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
(a) The head (charged part \(\mathrm{COO}^{-} \mathrm{Na}^{+}\)) of sodium stearate is hydrophilic and more likely to be solvated by water. (b) The tail (hydrocarbon part \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{16}\)) of sodium stearate is hydrophobic and more likely to bind to grease. (c) Emulsion formation is due to hydrogen bonding and ion-dipole interactions between sodium stearate's head and water molecules, and London dispersion forces between sodium stearate's tail and grease molecules.

Step by step solution

01

Understanding Sodium Stearate Structure

Sodium stearate has a hydrocarbon part (tail) \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{16}\) and a charged part (head) \(\mathrm{COO}^{-} \mathrm{Na}^{+}\). The tail is a long-chain hydrocarbon, which is hydrophobic (non-polar), while the head is ionic, consisting of a carboxylate group and a sodium ion, and it is hydrophilic (polar).
02

Solvation by water

(a) Solvation mainly depends on the compatibility of substances. Water is a polar solvent; thus, it solvates polar or hydrophilic parts of a compound. In the case of sodium stearate, its head (the charged part \(\mathrm{COO}^{-} \mathrm{Na}^{+}\)) is hydrophilic and more likely to be solvated by water.
03

Grease binding

(b) Grease, a complex mixture of hydrophobic compounds, will interact more with hydrophobic compounds or non-polar parts of a molecule. In sodium stearate, the tail (the hydrocarbon part \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{16}\)) is hydrophobic and more likely to bind to grease.
04

Emulsion formation

(c) Introducing sodium stearate into a mixture of grease and water creates an emulsion, in which small droplets of grease are surrounded by sodium stearate molecules. The intermolecular interactions responsible for this process are: 1. Hydrogen bonding and ion-dipole interaction between the sodium stearate's head and water molecules. 2. London dispersion force between the sodium stearate's tail and grease molecules. These interactions result in the formation of an emulsion, thus allowing the removal of grease while washing with water.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider two solutions, one formed by adding \(150 \mathrm{~g}\) of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) to $1 \mathrm{~L}\( of water and the other formed by adding \)150 \mathrm{~g}$ of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) to $1 \mathrm{~L}$ of water. Calculate the vapor pressure for each solution at \(25^{\circ} \mathrm{C} ;\) the vapor pressure of pure water at this temperature is \(3.17 \mathrm{kPa}\).

A sulfuric acid solution containing \(697.6 \mathrm{~g}\) of $\mathrm{H}_{2} \mathrm{SO}_{4}\( per liter of solution has a density of \)1.395 \mathrm{~g} / \mathrm{cm}^{3} .$ Calculate (a) the mass percentage, \((\mathbf{b})\) the mole fraction, (c) the molality, \((\mathbf{d})\) the molarity of $\mathrm{H}_{2} \mathrm{SO}_{4}$ in this solution.

Consider water and glycerol, $\mathrm{CH}_{2}(\mathrm{OH}) \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{OH}$. (a) Would you expect them to be miscible in all proportions? (b) List the intermolecular attractions that occur between a water molecule and a glycerol molecule.

If you compare the solubilities of the noble gases in water, you find that solubility increases from smallest atomic weight to largest, \(\mathrm{Ar}<\mathrm{Kr}<\mathrm{Xe}\). Which of the following statements is the best explanation? [Section 13.3] (a) The heavier the gas, the more it sinks to the bottom of the water and leaves room for more gas molecules at the top of the water. (b) The heavier the gas, the more dispersion forces it has, and therefore the more attractive interactions it has with water molecules. (c) The heavier the gas, the more likely it is to hydrogenbond with water. (d) The heavier the gas, the more likely it is to make a saturated solution in water.

A dilute aqueous solution of fructose in water is formed by dissolving $1.25 \mathrm{~g}\( of the compound in water to form \)0.150 \mathrm{~L}$ of solution. The resulting solution has an osmotic pressure of \(112.8 \mathrm{kPa}\) at \(20^{\circ} \mathrm{C}\). Assuming that the organic compound is a nonelectrolyte, what is its molar mass?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free