Problem 76 - Chemistry An Atoms First Approach

Question

A 10.00 -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?

Step-by-Step Solution

Answer

The molarity of acetic acid in the vinegar is 0.839276 M, and the mass percent of acetic acid in the vinegar is 5.00%.

Step 1: Write the balanced chemical equation for the reaction

The reaction between acetic acid (HC2H3O2) and sodium hydroxide (NaOH) can be represented by the following balanced chemical equation: \( HC_{2}H_{3}O_{2} + NaOH \rightarrow CH_{3}COONa + H_{2}O \)

Step 2: Calculate the moles of NaOH used

Given the volume and molarity of NaOH, we can find the moles of NaOH used in the reaction as follows: Moles = Molarity x Volume in liters Moles of NaOH = \( 0.5062\,M \times \frac{16.58\,mL}{1000} \,L \) Moles of NaOH = 0.00839276 mol

Step 3: Determine the moles of acetic acid in vinegar

From the balanced chemical equation, the stoichiometry of the reaction between acetic acid and NaOH is 1:1. Therefore, the moles of acetic acid in the vinegar are equal to the moles of NaOH used in the titration: Moles of acetic acid = Moles of NaOH = 0.00839276 mol

Step 4: Calculate the molarity of acetic acid

Now that we have the moles of acetic acid, we can determine the molarity using the formula: Molarity = Moles / Liters The given volume of acetic acid in the vinegar is 10 mL, so in liters it would be: \( \frac{10\,mL}{1000} = 0.01\,L \) Molarity of acetic acid = \( \frac{0.00839276\,mol}{0.01\,L} \) Molarity of acetic acid = 0.839276 M (Answer to part a)

Step 5: Calculate the mass of acetic acid in the sample

First, we need to find the mass of acetic acid in the sample: Mass = moles × molar mass The molar mass of acetic acid is \( 12.01\,(C) + 1.01\,(H) + 2(1.01\,(C) + 12.01\,(H) + 16.00\,(O)) = 60.05\,g\,mol^{-1}\) Mass of acetic acid = \( 0.00839276\,mol \times 60.05\,g\,mol^{-1} = 0.5036\,g \)

Step 6: Calculate the mass percent of acetic acid in the vinegar

We have the mass of acetic acid in 10 mL of vinegar. Given the density of the vinegar (1.006 g/cm³), we can find the mass of the vinegar sample: Mass = Volume × Density Mass of vinegar = \(10\,mL \times \frac{1.006\,g}{1\,cm^{3}} \) Mass of vinegar = 10.06 g Now we can calculate the mass percent of acetic acid in the vinegar: Mass percent of acetic acid = \( \frac{Mass\,of\,acetic\,acid}{Mass\,of\,vinegar} \times 100 \) Mass percent of acetic acid = \( \frac{0.5036\,g}{10.06\,g} \times 100 = 5.00\% \) (Answer to part b)

Key Concepts

Molarity Calculation

Understanding the molarity of a solution is vital for many chemical reactions, including titrations. Molarity, often denoted by the letter M, is a measure of the concentration of a solute in a solution. To calculate it, you divide the number of moles of solute by the volume of the solution in liters.
For instance, if you're given the amount of solute in moles and the volume of the solvent used to create the solution, just plug these values into the formula:
\[\begin{equation}Molarity (M) = \frac{moles\text{ of solute}}{liters\text{ of solution}}\end{equation}\]In the context of our vinegar titration exercise, we first determined the amount of sodium hydroxide (NaOH) used to reach the equivalence point. Using the molarity and volume of NaOH, we calculated the moles of NaOH that reacted with the acetic acid. Since the stoichiometry of acetic acid to NaOH is 1:1, this also gave us the moles of acetic acid. Lastly, these moles were used along with the volume of vinegar in liters to find the molarity of acetic acid in vinegar.

Stoichiometry

Stoichiometry is the art of balancing chemical reactions and is rooted in the law of conservation of mass. In stoichiometry, we use the coefficients of a balanced chemical equation to understand the ratio in which chemicals react.
For the titration of acetic acid with sodium hydroxide, the balanced equation is crucial. It tells us that one mole of acetic acid reacts with one mole of sodium hydroxide to produce one mole of sodium acetate and one mole of water. This 1:1:1:1 ratio allows us to infer that if we know the number of moles of one reactant (or product), we can find the moles of any other substance in the reaction.
Therefore, by calculating the moles of NaOH (the titrant) that reacted, which we did in the molarity calculation section, we already know the moles of acetic acid (the analyte) present in our vinegar sample, thanks to this stoichiometric relationship.

Mass Percent Calculation

The mass percent (also known as weight percent or mass fraction) is a way to express the concentration of a component in a mixture or a solution. It is calculated by dividing the mass of the component by the total mass of the mixture, and then multiplying the result by 100 to get a percentage.
\[\begin{equation}Mass\text{ percent }(\%) = \left( \frac{mass\text{ of component}}{total\text{ mass of mixture}} \right) \times 100\end{equation}\]In our vinegar example, to find the mass percent of acetic acid, we first needed the mass of acetic acid, which we obtained by multiplying its moles from the titration by its molar mass. We also needed the mass of the vinegar, which was calculated using the volume of the vinegar sample and its density. With these two masses, we applied the formula above to determine that the mass percent of acetic acid in the vinegar was 5.00%.
Using mass percent is especially valuable when dealing with solutions where the components are not easily separated or where we're interested in the composition of mixtures, such as food products or chemical mixtures like the vinegar in our example. It’s widely used in chemistry to communicate the concentration of various substances.