Problem 134 - Chemistry The Science in Context

Question

A food chemist determines the concentration of acetic acid in a sample of apple cider vinegar (see Problem 4.133 ) by acid-base titration. The density of the sample is \(1.01 \mathrm{g} / \mathrm{m} \mathrm{L}\) The titrant is \(1.002 M\) aOH. The average volume of titrant required to titrate \(25.00 \mathrm{mL}\) aliquots of the vinegar is \(20.78 \mathrm{mL} .\) What is the concentration of acetic acid in the vinegar? Express your answer the way a food chemist probably would: as percent by mass.

Step-by-Step Solution

Answer

Answer: Follow the step-by-step solution provided above to calculate the concentration of acetic acid in the vinegar sample as a percentage by mass.

Step 1: Determine the moles of OH- ions used in titration.

To find the moles of OH- ions used in the titration, we will use the formula: moles = molarity × volume. The molarity of the titrant is 1.002 M and the volume is 20.78 mL, which should be converted to liters (L). moles of OH- ions = (1.002 M)(20.78 mL × (1 L / 1000 mL))

Step 2: Calculate moles of acetic acid in the vinegar.

Since acetic acid and OH- ions react in a 1:1 ratio during titration, the moles of acetic acid present in the vinegar will be equal to the moles of OH- ions used in the titration. Hence, moles of acetic acid = moles of OH- ions

Step 3: Determine the mass of acetic acid in 25.00 mL of the vinegar.

To find the mass of acetic acid, we multiply the moles of acetic acid by its molar mass. The molar mass of acetic acid, CH3COOH, is 12.01 (C) + 4.03 (H) + 16.00 (O) + 16.00 (O) + 1.01 (H) = 60.05 g/mol. mass of acetic acid = moles of acetic acid × molar mass of acetic acid

Step 4: Calculate the mass of 25.00 mL of vinegar.

The mass of a given volume of the vinegar can be calculated by multiplying the volume by its density. The density of the sample is 1.01 g/mL. mass of 25.00 mL vinegar = 25.00 mL × 1.01 g/mL

Step 5: Calculate the percentage by mass of acetic acid in the vinegar.

Finally, to express the concentration of acetic acid as a percentage by mass, we will use the following formula: percent by mass = (mass of acetic acid / mass of vinegar) × 100 By calculating each of the previous steps, we can find the mass percentage of acetic acid in the vinegar sample.

Key Concepts

Molarity

Understanding molarity is key to working with solutions in chemistry, particularly in titration exercises. Molarity is a measure of the concentration of a solute in a solution and is denoted by the symbol 'M'. It is defined as the number of moles of solute divided by the volume of the solution in liters. When we express the concentration of a solution, we are allowing ourselves to predict how a solute will react given the volume of the solution. For our acetic acid example, knowing that the molarity of sodium hydroxide (NaOH) is 1.002 M helps us understand the amount of solute in a given volume of liquid and is instrumental in determining how much titrant is needed to neutralize the acid.

In simpler terms, when we talk about 1.002 M NaOH, it implies that in every liter of the sodium hydroxide solution, there is 1.002 moles of NaOH present. This could also be translated into millimoles per milliliter, which is often more practical in laboratory settings where volumes are smaller.

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. In the context of acid-base titration, stoichiometry dictates the mole-to-mole ratios in which reactants combine, allowing us to predict the outcomes of reactions based on the balanced chemical equation. For example, in the titration of acetic acid with sodium hydroxide, the stoichiometry is a straightforward 1:1 ratio; for every mole of acetic acid (CH3COOH), one mole of hydroxide (OH-) is required to neutralize it. Understanding this core concept is fundamental because it allows us to directly relate the amount of NaOH used to the amount of acetic acid in the vinegar sample. Without a correct stoichiometric interpretation of the reaction, calculating the concentration of acetic acid would be impossible.

Molar Mass

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It's the sum of the atomic masses of all the atoms in a molecule and is crucial for converting between moles and grams, a common task in chemistry problems like the one in question.

Calculating Molar Mass

For acetic acid (CH3COOH), we calculate its molar mass by adding the atomic masses of each atom: 12.01 g/mol for carbon (C), 1.01 g/mol for hydrogen (H), and 16.00 g/mol for oxygen (O). This allows us to determine the mass in grams of one mole of acetic acid, which is necessary for converting moles to grams when finding the mass percent of acetic acid in a solution.

Percent by Mass

Percent by mass, also known as weight percent or mass percent, is a way of expressing the concentration of a component in a mixture or solution. It is defined as the mass of the solute divided by the mass of the solution, then multiplied by 100 to get a percentage.

Finding Percent by Mass

In the titration problem, once the mass of acetic acid and the mass of the vinegar sample have been calculated, these values are used to determine the acetic acid's percentage by mass. This is commonly used in food chemistry to express the concentration of ingredients. In our example, to calculate the percent by mass of acetic acid in apple cider vinegar, divide the mass of acetic acid by the total mass of the vinegar sample and multiply by 100. This gives a direct and understandable measure of acetic acid concentration, which is particularly useful in applications like food chemistry, where ingredients must be quantified for consistency and regulation purposes.