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When performing a titration involving acetic acid, the first step is understanding the balanced chemical equation at play. In this exercise, we are dealing with the reaction between acetic acid (HC₂H₃O₂) and sodium hydroxide (NaOH). The balanced chemical equation is:

• HCâ‚‚H₃Oâ‚‚(aq) + NaOH(aq) → NaCâ‚‚H₃Oâ‚‚(aq) + Hâ‚‚O(l)

This equation shows a 1:1 mole ratio between acetic acid and sodium hydroxide. This indicates that each mole of NaOH neutralizes one mole of HC₂H₃O₂.
Understanding this relationship is key to calculating the exact amount of acetic acid present in the solution. You take the volume and molarity of NaOH used in the titration to determine the moles of sodium hydroxide, which, due to the 1:1 ratio, will equal the moles of acetic acid in the solution.

To determine the molarity of acetic acid in the vinegar, you begin with the moles of sodium hydroxide calculated from the titration. Given that the chemical reaction between acetic acid and sodium hydroxide follows a 1:1 molar ratio, the moles of NaOH used equals the moles of acetic acid present.
Once you have the moles of acetic acid, calculating molarity becomes straightforward. Molarity is the number of moles of solute per liter of solution. For this example, you determine:

• Moles of acetic acid = 0.00839 mol
• Volume of vinegar = 0.01000 L
• Molarity = \( \frac{0.00839 \text{ mol}}{0.01000 \text{ L}} = 0.839 \text{ mol/L} \)

This molarity calculation provides the concentration of acetic acid in the vinegar sample.

The mass percent of a solute in a solution tells you the concentration as a mass-to-mass ratio. To calculate the mass percent of acetic acid in the vinegar, you first need the mass of acetic acid. From earlier calculations brought forward:

• Moles of acetic acid = 0.00839
• Using the molar mass of acetic acid (60.05 g/mol), convert moles to grams:
• Mass of acetic acid = \(0.00839 \text{ mol} \times 60.05 \text{ g/mol} = 0.5038 \text{ g} \)

Following this, you'll need to find the mass of the vinegar sample provided. With a volume of 10.00 mL and density of 1.006 g/mL:

• Mass of vinegar = \( 10.00 \text{ mL} \times 1.006 \text{ g/mL} = 10.06 \text{ g} \)

Finally, calculate the mass percent:

• Mass percent of acetic acid = \( \frac{0.5038 \text{ g}}{10.06 \text{ g}} \times 100 = 5.01\% \)

This value demonstrates the amount of acetic acid in relation to the entire vinegar sample.

Chemical reaction equations are a fundamental part of chemistry. They provide a concise way to describe how substances interact during chemical reactions. In this titration problem, we had:

• HCâ‚‚H₃Oâ‚‚(aq) + NaOH(aq) → NaCâ‚‚H₃Oâ‚‚(aq) + Hâ‚‚O(l)

Each component in this equation gives insight into the reaction's progress. The left side denotes the reactants (acetic acid and sodium hydroxide), while the right side shows the products (sodium acetate and water). The balanced equation, which states equal moles of reactants and products, is crucial because:

• It respects the law of conservation of mass.
• It allows the calculation of reactant or product quantities needed or produced during the reaction.

Understanding and writing these reactions accurately aid in solving complex problems like titrations, as they guide the stoichiometric calculations necessary to find molarity and mass percent of substances involved.