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 \(16.58 \mathrm{~mL}\) 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

The molarity of acetic acid in the vinegar is \(0.839\,\text{M}\), and the mass percent of acetic acid in the vinegar is \(5.01\%\).

Step-by-step solution

Write the balanced chemical equation for the titration reaction

The balanced equation for the reaction between acetic acid and sodium hydroxide is: \( HC_2H_3O_2(aq) + NaOH(aq) \rightarrow H_2O(l) + NaC_2H_3O_2(aq) \)

Calculate moles of NaOH used in the titration

We will use the concentration and volume of NaOH to determine the moles of NaOH used in the titration. Moles of NaOH = Molarity × Volume Volume of NaOH = 16.58 mL = 0.01658 L Moles of NaOH = \(0.5062\,\text{M} \times 0.01658\,\text{L} = 0.00839\,\text{mol}\)

Determine moles of acetic acid in the vinegar

Using the stoichiometry of the balanced equation, we can determine the moles of acetic acid in the vinegar sample. 1 mol HC_2H_3O_2 reacts with 1 mol NaOH Therefore, moles of HC_2H_3O_2 = moles of NaOH = 0.00839 mol

Calculate the molarity of the acetic acid

Since molarity is defined as the moles of solute per liter of solution, we can use the moles of acetic acid and the volume of the vinegar to find the molarity of acetic acid. Molarity of HC_2H_3O_2 = \(\frac{\text{moles of HC}_{2}\text{H}_{3}\text{O}_{2}}{\text{volume of vinegar (L)}}\) Molarity of HC_2H_3O_2 = \(\frac{0.00839\,\text{mol}}{0.01000\,\text{L}} = 0.839\,\text{M}\) b. Determining the mass percent of acetic acid in the vinegar

Calculate the mass of acetic acid in the vinegar

To find the mass of acetic acid, multiply the moles of acetic acid by its molar mass. Molar mass of HC_2H_3O_2 = 12.01 + (4 × 1.01) + (2 × 16.00) = 60.06 g/mol Mass of HC_2H_3O_2 = moles × molar mass = 0.00839 mol × 60.06 g/mol = 0.5037 g

Calculate the mass of the vinegar sample

We are given the density of the vinegar and its volume. We can use this information to find the mass of the vinegar sample. Density = \(\frac{\text{mass}}{\text{volume}}\) Mass = Density × Volume Volume of the vinegar = 10.00 mL = 10.00 cm³ Density of the vinegar = 1.006 g/cm³ Mass of the vinegar = \(1.006\,\frac{\text{g}}{\text{cm}^{3}} \times 10.00\,\text{cm}^{3} = 10.06\,\text{g}\)

Calculate the mass percent of acetic acid in the vinegar

Mass percent is defined as the ratio of the mass of the solute (acetic acid) to the mass of the solution (vinegar), multiplied by 100. Mass percent of HC_2H_3O_2 = \(\frac{\text{mass of HC}_{2}\text{H}_{3}\text{O}_{2}}{\text{mass of vinegar}} \times 100\%\) Mass percent of HC_2H_3O_2 = \(\frac{0.5037\,\text{g}}{10.06\,\text{g}} \times 100\% = 5.01\%\) The molarity of acetic acid in the vinegar is 0.839 M, and the mass percent of acetic acid in the vinegar is 5.01%.

Key concepts

Molarity Calculation

Understanding molarity is fundamental in chemistry, especially in the context of titration. Molarity, often denoted as M, is a measure of concentration that indicates the number of moles of a solute per liter of solution. To calculate molarity, you need to know the moles of the solute and the volume of the solution in liters. In titration problems, it's common to begin with the molarity of one solution to find the molarity of the other after the equivalence point is reached.

For example, if a titration uses a known volume and concentration of sodium hydroxide (NaOH) to neutralize acetic acid (HC₂H₃O₂), you would first calculate the moles of NaOH using its molarity and volume. The balanced equation informs us that NaOH and acetic acid react in a 1:1 ratio; thus, the moles of NaOH are equal to the moles of acetic acid. With the moles of acetic acid and the volume of the acetic acid solution, the molarity of acetic acid can then be calculated. This step is often where students may struggle, and extra practice using various molarity calculations can enhance comprehension.

Stoichiometry

At the heart of titration is stoichiometry, the quantitative relationship between reactants and products in a chemical reaction. In the context of the provided problem, stoichiometry tells us how moles of acetic acid and sodium hydroxide react with each other. For every mole of acetic acid, there is an equal mole of sodium hydroxide required for complete neutralization, indicating a 1:1 ratio from the balanced chemical equation.

Stoichiometry is not only about ratios; it's about mass relationships as well. By understanding the molar mass of the compounds involved, you can convert between mass and moles, a key step in finding the mass percent of a solute in a solution. In our case, to find the mass percent of acetic acid in vinegar, you'd first find the mass of acetic acid by multiplying its moles by its molar mass. It's crucial to navigate these calculations accurately and students should practice converting between grams, moles, and liters to strengthen their stoichiometric skills.

Mass Percent

The mass percent, also known as mass fraction, is a way to express the concentration of a solution. It's defined as the mass of the solute divided by the total mass of the solution, multiplied by 100 to get a percentage. Mass percent can give you an understanding of the strength of the solution.

In a titration context, once you've calculated the mass of the solute, acetic acid in this case, and the total mass of the vinegar solution, finding the mass percent is straightforward. To accurately determine the mass, density and volume of the vinegar are utilized. This calculation can often be a source of confusion if the concept of density isn't well-understood, as both mass and volume are involved in its determination. However, the mass percent is a useful way to represent solution concentration, and knowing how to calculate it is essential for students exploring chemistry and its applications in real-world scenarios.