(Question 2) A 15.0 mL sample of vinegar was analyzed via titration in the same
ID: 1042321 • Letter: #
Question
(Question 2) A 15.0 mL sample of vinegar was analyzed via titration in the same manner as you will use in this experiment. The standardized solution of NaOH was 1.20 M. A total of 10.5 mL of the sodium hydroxide were required to reach the pink endpoint. What was the concentration of the acetic acid in the vinegar solution?
(Q3) Determine how many moles of acetic acid would be present in 500.0 mL of the vinegar solution you found the concentration of in question 2.
(Q4) Determine how many grams of acetic acid are present in the number of moles you found in question 3.
(Q5) A vinegar solution has a density of 1.005 g/mL. How many grams are present in 500.0 mL of a vinegar solution?
(Q6) Using your answers from questions 4 and 5, determine the mass percent of acetic acid in vinegar.
Explanation / Answer
Question 2
Molarity of NaOH = 1.2 M
Volume of NaOH = 10.5 ml
MOles of NaOH = 1.2 M x 0.0105 L
= 0.0126 moles
Moles of NaOH = Moles of Acetic acid
Moles of acetic acid = 0.0126
Concentration of acetic acid = 0.0126 / 0.015
= 0.84 M
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