Question

A \(10.00-\mathrm{mL}\) . sample of vinegar, an aqueous solution of acetic acid \(\left(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right),\) is titrated with \(0.5062 \mathrm{M} \mathrm{NaOH},\) and 16.58 mL is required to reach the equivalence point. a. What is the molarity of the acetic acid? b. If the density of the vinegar is \(1.006 \mathrm{g} / \mathrm{cm}^{3},\) what is the mass percent of acetic acid in the vinegar?

Short answer

a) The molarity of acetic acid in the vinegar is \(0.839 \: \text{mol/L}\). b) The mass percent of acetic acid in the vinegar is \(5.01\%\).

Step-by-step solution

Write the balanced chemical equation for the titration

The reaction between acetic acid and sodium hydroxide in the titration is as follows: HC₂H₃O₂(aq) + NaOH(aq) → NaC₂H₃O₂(aq) + H₂O(l) This equation is already balanced.

Calculate the moles of sodium hydroxide

Use the given volume and molarity of sodium hydroxide solution to find the moles of sodium hydroxide used in the titration: moles of NaOH = (volume of NaOH)(molarity of NaOH) moles of NaOH = (16.58 mL)(0.5062 mol/L) Make sure to convert the volume from mL to L: moles of NaOH = (0.01658 L)(0.5062 mol/L) = 0.00839 mol

Calculate moles of acetic acid

As per the balanced chemical equation in Step 1, the mole ratio between acetic acid and sodium hydroxide is 1:1. Therefore, the moles of acetic acid present in the 10.00-mL vinegar solution are equal to the moles of sodium hydroxide: moles of HC₂H₃O₂ = 0.00839 mol

Calculate the molarity of acetic acid

Now that we have the moles of acetic acid, we can find the molarity (moles of solute per liter of solution): molarity of HC₂H₃O₂ = (moles of HC₂H₃O₂) / (volume of vinegar in L) molarity of HC₂H₃O₂ = (0.00839 mol) / (0.01000 L) = 0.839 mol/L a) The molarity of acetic acid in the vinegar is 0.839 mol/L.

Calculate the mass of acetic acid in the vinegar sample

First, find the moles of acetic acid in the 10.00 mL sample: moles of HC₂H₃O₂ = (0.01000 L)(0.839 mol/L) = 0.00839 mol Next, convert moles of acetic acid to grams using the molar mass of acetic acid, which is 60.05 g/mol: mass of HC₂H₃O₂ = (0.00839 mol)(60.05 g/mol) = 0.5038 g

Calculate the mass percent of acetic acid

Use the density of the vinegar and the volume of the sample to find the mass of the vinegar: mass of vinegar = (volume of vinegar)(density of vinegar) mass of vinegar = (10.00 mL)(1.006 g/mL) = 10.06 g Next, calculate the mass percent of acetic acid in the vinegar: mass percent of HC₂H₃O₂ = [(mass of HC₂H₃O₂) / (mass of vinegar)] × 100 mass percent of HC₂H₃O₂ = [(0.5038 g) / (10.06 g)] × 100 = 5.01% b) The mass percent of acetic acid in the vinegar is 5.01%.

Key concepts

Acetic Acid Calculation

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:

• HC₂H₃O₂(aq) + NaOH(aq) → NaC₂H₃O₂(aq) + H₂O(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.

Molarity Determination

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.

Mass Percent Calculation

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 Equation

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:

• HC₂H₃O₂(aq) + NaOH(aq) → NaC₂H₃O₂(aq) + H₂O(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.