Question

The distinctive odor of vinegar is due to acetic acid, \(\mathrm{CH}_{3} \mathrm{COOH}\), which reacts with sodium hydroxide according to: $$ \mathrm{CH}_{3} \mathrm{COO}(a q)+\mathrm{NaOH}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{NaCH}_{3} \mathrm{OO}(a q) $$ If \(3.45 \mathrm{~mL}\) of vinegar needs \(42.5 \mathrm{~mL}\) of \(0.115 \mathrm{M} \mathrm{NaOH}\) to reach the equivalence point in a titration, how many grams of acetic acid are in a 1.00-qt sample of this vinegar?

Short answer

There are approximately 79.86 grams of acetic acid in the 1.00-qt sample of vinegar.

Step-by-step solution

Calculate the moles of NaOH used in titration

Using the molarity and volume of NaOH, we can find the number of moles of NaOH used in the reaction. Moles of NaOH = Molarity of NaOH × Volume of NaOH Moles of NaOH = \(0.115 \, mol/L \times 0.0425 \, L = 0.00484 \, mol\) Note that we have converted the volume of NaOH to liters (from mL) for the calculation.

Calculate the moles of acetic acid in the vinegar

From the balanced equation, we know that 1 mole of acetic acid reacts with 1 mole of sodium hydroxide. So, the moles of acetic acid present in the sample are equal to the moles of NaOH required to neutralize it. Moles of acetic acid = Moles of NaOH = 0.00484 mol

Calculate the concentration of acetic acid in the vinegar

Now we can find the molarity of the acetic acid in the vinegar. Molarity of acetic acid = Moles of acetic acid / Volume of vinegar Molarity of acetic acid = \(\frac{0.00484 \, mol}{0.00345 \, L} = 1.4029 \, mol/L\)

Calculate the mass of acetic acid in 1.00-qt sample

First, we need to convert the volume of the 1.00-qt sample to liters. 1 qt = 0.946353 L (1 quart is approximately 0.946353 liters) Now we can calculate the moles of acetic acid in the 1.00-qt sample. Moles of acetic acid in 1.00-qt sample = molarity of acetic acid × volume of the 1.00-qt sample = \(1.4029 \, mol/L \times 0.946353 \, L = 1.3293 \, mol\) Finally, we calculate the mass of acetic acid using the molar mass of acetic acid (CH3COOH), which is approximately 60.05 g/mol. Mass of acetic acid = moles of acetic acid × molar mass of acetic acid = \(1.3293 \, mol \times 60.05 \, g/mol = 79.86 \, g\) Therefore, there are approximately 79.86 grams of acetic acid in the 1.00-qt sample of vinegar.

Key concepts

Stoichiometry

Stoichiometry is a section of chemistry that involves using balanced chemical equations to calculate the relative quantities of reactants and products involved in a chemical reaction. In an acid-base titration involving acetic acid and sodium hydroxide, stoichiometry helps us understand the simple 1:1 molar relationship between the acid and the base.

Given the balanced equation:
\[ \mathrm{CH}_{3}\mathrm{COOH} + \mathrm{NaOH} \rightarrow \mathrm{H}_{2}\mathrm{O} + \mathrm{NaCH}_{3}\mathrm{COO} \]

This equation tells us that one mole of acetic acid reacts with one mole of sodium hydroxide to produce one mole of water and one mole of sodium acetate. When we carry out the titration, we measure the amount of base needed to neutralize the acid, which corresponds to the amount of acid originally present in the solution. By applying stoichiometry, we can then calculate the mass of acetic acid in any given volume of vinegar.

Molarity

Molarity, often represented by the symbol \( M \), is a measure of concentration that tells us the number of moles of a substance present in one liter of solution. It is a crucial concept for understanding solutions and their reactions, including titrations.

For example, if the molarity of a sodium hydroxide solution is \( 0.115 M \), it means that there are \( 0.115 \) moles of \( \mathrm{NaOH} \) dissolved in every one liter of the solution. In our titration problem, we use the molarity of the \( \mathrm{NaOH} \) solution (along with the volume of the solution used in the titration) to calculate the moles of \( \mathrm{NaOH} \) and then the moles of acetic acid in the vinegar sample. Understanding molarity is essential to determining the concentration of the acetic acid solution and, consequently, the quantity of acetic acid in a given volume of vinegar.

Acid-Base Titration

Acid-base titration is a process used to determine the concentration of an acid or a base in a solution. During a titration, one solution (titrant) is slowly added to another solution (analyte) until the chemical reaction between the two solutes is complete. This point of completion is known as the equivalence point.

In the case of acetic acid titration, the titrant is the \( \mathrm{NaOH} \) solution. We use an indicator or a pH meter to observe when the acetic acid is completely neutralized, which occurs at the equivalence point. The volume of \( \mathrm{NaOH} \) required to reach this point allows us to calculate the molarity and then the mass of acetic acid in the vinegar sample using stoichiometry and molarity concepts.

Mole Concept

The mole concept is a fundamental principle that links the mass of substances to the number of particles, atoms, or molecules they contain. One mole contains Avogadro's number (approximately \( 6.022 \times 10^{23} \)) of entities, whether they're atoms, molecules, ions, or electrons.

In the context of acetic acid titration, we first find the number of moles of sodium hydroxide used in the titration. Since the reaction between acetic acid and sodium hydroxide is a 1:1 molar ratio, we know that the moles of acetic acid in the vinegar will be equal to the moles of sodium hydroxide used. To calculate the mass of acetic acid from the number of moles, we multiply by acetic acid's molar mass (approximately \( 60.05 g/mol \)). This demonstrates the mole concept's critical role in connecting the volume and concentration of a solution to the mass of a specific substance within that solution.