Final answer:
To find the molar concentration of acetic acid in vinegar, convert the NaOH volume used in titration to moles, then use the 1:1 stoichiometry of the reaction with acetic acid to determine the moles of acetic acid and divide by the volume of the vinegar solution to get the molarity, which is 0.5533 M.
Step-by-step explanation:
If you require 30.99 ml of 0.1786 M NaOH solution to titrate 10.0 ml of HC₂H₃O₂ solution, the molar concentration of acetic acid in the vinegar can be calculated using a stoichiometric analysis of the neutralization reaction occurring during titration. In this reaction, sodium hydroxide (NaOH) and acetic acid (HC₂H₃O₂) react in a 1:1 molar ratio as indicated by the balanced chemical equation:
NaOH + HC₂H₃O₂ → NaC₂H₃O₂ + H₂O
First, you need to convert the volume of NaOH solution used in the titration to moles using its molarity. The number of moles of NaOH is equal to:
Moles of NaOH = Volume of NaOH (L) x Molarity of NaOH (mol/L)
Moles of NaOH = 0.03099 L x 0.1786 mol/L = 0.005533 moles
Since the molar ratio between NaOH and HC₂H₃O₂ is 1:1, this is also the number of moles of acetic acid that were present in the 10.0 ml vinegar solution. To find the molarity of the acetic acid, divide the moles of acetic acid by the volume of the vinegar solution in liters:
Molarity of acetic acid = Moles of acetic acid / Volume of vinegar (L)
Molarity of acetic acid = 0.005533 moles / 0.010 L = 0.5533 M
Therefore, the molar concentration of acetic acid in the vinegar is 0.5533 M.