Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Hydrogen chloride (HCl) gas is much more soluble than propane gas \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) in water, even though \(\mathrm{HCl}\) has a lower boiling point. Explain.

Short Answer

Expert verified
\(\mathrm{HCl}\) is more soluble in water than \(\mathrm{C}_3\mathrm{H}_8\) because \(\mathrm{HCl}\) is polar and can form hydrogen bonds with water molecules.

Step by step solution

01

- Identify the Properties of \texttt{HCl}

Hydrogen chloride (\(\mathrm{HCl}\)) is a polar molecule. It can form hydrogen bonds with water due to the presence of hydrogen atoms and has permanent dipole-dipole attractions.
02

- Identify the Properties of \(\mathrm{C}_3\mathrm{H}_8\)

Propane ( \(\mathrm{C}_3\mathrm{H}_8\) ) is a nonpolar molecule. It cannot form hydrogen bonds and only exhibits weak van der Waals (dispersion) forces.
03

- Discuss Solubility Mechanisms

Water is a polar solvent, meaning it can dissolve polar substances more effectively through dipole-dipole interactions and hydrogen bonding. Nonpolar substances do not interact as strongly with water molecules.
04

- Conclusion Based on Polarity and Interactions

The greater solubility of \(\mathrm{HCl}\) compared to \(\mathrm{C}_3\mathrm{H}_8\) is due to its ability to form hydrogen bonds and engage in stronger dipole-dipole interactions with water molecules, despite \(\mathrm{HCl}\)'s lower boiling point.

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!

Key Concepts

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

Polar Molecules
Polar molecules have areas of positive and negative charge, which means they have a dipole moment. This happens because the electrons in the molecule are not evenly distributed. Water (H₂O) and hydrogen chloride (HCl) are both examples of polar molecules.
In polar molecules, one end of the molecule is slightly positive, while the other end is slightly negative. This difference in charge allows polar molecules to interact strongly with each other and with other polar substances.
For example:
  • Oxygen in H₂O is more electronegative than hydrogen, causing a partial negative charge on the oxygen atom.
  • Similarly, chlorine in HCl is more electronegative than hydrogen, causing a partial negative charge on the chlorine atom.
In the case of solubility, water is a polar solvent, which makes it great at surrounding and interacting with other polar molecules like HCl, but it is less effective with nonpolar molecules such as propane (C₃H₈).
Hydrogen Bonding
Hydrogen bonding is a strong type of dipole-dipole interaction that occurs between molecules where hydrogen is directly bonded to a more electronegative atom like oxygen, nitrogen, or fluorine. This bond causes hydrogen to have a slightly positive charge.
In water, hydrogen bonds play a crucial role in its properties. Each hydrogen atom of a water molecule can form a hydrogen bond with the oxygen atoms of neighboring water molecules.
Molecules like HCl can form hydrogen bonds with water because they have a hydrogen atom and a highly electronegative element (chlorine). These hydrogen bonds are stronger than other types of intermolecular forces, making the molecule highly soluble in polar solvents like water.
However, C₃H₈ cannot form such bonds due to its nonpolar nature, which makes it much less soluble in water. Thus, the ability to form hydrogen bonds significantly increases the solubility of substances in water.
Dipole-Dipole Interactions
Dipole-dipole interactions occur between polar molecules, where the positive end of one molecule is attracted to the negative end of another. These interactions are weaker than hydrogen bonds but stronger than Van der Waals forces.
When considering solubility, these interactions are critical. Water, being polar, has a high ability to interact with other polar molecules through dipole-dipole forces.
For example, because hydrogen chloride (HCl) is a polar molecule, it experiences strong dipole-dipole interactions with water molecules. As a result, it is highly soluble in water.
In contrast, propane (C₃H₈) is nonpolar and does not have a permanent dipole. Instead, it only exhibits weak dispersion forces, which do not interact strongly with water molecules. This is why propane is much less soluble in water compared to hydrogen chloride.
To summarize, dipole-dipole interactions significantly enhance the solubility of polar substances in polar solvents like water.

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

How many moles of solute particles are present in \(1 \mathrm{~L}\) of each of the following aqueous solutions? (a) \(0.3 M \mathrm{KBr}\) (b) \(0.065 \mathrm{M} \mathrm{HNO}_{3}\) (c) \(10^{-4} M \mathrm{KHSO}_{4}\) (d) \(0.06 M\) ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\)

A saturated \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) solution is prepared, and a small excess of solid is present (white pile in beaker). A seed crystal of \(\mathrm{Na}_{2}{ }^{14} \mathrm{CO}_{3}\left({ }^{14} \mathrm{C}\right.\) is a radioactive isotope of \({ }^{12} \mathrm{C}\) ) is added ( small red piece), and the radioactivity is measured over time. (a) Would you expect radioactivity in the solution? Explain. (b) Would you expect radioactivity in all the solid or just in the seed crystal? Explain.

The release of volatile organic compounds into the atmosphere is regulated to limit ozone formation. In a laboratory simulation, \(5 \%\) of the ethanol in a liquid detergent is released. Thus, a "down-the-drain" factor of 0.05 is used to estimate ethanol emissions from the detergent. The \(k_{\mathrm{H}}\) values for ethanol and 2-butoxyethanol \(\left(\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{OCH}_{2} \mathrm{CH}_{2} \mathrm{OH}\right)\) are \(5 \times 10^{-6} \mathrm{~atm} \cdot \mathrm{m}^{3} / \mathrm{mol}\) and \(1.6 \times 10^{-6} \mathrm{~atm} \cdot \mathrm{m}^{3} / \mathrm{mol}\), respectively. (a) Estimate a "down-the-drain" factor for 2 -butoxyethanol in the detergent. (b) What is the \(k_{\mathrm{H}}\) for ethanol in units of \(\mathrm{L} \cdot \mathrm{atm} / \mathrm{mol} ?\) (c) Is the value found in part (b) consistent with a value given as \(0.64 \mathrm{~Pa} \cdot \mathrm{m}^{3} / \mathrm{mol} ?\)

What are the mole fraction and the mass percent of a solution made by dissolving \(0.30 \mathrm{~g}\) of \(\mathrm{KI}\) in \(0.400 \mathrm{~L}\) of water \((d=1.00 \mathrm{~g} / \mathrm{mL}) ?\)

Why does the solubility of any gas in water decrease with rising temperature?

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