Question

For each of the following, choose the pair of substances you would expect to give the most ideal solution. Explain your choices. a. \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{6} \mathrm{CH}_{3}\) and \(\mathrm{CH}_{3}\left(\mathrm{CH}_{25} \mathrm{CH}_{3} \text { or } \mathrm{H}_{2} \mathrm{O} \text { and }\right.\) \(\mathrm{CH}_{3}-\mathrm{O}-\mathrm{CH}_{3}\) b. \(\mathrm{CH}_{3} \mathrm{OH}\) and \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) or \(\mathrm{CH}_{3} \mathrm{F}\) and \(\mathrm{HF}\)

Short answer

The most ideal solutions would be formed between: a) CH3(CH2)6CH3 and CH3(CH2)25CH3, as they have similar London dispersion forces due to their similar nonpolar structures. b) CH3OH and CH3CH2OH, as they have similar intermolecular forces, including hydrogen bonding, due to their similar molecular structures.

Step-by-step solution

a) Analyze the molecules

For the first pair, we have a saturated hydrocarbon (C10H22) and a larger saturated hydrocarbon (C26H54). For the second pair, we have water (H2O) and dimethyl ether (CH3-O-CH3).

a) Compare intermolecular forces

The two saturated hydrocarbons only have London dispersion forces between their molecules since they are nonpolar. As the London dispersion forces increase with the size of the molecules, the two substances will have similar intermolecular forces, making them more likely to form an ideal solution. In the second pair, water molecules form strong hydrogen bonds, while dimethyl ether molecules only have dipole-dipole interactions and induce-dipole forces. These two substances have significantly different intermolecular forces, making them less likely to form an ideal solution.

a) Choose the most ideal solution

Based on the similarities in the intermolecular forces, the pair of substances that would likely form the most ideal solution would be CH3(CH2)6CH3 and CH3(CH2)25CH3.

b) Analyze the molecules

For the first pair, we have methanol (CH3OH) and ethanol (CH3CH2OH). For the second pair, we have methyl fluoride (CH3F) and hydrogen fluoride (HF).

b) Compare intermolecular forces

Methanol and ethanol have similar molecular structures, and both are capable of forming hydrogen bonds; ethanol has an additional carbon and two hydrogens. Thus, both substances in the first pair have similar intermolecular forces, making them more likely to form an ideal solution. In the second pair, methyl fluoride has only dipole-dipole forces, while hydrogen fluoride forms strong hydrogen bonds. These two substances have significantly different intermolecular forces, making them less likely to form an ideal solution.

b) Choose the most ideal solution

Based on the similarities in the intermolecular forces, the pair of substances that would likely form the most ideal solution would be CH3OH and CH3CH2OH.

Key concepts

Intermolecular Forces

Intermolecular forces are the forces of attraction or repulsion between neighboring molecules. These forces play a crucial role in determining the physical properties of substances, such as boiling and melting points, and their ability to dissolve in each other to form solutions. The strength and nature of these forces depend on the type of molecular interactions present.

There are several types of intermolecular forces including:

• London Dispersion Forces: These are weak forces that result from temporary fluctuations in electron density in nonpolar molecules or noble gases. They increase with molecular size or mass.
• Dipole-Dipole Interactions: Occurring in polar molecules with permanent dipoles, these forces are stronger than dispersion forces but weaker than hydrogen bonds.
• Hydrogen Bonding: A special type of dipole-dipole interaction that occurs when hydrogen is covalently bonded to a highly electronegative atom, like oxygen, nitrogen, or fluorine.

Intermolecular forces influence how well different substances mix. In an ideal solution, the intermolecular forces between the different substances are similar, leading to homogeneous mixtures without significant changes in properties from the pure substances. This is why understanding these forces is key in predicting the behavior of mixtures.

Hydrogen Bonding

Hydrogen bonding is a special type of intermolecular force that is particularly strong compared to other dipole-dipole interactions. This force occurs when a hydrogen atom, which is covalently bonded to an electronegative atom such as nitrogen, oxygen, or fluorine, interacts with another electronegative atom in a nearby molecule.

Characteristics of hydrogen bonds include:

• High Directionality: They have a specific orientation leading to the alignment of the molecules.
• Significantly enhance boiling point: Due to their strength, substances with hydrogen bonds tend to have higher boiling points than those with mere dipole interactions or dispersion forces.
• Influence Solubility: Molecules capable of hydrogen bonding often dissolve in water, an excellent hydrogen-bonding solvent.

For example, in methanol and ethanol, hydrogen bonds play a major role. Both these alcohols can form hydrogen bonds due to the presence of the -OH (hydroxyl) group, allowing them to mix well and form ideal solutions with each other.

London Dispersion Forces

London dispersion forces are the weakest of all intermolecular forces but universally present between all atoms and molecules. They are also known as Van der Waals forces and arise due to temporary dipoles formed when the electron clouds of two adjacent atoms become distorted.

Key points about London dispersion forces:

• Depend on Molecular Size: These forces become more significant as the size and shape of a molecule increase. Larger molecules have more electrons that can be polarized, leading to stronger dispersion forces.
• Present in Nonpolar Molecules: While polar molecules have stronger interactions overall, nonpolar molecules like hydrocarbons rely primarily on dispersion forces to attract each other.
• Temperature Sensitivity: As temperature increases, the effect of these forces decreases because increased kinetic energy overcomes the attractions.

In solutions, substances like long-chain alkanes that only exhibit London dispersion forces can still form ideal solutions with each other if their sizes and shapes are similar, as seen with the pairing of C10H22 and C26H54 in the exercise example.