Question

Acetic acid is a polar molecule and can form hydrogen bonds with water molecules. Therefore, it has a high solubility in water. Yet acetic acid is also soluble in benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\), a nonpolar solvent that lacks the ability to form hydrogen bonds. A solution of \(3.8 \mathrm{~g}\) of \(\mathrm{CH}_{3} \mathrm{COOH}\) in \(80 \mathrm{~g} \mathrm{C}_{6} \mathrm{H}_{6}\) has a freezing point of \(3.5^{\circ} \mathrm{C}\). Calculate the molar mass of the solute, and suggest what its structure might be. (Hint: Acetic acid molecules can form hydrogen bonds between themselves.)

Short answer

The molar mass suggests the solute is likely a dimer of acetic acid.

Step-by-step solution

Determine the Freezing Point Depression

The freezing point of pure benzene is known to be around 5.5°C. Given that the freezing point of the benzene solution is 3.5°C, we can find the depression in the freezing point using: \[ \Delta T_f = T_f^ \circ - T_f \]where \(T_f^ \circ = 5.5\,^{\circ} \mathrm{C}\), and \(T_f = 3.5\,^{\circ} \mathrm{C}\). Thus, \(\Delta T_f = 5.5 - 3.5 = 2.0\,^{\circ} \mathrm{C}\).

Use the Freezing Point Depression Equation

The freezing point depression can be calculated using the formula:\[ \Delta T_f = i \cdot K_f \cdot m \]where \(i\) is the van 't Hoff factor, \(K_f\) is the freezing point depression constant for benzene (5.12 °C·kg/mol), and \(m\) is the molality of the solution.

Calculate Molality

Molality \(m\) is calculated using:\[ m = \frac{\text{moles of solute}}{\text{kg of solvent}} \]Since we need to find the molar mass first, express the moles of solute in terms of the molar mass \(M\):\[ m = \frac{3.8}{M \times 0.080} \]

Substitute Values and Solve for Molar Mass

Substitute \(m\) from the previous step into the freezing point depression formula:\[ 2.0 = i \times 5.12 \times \frac{3.8}{M \times 0.080} \]Assume \(\(i = 1\)\) for monomeric acetic acid, simplify and solve for \(M\):\[ 2.0 = 5.12 \times \frac{3.8}{0.080M} \]Calculate \(M\) to find the molar mass.

Suggest the Structure

Calculate the molar mass using the equation. If the calculated molar mass matches approximately the dimer's mass of acetic acid (due to hydrogen bonding), suggest that acetic acid forms dimers in benzene.

Key concepts

Acetic Acid Solubility

Acetic acid (\(\text{CH}_3\text{COOH}\)) is known for its ability to dissolve well in water, a highly polar solvent. This is primarily due to its polar nature and the capability to form hydrogen bonds with water molecules. Large solubility is driven by:

• Hydrogen bonding between acetic acid and water.
• Polarity of acetic acid complementing water's properties.

Interestingly, acetic acid is also soluble in benzene (\(\text{C}_6\text{H}_6\)), a nonpolar solvent lacking hydrogen bonding capabilities. How does acetic acid dissolve in benzene then? Despite being polar, acetic acid tends to exist as dimers when in nonpolar solvents. These acetic acid dimer molecules reduce polarity, making them compatible with benzene's nonpolar environment. Therefore, even without conventional hydrogen bonding with benzene, acetic acid maintains solubility through intermolecular attractions within its dimer structure.

Hydrogen Bonding

Hydrogen bonding is a powerful intermolecular force that significantly affects the properties of molecules, such as solubility and boiling points. In acetic acid, hydrogen bonding occurs when the hydrogen from the -OH group gets attracted to the electronegative oxygen of another molecule's carbonyl group (\(\text{C}=\text{O}\)). This results in:

• Formation of acetic acid dimers.
• Stabilization of the liquid structure.

When in water, acetic acid forms hydrogen bonds with water molecules, enhancing solubility. However, in benzene, acetic acid can't form hydrogen bonds with the solvent but establishes intra-molecular hydrogen bonding by pairing up with other acetic acid molecules to form dimers. These dimers have a reduced net polarity, making them compatible with the nonpolar benzene. This self-association is a classic demonstration of hydrogen bonding's role in modifying molecule interactions and states.

Molar Mass Calculation

A crucial part of understanding chemical solutions involves calculating the molar mass of solutes within them. When faced with problems involving freezing point depression, it's necessary to first determine how much the freezing point changes. In this exercise:

• Determine \(\Delta T_f\), the change in freezing point.
• Use \(\Delta T_f = i \cdot K_f \cdot m\), where \(i\) is the van 't Hoff factor, \(K_f\) is the solvent's freezing point depression constant, and \(m\) is molality.

Molality is key here, defined as moles of solute per kilogram of solvent. To find moles, solve for molar mass: express moles in terms of mass of solute and unknown molar mass (\(M\)). Substitute and solve the equation to find the molar mass. In this scenario, if acetic acid forms dimers in benzene, our calculated molar mass will likely reflect the combined mass of these dimer structures rather than the monomer, echoing acetic acid's unique intermolecular behavior.