Question

Suppose you are titrating vinegar, which is an acetic acid solution of unknown concentration, with a sodium hydroxide solution according to the equation HC₂H₃O₂ + NaOH ⟶H₂O + NaC₂H₃O₂. if you require 33.04 ml of 0.1269 m n a o h solution to titrate 10.0 ml of HC₂H₃O₂ solution, what is the molar concentration of acetic acid in the vinegar?

Answer 1

The molarity of acetic acid in the vinegar is calculated to be 0.41924 M by using the amount of NaOH required to titrate a known volume of vinegar and the 1:1 molar ratio of the reaction between acetic acid and NaOH.

Step-by-step explanation:

To calculate the molar concentration of acetic acid in the vinegar, we can use the titration equation provided, which shows that acetic acid (HC₂H₃O₂) reacts with sodium hydroxide (NaOH) in a 1:1 molar ratio. Since 33.04 ml (or 0.03304 L) of 0.1269 M NaOH is required to neutralize 10.0 ml of acetic acid, we can calculate the moles of NaOH used in the titration: moles of NaOH = volume (L) × molarity (M). Subsequently, the number of moles of acetic acid is the same due to the 1:1 ratio. Thus, the molarity of acetic acid is the moles of acetic acid divided by the volume of the acetic acid solution in liters.

Here's the step-by-step calculation:

1. Calculate moles of NaOH: (0.03304 L) × (0.1269 M) = 0.004192416 moles of NaOH
2. Since the molar ratio of HC₂H₃O₂ to NaOH is 1:1, moles of acetic acid = moles of NaOH = 0.004192416
3. Calculate the molarity of acetic acid: Molarity = moles/volume (L) = 0.004192416 moles / 0.010 L = 0.41924 M

Therefore, the molarity of acetic acid in the vinegar is 0.41924 M.