Question

The distinctive odor of vinegar is due to acetic acid, \(\mathrm{CH}_{3} \mathrm{COOH},\) which reacts with sodium hydroxide according to: \(\mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{NaOH}(a q) \longrightarrow\) $$ \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{NaCH}_{3} \mathrm{COO}(a q) $$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 80.48 grams of acetic acid in a 1.00-quart sample of vinegar.

Step-by-step solution

Determine moles of NaOH

First, calculate the number of moles of \( \text{NaOH} \) used in the titration. We know the volume of \( \text{NaOH} \) is \( 42.5 \ \text{mL} \) or \( 0.0425 \ \text{L} \), and the concentration is \( 0.115 \ \text{M} \). Use the formula: \[ \text{moles} = V \times M \]\[ \text{moles of NaOH} = 0.0425 \ \text{L} \times 0.115 \ \text{mol/L} = 0.0048875 \ \text{mol} \]

Determine moles of acetic acid

Since the balanced equation shows a 1:1 molar ratio between \( \text{CH}_3\text{COOH} \) and \( \text{NaOH} \), the moles of acetic acid \( \text{CH}_3\text{COOH} \) are equal to the moles of \( \text{NaOH} \).\[ \text{moles of CH}_3\text{COOH} = 0.0048875 \text{ mol} \]

Calculate mass of acetic acid in the solution

Next, find the mass of acetic acid in the vinegar sample using its molar mass, which is approximately \( 60.05 \ \text{g/mol} \).\[ \text{mass of CH}_3\text{COOH} = 0.0048875 \ \text{mol} \times 60.05 \ \text{g/mol} = 0.293599 \ \text{g} \]

Scale up the mass to the entire sample

To find the mass of acetic acid in a 1.00-quart (qt) sample, relate it to the titration sample size. Convert the quart to milliliters: 1 qt = 946.353 mL. Use the proportion:\[ \frac{0.293599 \ \text{g}}{3.45 \ \text{mL}} = \frac{\text{mass in 1.00 qt}}{946.353 \ \text{mL}} \]\[ \text{mass in 1.00 qt} = \frac{0.293599 \ \text{g} \times 946.353 \ \text{mL}}{3.45 \ \text{mL}} = 80.48 \ \text{g} \]

Key concepts

Molarity Calculations

Molarity is an essential concept in chemistry that indicates the concentration of a solution. It is defined as the number of moles of solute divided by the liters of solution. For example, in our acetic acid titration problem, the molarity of sodium hydroxide (\(\mathrm{NaOH}\)) is given as 0.115 M (mol/L). Knowing this molarity allows us to calculate how many moles of \(\mathrm{NaOH}\) were needed during the titration. The formula, \( \text{moles} = V \times M \) (where \(V\) is volume in liters and \(M\) is molarity), was used to find that \(0.0425\,\text{L} \times 0.115\,\text{mol/L} = 0.0048875\,\text{mol}\).

This straightforward calculation shows the power of using molarity. It grants us a simple pathway from known volumes and concentrations to the number of moles, a critically useful measure in chemistry. Molar calculations help compare and use various concentrations of different solutions uniformly, regardless of the solute or the volume size.

Chemical Reactions and Stoichiometry

Stoichiometry is the steppingstone to understanding chemical reactions. It involves analyzing the proportion between reactants and products in a chemical equation. In our example, the reaction between acetic acid (\(\mathrm{CH}_{3}\mathrm{COOH}\)) and sodium hydroxide (\(\mathrm{NaOH}\)) follows a one-to-one molar ratio, as presented in the balanced equation:
\[\mathrm{CH}_{3} \mathrm{COOH}(aq) + \mathrm{NaOH}(aq) \rightarrow \mathrm{H}_{2} \mathrm{O}(l) + \mathrm{NaCH}_{3} \mathrm{COO}(aq)\]

Due to this simple 1:1 relationship, the number of moles of acetic acid neutralized equals the moles of hydrogen ion (\(\mathrm{H}^+\)) required from \(\mathrm{NaOH}\). From a stoichiometric perspective, this tells us that for every mole of \(\mathrm{NaOH}\), one mole of \(\mathrm{CH}_{3}\mathrm{COOH}\) is consumed, forming water and sodium acetate (\(\mathrm{NaCH}_{3}\mathrm{COO}\)).Understanding stoichiometry simplifies chemical analysis by emphasizing proportions and conservation of mass. It allows chemists to predict the quantities of products from given reactants accurately, a necessary skill in formulating reactions and understanding chemical processes.

Solution Preparation and Dilution Calculations

Solutions and dilution calculations involve preparing specific concentrations of a solution from either a pure solute or a more concentrated solution. Here, the concept is crucial when scaling lab results to practical applications. For example, in our problem, we've calculated the mass of acetic acid present in a 3.45 mL sample of vinegar. We then scaled this finding to determine the mass found in a one quart (roughly 946 mL) sample.

To do this, the mass of acetic acid in the small sample was related to the larger volume using a proportion:

• Mass in small sample: 0.293599 g
• Volume of small sample: 3.45 mL
• Volume of large sample: 946.353 mL

The calculation follows:\[\text{mass in 1.00 qt} = \frac{0.293599 \ \text{g} \times 946.353 \ \text{mL}}{3.45 \ \text{mL}}\]Resulting in about 80.48 grams of acetic acid per quart.This illustrates how dilution calculations can transfer lab-based data to real-world scenarios. Such scaled calculations are fundamental across many scientific fields, supporting precise adjustments to solution concentrations needed for various experiments and applications.