A 15.0 ml sample of vinegar was analyzed via titration in the same manner as you

Question

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? 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?

Explanation / Answer

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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