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

Which salt, \(\mathrm{Li}_{2} \mathrm{SO}_{4}\) or \(\mathrm{Cs}_{2} \mathrm{SO}_{4}\), is expected to have the more exothermic heat of hydration? Explain briefly.

Short Answer

Expert verified
Li_2SO_4 has a more exothermic heat of hydration than Cs_2SO_4 due to the smaller size of Li^+ ions.

Step by step solution

01

Understand the Concept of Heat of Hydration

The heat of hydration is the energy change when ions are dissolved in water. It is typically exothermic, meaning energy is released. Smaller and more highly charged ions have a more exothermic heat of hydration because they interact more strongly with water molecules.
02

Analyze Cation Characteristics

Lithium ( Li^+ ) and cesium ( Cs^+ ) are both alkali metals, and their respective ions are part of Li_2SO_4 and Cs_2SO_4 salts. Li^+ ions are smaller in radius compared to Cs^+ ions, leading to stronger interactions with water molecules.
03

Compare Exothermicity

Due to the smaller size of Li^+ compared to Cs^+ , Li_2SO_4 is expected to have a more exothermic heat of hydration than Cs_2SO_4 because smaller ions are hydrated more exothermically.

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.

Li2SO4 vs Cs2SO4 comparison
When comparing \( \text{Li}_2\text{SO}_4 \) and \( \text{Cs}_2\text{SO}_4 \), it's crucial to look at the ions involved. Both are composed of a sulfate ion \( (\text{SO}_4^{2-}) \) paired with alkali metal ions—lithium \( (\text{Li}^+) \) for \( \text{Li}_2\text{SO}_4 \) and cesium \( (\text{Cs}^+) \) for \( \text{Cs}_2\text{SO}_4 \). The primary difference is in the size and charge density of these ions.

\( \text{Li}^+ \), being a smaller ion, possesses a high charge density. This means its positive charge is concentrated over a smaller area, leading to stronger interactions with the surrounding water molecules. On the other hand, \( \text{Cs}^+ \) has a much larger ionic radius and thus a lower charge density. This results in weaker interaction with water.

Consequently, \( \text{Li}_2\text{SO}_4 \) is expected to have a more exothermic heat of hydration compared to \( \text{Cs}_2\text{SO}_4 \) because smaller ions typically release more energy when hydrated due to their higher charge-to-size ratio.
factors affecting exothermicity
The exothermicity of a salt's heat of hydration is influenced by several factors. Understanding these provides insights into why certain salts release more energy during hydration.

Here are key factors to consider:
  • Ion Size: Smaller ions, like \( \text{Li}^+ \), interact more strongly with water molecules because of their tight proximity. This results in higher hydration energy.
  • Charge Density: Ions with high charge density due to their small size or higher charge will have more exothermic hydration. \( \text{Li}^+ \)'s smaller size gives it a high charge density.
  • Polarizability: Although less significant here, polarizability can affect solvation. Larger ions, more easily polarized, might interact differently but often weakly compared to smaller, less polarizable ions.
In summary, smaller and more densely charged ions form more stable interactions with water, leading to a greater release of energy.
interaction with water molecules
The interaction between ions and water molecules is pivotal in determining the extent of energy released during hydration. Water is a polar molecule, meaning it has a partial positive charge near the hydrogen atoms and a partial negative charge near the oxygen atom. This polarity enables water to interact effectively with ions.

When an ion like \( \text{Li}^+ \) enters water, the negative ends of water molecules are attracted to its positive charge. As a result, these water molecules surround and effectively "solvate" the ion. This leads to the formation of a hydration shell. The formation of this shell is exothermic, releasing energy as bonds form between the water's oxygen atoms and the lithium ions.

In contrast, \( \text{Cs}^+ \) ions form weaker interactions due to their larger size and lower charge density. This results in a less tightly bound hydration shell, hence a less exothermic process. The energy dynamics in such interactions play a crucial role in understanding the solvation process and predicting the behavior of different salts in aqueous solutions.

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

Which pairs of liquids will be miscible? (a) \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) (b) \(\mathrm{C}_{6} \mathrm{H}_{6}\) (benzene) and \(\mathrm{CCl}_{4}\) (c) \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\)

Anthracene, a hydrocarbon obtained from coal, has an empirical formula of \(\mathrm{C}_{7} \mathrm{H}_{5} .\) To find its molecular formula you dissolve \(0.500 \mathrm{g}\) in \(30.0 \mathrm{g}\) of benzene. The boiling point of the pure benzene is \(80.10^{\circ} \mathrm{C},\) whereas the solution has a boiling point of \(80.34^{\circ} \mathrm{C} .\) What is the molecular formula of anthracene?

A \(35.0-\mathrm{g}\) sample of ethylene glycol, \(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH},\) is dissolved in \(500.0 \mathrm{g}\) of water. The vapor pressure of water at \(32^{\circ} \mathrm{C}\) is \(35.7 \mathrm{mm}\) Hg. What is the vapor pressure of the water-ethylene glycol solution at \(32^{\circ} \mathrm{C} ?\) (Ethylene glycol is nonvolatile.)

Calculate the osmotic pressure of a \(0.0120 \mathrm{M}\) solution of NaCl in water at \(0^{\circ}\) C. Assume the van't Hoff factor, \(i\), is 1.94 for this solution.

You make a saturated solution of \(\mathrm{NaCl}\) at \(25^{\circ} \mathrm{C} .\) No solid is present in the beaker holding the solution. What can be done to increase the amount of dissolved \(\mathrm{NaCl}\) in this solution? (See Figure 14.12.) (a) Add more solid NaCl. (b) Raise the temperature of the solution. (c) Raise the temperature of the solution and add some NaCl. (d) Lower the temperature of the solution and add some NaCl.

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