A 5.00 mL sample of vinegar, an aqueous solution of acetic acid (HC2H3O2), is titrated with 0.0599 M Ca(OH)2, and 33.93 mL of the Ca(OH)2 solution is required to reach the equivalence point. What is the molarity of the acetic acid

[CH₃COOH] = 0.813 M

Explanation:

First of all we need to know the base of any titration.

In any titration, when you reach the equivalence point we say that:

mmoles of acid = mmoles of base

mmoles of acetic acid = mmoles of Ca(OH)₂

We can know the mmoles of the hydroxide with the given information.

33.93 mL . 0.0599M = 2.03 mmoles of base

The thing is that the calcium hydroxide release 2 OH, then to neutralize all the acetic acid, we need double of hydroxides.

 Ca(OH)₂      →     Ca²⁺     +    2OH⁻

2.03 mmol                           4.06 mmoles

These are the moles required to neutralize the acetic acid, because we said that in the equivalence point, mmoles of acid = mmoles of base

So, as Molarity can be defined as mmoles / mL

M = 4.06 mmoles/5mL → 0.813 M