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

a. The molarity of the acetic acid solution is \(0.839556 \ \mathrm{M}\). b. The mass percent of acetic acid in the vinegar is \(5.007 \% \).

Step-by-step solution

The equivalence point is when the moles of the acid equal the moles of the base, so let's first calculate how many moles of NaOH we have: moles of NaOH = volume of NaOH solution * concentration of the NaOH solution moles of NaOH = 0.01658 L * 0.5062 M = 0.00839556 mol Here, 16.58 mL of NaOH solution is converted into L by dividing by 1000. #Step 2: Calculate moles of acetic acid solutions at the equivalence point#

Since we've reached the equivalence point, we know that the moles of acetic acid (HC2H3O2) are equal to the moles of the base NaOH: moles of acetic acid = moles of NaOH moles of acetic acid = 0.00839556 mol #Step 3: Calculate the molarity of the acetic acid solution#

Now we can use the moles of the acetic acid and the initial volume of the solution to calculate the molarity of the acetic acid: molarity of acetic acid = moles of acetic acid / volume of acetic acid solution molarity of acetic acid = 0.00839556 mol / 0.01 L = 0.839556 M a. The molarity of the acetic acid solution is 0.839556 M. #Step 4: Calculate the mass of acetic acid in the vinegar#

Knowing the molarity of acetic acid and the volume of the vinegar, we can find the mass of acetic acid: mass of acetic acid = moles of acetic acid * molar mass of acetic acid mass of acetic acid = 0.00839556 mol * 60 g/mol = 0.5037336 g Here, the molar mass of acetic acid (HC2H3O2) is 60 g/mol, calculated as (1 * 2 + 12 * 2 + 16 * 2 + 1 * 3). #Step 5: Calculate the mass of the vinegar#

Using the density and the volume of the vinegar, we can find the mass of the vinegar: mass of vinegar = density of vinegar * volume of vinegar * 1000 mass of vinegar = 1.006 g/cm³ * 10 cm³ = 10.06 g Here, 10 mL of vinegar is converted into cm³ as 1 mL = 1 cm³ #Step 6: Calculate the mass percent of acetic acid in the vinegar#

Finally, we can use the mass of acetic acid and the mass of vinegar to calculate the mass percent of acetic acid in the vinegar: mass percent of acetic acid = (mass of acetic acid / mass of vinegar) * 100 mass percent of acetic acid = (0.5037336 g / 10.06 g) * 100 = 5.007 % b. The mass percent of acetic acid in the vinegar is 5.007 %.

Key concepts

Equivalence Point

The equivalence point in a titration is a crucial concept. It is the stage at which the amount of acid in a solution is completely neutralized by the base added. This means that the number of moles of hydrogen ions from the acid exactly equals the number of moles of hydroxide ions from the base. For example, during a titration of acetic acid with sodium hydroxide \(\text{(NaOH)}\), the equivalence point is reached when the reaction \(\text{HC}_2\text{H}_3\text{O}_2 + \text{NaOH} \rightarrow \text{NaC}_2\text{H}_3\text{O}_2 + \text{H}_2\text{O}\) is complete.
To determine how much titrant (NaOH in this case) you need to reach the equivalence point, you must measure precisely. In our example, 16.58 mL of NaOH is used. At the equivalence point, the amount of \(\text{NaOH}\) is equal to the amount of acetic acid present initially in the solution.

Molarity Calculation

Molarity is a way of expressing the concentration of a solution in terms of moles of solute per liter of solution. It is denoted by the unit \(\text{M}\). Calculating the molarity involves finding the amount of original solute in the solution. For example, to find the molarity of acetic acid in vinegar using titration data:
1. Calculate the number of moles of NaOH used: \(\text{moles of NaOH} = 0.01658 \text{ L} \times 0.5062 \text{ M} = 0.00839556 \text{ mol}\).
2. At equivalence, these moles are equal to the moles of acetic acid present.
3. Finally, divide the number of moles of acetic acid by the volume of the solution in liters to determine the molarity: \(0.00839556 \text{ mol} / 0.01 \text{ L} = 0.839556 \text{ M}\).
This means that the vinegar solution initially contains 0.839556 moles of acetic acid per liter.

Mass Percent

Mass percent is a way of expressing the concentration of an element in a compound or a component in a mixture. Specifically, it measures the mass of the solute divided by the total mass of the solution, multiplied by 100 to get a percentage.
Here's how you can calculate the mass percent of acetic acid in vinegar:
- First, the mass of acetic acid is determined from its moles using its molar mass (60 g/mol): \(0.00839556 \text{ mol} \times 60 \text{ g/mol} = 0.5037336 \text{ g}\).
- Next, calculate the mass of the vinegar using its density (1.006 g/cm³) and volume (10 mL): \(1.006 \text{ g/cm}^3 \times 10 \text{ cm}^3 = 10.06 \text{ g}\).
- The mass percent of acetic acid is then \(\left( \frac{0.5037336 \text{ g}}{10.06 \text{ g}} \right) \times 100 = 5.007\%\).
This value indicates that 5.007% of the vinegar's mass is due to acetic acid.

Acetic Acid

Acetic acid, known chemically as \(\text{HC}_2\text{H}_3\text{O}_2\), is a weak organic acid that gives vinegar its characteristic taste and smell. It plays a significant role in chemical reactions such as titrations because of its weak acid properties. In a titration, acetic acid reacts with a strong base, like sodium hydroxide, to determine its concentration.
Acetic acid has the following characteristics:

• Structure: It consists of a carboxyl group \(\text{(COOH)}\) bonded to a methyl group \(\text{(CH}_3\text{)}\).
• Molar Mass: 60 g/mol.
• Properties: It is a weak acid, does not completely dissociate in water.

Understanding these properties helps in conducting more accurate titration experiments and calculating chemical reactions involving acetic acid.