Question

A 18.0 mL sample of vinegar, which is an aqueous solution of acetic acid, CH₃COOH, requires 24.5 mL of 0.600 M NaOH to reach the endpoint in a titration.

CH₃COOH(aq) + NaOH(aq) → CH₃COONa(aq) + H₂O(l)

What is the molarity of the acetic acid solution?

Answer 1

To find the molarity of acetic acid in vinegar, one calculates the moles of NaOH used in titration and applies that number, given the 1:1 reaction stoichiometry, to find the moles of acetic acid. Then, divide by the volume of the vinegar sample in liters to find the molarity.

Step-by-step explanation:

The molarity of the acetic acid solution can be determined using the titration data provided. Knowing that the reaction between acetic acid (CH₃COOH) and sodium hydroxide (NaOH) goes to completion and is a 1:1 stoichiometric reaction, we can use the information about the volume and molarity of NaOH to find the moles of NaOH, which will be equal to the moles of CH₃COOH that reacted.

To find the molarity (M) of the acetic acid solution, you would use the formula:

Molarity (M) = Moles of solute / Volume of solution (L)

First, we calculate the moles of NaOH:

Moles of NaOH = Molarity of NaOH (0.600 M) × Volume of NaOH (0.0245 L)

Then, since the stoichiometry is 1:1, these moles are equal to the moles of acetic acid in the vinegar sample. Next, we calculate the molarity of acetic acid:

Molarity of CH₃COOH = Moles of CH₃COOH / Volume of CH₃COOH solution (0.0180 L)

By plugging the numbers into these equations, the student will find the molarity of the acetic acid solution in the vinegar.