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Chapter 13: Problem 16

Indicate the principal type of solute-solvent interaction in each of the following solutions and rank the solutions from weakest to strongest solute- solvent interaction: (a) KCl in water, (b) \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) in benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right),\) (c) methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) in water.

Short Answer

Expert verified
The principal type of solute-solvent interaction for each solution is: (a) KCl in water: Ion-dipole interaction (b) \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) in benzene: London Dispersion Forces (LDFs) (c) methanol in water: Hydrogen bonding The solutions are ranked from weakest to strongest solute-solvent interaction as: \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) in benzene (LDFs) < methanol in water (hydrogen bonding) < KCl in water (ion-dipole).

Step by step solution

01

Identify the principal type of solute-solvent interaction for each solution

The three solutions given are: (a) KCl in water (b) \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) in benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) (c) methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) in water For (a), KCl is an ionic compound and water is a polar solvent. The principal interaction in this case is ion-dipole interaction. For (b), \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) and benzene are nonpolar compounds, so the principal interaction in this case is London Dispersion Forces (LDFs). For (c), methanol and water are both polar molecules, and both can form hydrogen bonds. Thus, the principal interaction in this case is hydrogen bonding.
02

Rank the solutions based on solute-solvent interaction strength

We have three types of interactions in the given solutions: 1. Ion-Dipole interaction (KCl in water) 2. London Dispersion Forces (LDFs) (\(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) in benzene) 3. Hydrogen Bonding (methanol in water) In general, hydrogen bonding is stronger than dipole-dipole interactions, and dipole-dipole interactions are stronger than London Dispersion Forces. Ion-dipole interactions are even stronger than hydrogen bonds. So, we can rank them as follows: Weakest interaction: \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) in benzene (LDFs) Middle-strength interaction: methanol in water (hydrogen bonding) Strongest interaction: KCl in water (ion-dipole) Thus, the solutions are ranked from weakest to strongest solute-solvent interaction as: \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) in benzene < methanol in water < KCl in water.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ion-Dipole Interactions
Ion-dipole interactions are strong forces that occur between an ion and a polar molecule. These interactions are a key concept in understanding how ionic compounds dissolve in polar solvents, like water.

When an ionic compound such as potassium chloride (KCl) is added to water, the positive ends of the water molecules are attracted to the chloride ions (Cl鈦�), while the negative ends are attracted to the potassium ions (K鈦�).
  • This orientation creates a strong association because water, a polar molecule, has a permanent dipole due to its hydrogen and oxygen atoms.
  • The strength of ion-dipole interactions depends on the charge density of the ion and the polarity of the solvent molecule.
These interactions are crucial because they explain why many ionic compounds, which have fully charged ions, dissolve more readily in polar solvents than in nonpolar solvents.
London Dispersion Forces
London Dispersion Forces (LDFs) are the weakest type of intermolecular forces. Despite being weak, they are universal and act between all atoms and molecules, regardless of their polarity.

In solutions like dichloromethane (\(\mathrm{CH}_{2} \mathrm{Cl}_{2}\)) in benzene, LDFs are the primary source of interaction.
  • LDFs occur because electrons in atoms are constantly moving, creating temporary dipoles that induce dipoles in neighboring atoms or molecules.
  • The strength of these forces increases with larger and more massive atoms or molecules due to their increased polarizability.
Even though LDFs are weak compared to other forces like hydrogen bonding and ion-dipole interactions, they are significant for understanding phenomena like boiling points and solubility in nonpolar solvents.
Hydrogen Bonding
Hydrogen bonding is a special type of dipole-dipole attraction that occurs when hydrogen is bonded to a highly electronegative atom, such as oxygen, nitrogen, or fluorine. This creates a strong dipole due to the large difference in electronegativities of the hydrogen and its bonded partner.

In a solution like methanol (\(\mathrm{CH}_{3} \mathrm{OH}\)) in water, hydrogen bonds form between the hydrogen of methanol and the oxygen of water molecules.
  • Hydrogen bonds are significantly stronger than other dipole-dipole interactions, making them critical in biological systems and many chemical properties.
  • This interaction contributes to the high boiling point and solubility of substances like water and alcohols.
Understanding hydrogen bonding is essential, as it explores the way molecules interact at a macro level, affecting properties like solubility, melting points, and boiling points.

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Most popular questions from this chapter

Describe how you would prepare each of the following aqueous solutions: (a) 1.50 \(\mathrm{L}\) of 0.110 \(\mathrm{M}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) solution, starting with solid \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} ;\) (b) 225 \(\mathrm{g}\) of a solution that is 0.65 \(\mathrm{m}\) in \(\mathrm{Na}_{2} \mathrm{CO}_{3},\) starting with the solid solute; ( c ) 1.20 L of a solution that is 15.0\(\% \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) by mass (the density of the solution is 1.16 \(\mathrm{g} / \mathrm{mL}\) , starting with solid solute; (\boldsymbol{d} ) ~ a ~ 0.50 \(\mathrm{M}\) solution of HCl that would just neutralize 5.5 \(\mathrm{g}\) of \(\mathrm{Ba}(\mathrm{OH})_{2}\) starting with 6.0 \(\mathrm{M} \mathrm{HCl}\) .

The maximum allowable concentration of lead in drinking water is 9.0 ppb. (a) Calculate the molarity of lead in a 9.0- ppb solution. (b) How many grams of lead are in a swimming pool containing 9.0 ppb lead in 60 \(\mathrm{m}^{3}\) of water?

During a person's typical breathing cycle, the \(\mathrm{CO}_{2}\) concentration in the expired air rises to a peak of 4.6\(\%\) by volume. (a) Calculate the partial pressure of the CO \(_{2}\) in the expired air at its peak, assuming 1 atm pressure and a body temperature of \(37^{\circ} \mathrm{C}\) (b) What is the molarity of the \(\mathrm{CO}_{2}\) in the expired air at its peak, assuming a body temperature of \(37^{\circ} \mathrm{C} ?\)

The solubility of \(\mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3} \cdot 9 \mathrm{H}_{2} \mathrm{O}\) in water is 208 \(\mathrm{g}\) per 100 \(\mathrm{g}\) of water at \(15^{\circ} \mathrm{C}\) . A solution of \(\mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3} \cdot 9 \mathrm{H}_{2} \mathrm{O}\) in water at \(35^{\circ} \mathrm{C}\) is formed by dissolving 324 \(\mathrm{g}\) in 100 \(\mathrm{g}\) of water. When this solution is slowly cooled to \(15^{\circ} \mathrm{C},\) no precipitate forms. (a) Is the solution that has cooled down to \(15^{\circ}\) Cunsaturated, saturated, or supersaturated? (b) You take a metal spatula and scratch the side of the glass vessel that contains this cooled solution, and crystals start to appear. What has just happened? (c) At equilibrium, what mass of crystals do you expect to form?

Oil and water are immiscible. Which is the most likely reason? (a) Oil molecules are denser than water. (b) Oil molecules are composed mostly of carbon and hydrogen. (c) Oil molecules have higher molar masses than water. (d) Oil molecules have higher vapor pressures than water. (e) Oil molecules have higher boiling points than water.
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