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Soaps consist of compounds such as sodium stearate, \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{16} \mathrm{COO}-\mathrm{Na}^{+},\) that have both hydrophobic and hydrophilic parts. Consider the hydrocarbon part of sodium stearate to be the "tail" and the charged part to be the "head." (a) Which part of sodium stearate, head or tail, is more likely to be solvated by water? (b) Grease is a complex mixture of (mostly) hydrophobic compounds. Which part of sodium stearate, head or tail, is most likely to bind to grease? (c) If you have large deposits of grease that you want to wash away with water, you can see that adding sodium stearate will help you produce an emulsion. What intermolecular interactions are responsible for this?

Short Answer

Expert verified
In summary, the charged head of sodium stearate is more likely to be solvated by water, while the hydrophobic hydrocarbon tail is more likely to bind with grease. The formation of an emulsion occurs due to hydrophobic interactions between the tails and grease, and electrostatic (polar) interactions between the charged heads and water. This allows grease and water to mix, which is a key property of soaps.

Step by step solution

01

Part (a): Solvation of sodium stearate by water

To determine which part of sodium stearate is more likely to be solvated by water, we need to examine the hydrophobic and hydrophilic parts of the molecule. The hydrocarbon tail (methyl group) is hydrophobic, while the charged head (carboxylate group) is hydrophilic. Since water is a polar solvent, it will interact more favorably with the hydrophilic part of the molecule. Therefore, the charged head of sodium stearate is more likely to be solvated by water.
02

Part (b): Interaction between sodium stearate and grease

To determine which part of sodium stearate is more likely to bind to grease, we need to consider the properties of grease. Grease is a mixture of hydrophobic compounds. Thus, the hydrophobic part of sodium stearate - the hydrocarbon tail - is more likely to interact and bind with grease.
03

Part (c): Intermolecular interactions responsible for emulsion formation

When sodium stearate is added to a mixture of grease and water, it promotes the formation of an emulsion. This is due to the surfactant properties of sodium stearate. In an emulsion, the hydrocarbon tails of sodium stearate molecules interact with the grease, while the charged heads interact with water. This allows the grease and water to mix, leading to an emulsion. The intermolecular interactions responsible for this are hydrophobic interactions between the tails and grease, and electrostatic (polar) interactions between the charged heads and water.

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Key Concepts

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

Hydrophobic and Hydrophilic Interactions
Understanding the behavior of substances like soap in water involves exploring hydrophobic and hydrophilic interactions. Sodium stearate, a common soap component, is a molecule that has two distinct parts: a long hydrocarbon tail that is hydrophobic (water-fearing) and a charged head that is hydrophilic (water-loving).

When sodium stearate encounters water, the hydrophilic heads readily solvate, meaning they interact and mix with the water due to their polarity which is similar to that of water molecules. This attractive interaction between water and hydrophilic substances is largely due to hydrogen bonding and dipole-dipole forces.

In contrast, the hydrophobic tails repel water. They have little to no affinity for it because of their nonpolar nature. They prefer to bind to other nonpolar substances, such as oils and greases. This fundamental concept of hydrophobic and hydrophilic interactions is key to understanding how sodium stearate functions as a soap, which is crucial in the creation of emulsions, where the two types of substances are dispersed within each other.
Intermolecular Forces
The interactions between different parts of a soap molecule and their surrounding environment are governed by intermolecular forces. These forces are the attractions or repulsions between molecules that dictate how a substance behaves in different contexts. There are several kinds of intermolecular forces, with the most common being van der Waals forces, hydrogen bonds, and ionic or electrostatic interactions.

Van der Waals forces encompass various weak attractions, including London dispersion forces, which play a significant role in the hydrophobic interactions between the nonpolar tails of sodium stearate and grease. Hydrogen bonds are relatively stronger forces that can occur between the hydrophilic heads of sodium stearate and water molecules. Electrostatic interactions between positive and negative charges are what allow the ionic head of sodium stearate to dissolve, or solvate, in water.

These forces collectively enable substances with different properties to interact in ways that can be harnessed in practical applications, like cleaning, where the differing affinities for water and grease are essential.
Surfactants in Chemistry
Surfactants, or surface-active agents, serve as mediators between two substances that typically do not mix well, such as oil and water. Sodium stearate is a classic example of a surfactant due to its amphiphilic nature—it has both a hydrophilic head and a hydrophobic tail. The hydrophilic head gravitates toward water, whereas the hydrophobic tail prefers the company of oils and greases.

In an emulsion, surfactants like sodium stearate facilitate the dispersion of one substance within another by reducing the surface tension at the water–oil interface. This property is vital in a myriad of products, from detergents and shampoos to emulsifiers in food. Understanding surfactants and their role in chemistry is invaluable not only for tasks like removing grease but also for complex industrial processes, pharmaceuticals, and the creation of various nanotechnologies.

Through their unique molecular structures, surfactants can surround nonpolar substances like oil, allowing them to become suspended in polar solvents like water, hence forming stable emulsions—a principle that is crucial to the cleansing action of soaps.

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Most popular questions from this chapter

(a) Do colloids made only of gases exist? Why or why not? (b) In the 1850 s, Michael Faraday prepared ruby-red colloids of gold nanoparticles in water that are still stable today. These brightly colored colloids look like solutions.What experiment(s) could you do to determine whether a given colored preparation is a solution or colloid?

When ammonium chloride dissolves in water, the solution becomes colder. (a) Is the solution process exothermic or endothermic? (b) Why does the solution form?

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 \(\mathrm{M} \mathrm{HCl}\) .

(a) Would you expect stearic acid, \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{16} \mathrm{COOH},\) to be more soluble in water or in carbon tetrachloride? (b) Which would you expect to be more soluble in water, cyclohexane or dioxane?

Describe how you would prepare each of the following aqueous solutions, starting with solid KBr: (a) 0.75 L of \(1.5 \times 10^{-2} M \mathrm{KBr},\) (b) 125 \(\mathrm{g}\) of \(0.180 \mathrm{m} \mathrm{KBr},(\mathbf{c}) 1.85 \mathrm{L}\) of a solution that is 12.0\(\% \mathrm{KBr}\) by mass (the density of the solution is 1.10 \(\mathrm{g} / \mathrm{mL}\) , ( \(\mathrm{d}\) ) a 0.150 \(\mathrm{M}\) solution of \(\mathrm{KBr}\) that contains just enough KBr to precipitate 16.0 \(\mathrm{g}\) of AgBr from a solution containing 0.480 mol of \(\mathrm{AgNO}_{3} .\)

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