Problem 7 - Clinical Chemistry

Question

You must make \(1 \mathrm{~L}\) of \(0.2 \mathrm{M}\) acetic acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\). All you have available is concentrated glacial acetic acid (assay value, \(98 \%\); specific gravity, \(1.05\) \(\mathrm{g} / \mathrm{mL}\) ). It will take milliliters of acetic acid to make this solution. Assume a gram molecular weight of \(60.05\) grams. a. \(11.7\) b. \(1.029\) c. \(3.42\) d. \(12.01\)

Step-by-Step Solution

Answer

11.7

Step 1: Determine the moles of acetic acid needed

First, determine the moles of acetic acid needed to make 1 liter of a 0.2 M solution. Use the formula \[\text{Moles} = \text{Concentration} \times \text{Volume}\].Here, the concentration is 0.2 M and the volume is 1 L.\[\text{Moles} = 0.2 \text{ M} \times 1 \text{ L} = 0.2 \text{ moles} \]

Step 2: Calculate the mass of acetic acid required

Using the molecular weight of acetic acid (60.05 g/mol), calculate the mass needed.\[\text{Mass} = \text{Moles} \times \text{Molecular Weight}\]\[\text{Mass} = 0.2 \text{ moles} \times 60.05 \text{ g/mol} = 12.01 \text{ grams}\]

Step 3: Determine the volume of glacial acetic acid needed

Since glacial acetic acid is 98% pure, calculate the mass of pure acetic acid per mL of the glacial acetic acid. Using the density (1.05 g/mL), find how many grams of pure acetic acid there are in 1 mL.\[\text{Pure acetic acid} = 1.05 \text{ g/mL} \times 0.98 = 1.029 \text{ g/mL}\]

Step 4: Calculate the volume of glacial acetic acid needed

Divide the total mass of pure acetic acid required by the mass of pure acetic acid per mL of glacial acetic acid to find the volume needed.\[\text{Volume} = \frac{\text{Mass required}}{\text{Mass per mL of glacial acetic acid}} \]\[\text{Volume} = \frac{12.01 \text{ grams}}{1.029 \text{ g/mL}} = 11.67 \text{ mL}\]

Key Concepts

molarity calculations

Molarity is a measure of the concentration of a solute in a solution. It expresses the number of moles of a substance per liter of solution. The formula for calculating molarity (\textbf{M}) is \( \text{M} = \frac{\text{moles of solute}}{\text{liters of solution}} \). In this exercise, we wanted to make a specific solution: 1 liter of 0.2 M acetic acid. To find out how many moles of acetic acid we needed, we multiplied the concentration by the volume: \( \text{Moles} = \text{Concentration} \times \text{Volume} \). With a concentration of 0.2 M and a volume of 1 L, this gave us 0.2 moles of acetic acid. By understanding molarity calculations, you can prepare solutions with the exact concentration needed.

molecular weight

Molecular weight (also known as molar mass) is the mass of a given molecule. It is calculated by adding up the atomic masses of all the atoms in the molecule. For acetic acid (\textbf{CH}_3\text{COOH}), the molecular weight is calculated by summing the atomic weights of its constituent atoms: carbon (C), hydrogen (H), and oxygen (O). This is given as 60.05 grams per mole. In this exercise, we used the molecular weight to determine the mass of acetic acid needed: \( \text{Mass} = \text{Moles} \times \text{Molecular Weight} \). With 0.2 moles and a molecular weight of 60.05 g/mol, we calculated that we needed 12.01 grams of acetic acid. Understanding molecular weight is crucial for converting between moles and grams.

solution preparation

Solution preparation involves mixing the correct amounts of solute and solvent to achieve a desired concentration. In this exercise, after determining we needed 12.01 grams of acetic acid, we had to find out how much glacial acetic acid (which is 98% pure) was required. We first determined the mass of pure acetic acid per milliliter of the solution, using the density (1.05 g/mL) and purity (98%): \( 1.05 \text{ g/mL} \times 0.98 = 1.029 \text{ g/mL} \). Then, we divided the total mass needed (12.01 grams) by this value to find the volume of glacial acetic acid: \( \frac{12.01 \text{ grams}}{1.029 \text{ g/mL}} = 11.67 \text{ mL} \). This shows the importance of carefully measuring and calculating each component to prepare an accurate solution.