Acetic acid (CH \(_{3} \mathrm{COOH}\) ) is an important ingredient of vinegar. A sample of \(50.0 \mathrm{~mL}\) of a commercial vinegar is titrated against a \(1.00 \mathrm{M} \mathrm{NaOH}\) solution. What is the concentration (in \(M\) ) of acetic acid present in the vinegar if \(5.75 \mathrm{~mL}\) of the base are needed for the titration?

A neutralization reaction is a chemical process in which an acid reacts with a base to produce a salt and water. This type of reaction is important because it allows the determination of unknown concentrations of acids or bases when a standard solution of known concentration is used in a process known as titration. In our exercise, acetic acid ((CH_3COOH)) reacts with sodium hydroxide (NaOH) to form sodium acetate (CH_3COONa) and water (H_2O). The reaction is one-to-one, meaning for every mole of the acid, one mole of the base will react, forming one mole of salt and one mole of water. Understanding this equimolar relationship is crucial for the accuracy of the titration process and the calculation of the unknown concentration.

Stoichiometry involves the quantitative relationships between the reactants and products in a chemical reaction. It is based on the law of conservation of mass where the total mass of reactants equals the total mass of products. In the context of our exercise, stoichiometry uses the balanced equation of the neutralization reaction to determine the amount of acetic acid present in the vinegar sample. Since the exercise gives us a balanced chemical equation, we can conclude that the mole ratio of acetic acid to sodium hydroxide is 1:1. This allows us to use the moles of sodium hydroxide, which we obtained from titration, to directly find the moles of acetic acid. Stoichiometry is the bridge that connects qualitative information (the balanced equation) to quantitative data (the moles), which is then used to find the concentration.

Molarity (M)) is a measure of the concentration of a solution, defined as the number of moles of solute present in one liter of solution. It's one of the most commonly used units of concentration in chemistry. In our exercise, we used the molarity of the NaOH solution and the volume it took to neutralize the acetic acid to calculate the molarity of the acetic acid in the vinegar. The process involved calculating the moles of NaOH with the given molarity, and since the reaction has a 1:1 mole ratio, the same number of moles will be used for acetic acid. To obtain the concentration of acetic acid, the calculated moles are then divided by the vinegar's volume in liters. For clearer comprehension and application of the concept in similar exercises, it's crucial to keep in mind that molarity is an expression of the ratio of moles to volume, providing a direct method of comparing the strengths of different solutions.