5.00 mL

2. Molarity of NaOH used in each trial = 0.20M

3. Volume of NaOH used in trail#1 (Final – Initial buret reading) = __17.80______ mL

4. Volume of NaOH used in trail#2 (Final – Initial buret reading) = ___18.20_____ mL

5. Average Volume of NaOH used = (#3 + #4) / 2 = ___18______ mL

6. Calculate Molarity of Acetic acid in vinegar = _________ M ?????? (molarity of acetic acid x volume of acid) = (molarity of base x volume of base) ?

Answer 1

Molarity of acetic acid *volume of acid= Molarity of base*volume of base

Therefore, M1*5.00 ml= 18.00 ml *0.200 M

Therefore, M1= 18.00 ml*0.200 M/5.00 ml

Therefore, M1= 0.72 M

Therefore, the Molarity of acetic acid in vinegar is 0.72 M

Thank You!