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------------------------------------------------ If 13.51 mL of 0.1067 M NaOH was needed to titrate 2.00 mL of vinegar, how many moles of acetic acid are in the 2.00 mL vinegar sample?

The moles of acetic acid in the 2.00 mL vinegar sample can be calculated by determining the moles of NaOH used in the titration, since they react in a 1:1 ratio. The moles of NaOH is 0.00144, so the moles of acetic acid in the vinegar sample is also 0.00144.

Explanation:

This question is about calculating the number of moles of acetic acid in a vinegar sample. In titration, when the solution of sodium hydroxide (NaOH) completely neutralizes the solution of acetic acid in vinegar, we reach what's known as the equivalence point.

Acetic acid (CH3COOH) reacts with NaOH in a 1:1 stoichiometric ratio. Therefore, the moles of acetic acid will be equal to the moles of NaOH used to reach the equivalence point.

To find the number of moles of NaOH, use the formula: Moles = Molarity * Volume (in liters). The molarity of NaOH is given as 0.1067 M and the volume used is 13.51 mL (or 0.01351 L, because we need to convert mL to L).

Moles of NaOH = 0.1067 M * 0.01351 L = 0.00144 moles. Since NaOH and acetic acid react in a 1:1 ratio, the moles of acetic acid in the 2.00 mL vinegar sample will also be 0.00144 moles.

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