\(\mathrm{KBr}\) is relatively soluble in water, yet its enthalpy of solution is \(+19.8 \mathrm{kJ} / \mathrm{mol} .\) Which of the following statements provides the best explanation for this behavior? (a) Potassium salts are always soluble in water. (b) The entropy of mixing must be unfavorable. (c) The enthalpy of mixing must be small compared to the enthalpies for breaking up water-water interactions and K-Brionic interactions. (d) KBr has a high molar mass compared to other salts like NaCl.

Short Answer

Expert verified
The best explanation for KBr's behavior with positive enthalpy of solution while being soluble in water is given by statement (c): The enthalpy of mixing must be small compared to the enthalpies for breaking up water-water interactions and K-Br ionic interactions.

Step by step solution

01

Statement (a) explanation

Potassium salts are always soluble in water. This statement does not provide any explanation for the enthalpy of solution being positive (+19.8 kJ/mol). It only states KBr's solubility properties.
02

Statement (b) explanation

The entropy of mixing must be unfavorable. Entropy is a measure of disorder in a system. The dissolution process generally results in an increase in the disorder of the system due to mixing. This statement is contradictory since KBr is soluble in water, which indicates that the entropy of mixing must be favorable, not unfavorable.
03

Statement (c) explanation

The enthalpy of mixing must be small compared to the enthalpies for breaking up water-water interactions and K-Br ionic interactions. Enthalpy of solution is the sum of: 1. Enthalpy for breaking up water-water interactions (enthalpy of hydration for water). 2. Enthalpy for breaking up K-Br ionic interactions (lattice enthalpy). 3. Enthalpy for forming new interactions between water molecules and solute ions (enthalpy of hydration for solute ions). A positive enthalpy of solution means that the process is endothermic, which means it absorbs energy from its surroundings. This statement suggests that the favorable enthalpy of mixing is relatively small compared to the endothermic processes of breaking up water-water and K-Br ionic interactions. Therefore, the overall enthalpy of solution is positive. This statement provides a reasonable explanation for the behavior of KBr with a positive enthalpy of solution.
04

Statement (d) explanation

KBr has a high molar mass compared to other salts like NaCl. This statement doesn't directly explain the connection between KBr's solubility in water and its positive enthalpy of solution. Molar mass doesn't necessarily correlate with solubility or enthalpy of solution.
05

Conclusion

The best explanation for KBr's behavior with positive enthalpy of solution while being soluble in water is given by statement (c): The enthalpy of mixing must be small compared to the enthalpies for breaking up water-water interactions and K-Br ionic interactions.

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

Most fish need at least 4 ppm dissolved \(\mathrm{O}_{2}\) in water for survival. (a) What is this concentration in mol/L? (b) What partial pressure of \(\mathrm{O}_{2}\) above water is needed to obtain 4 \(\mathrm{ppm} \mathrm{O}_{2}\) in water at \(10^{\circ} \mathrm{C} ?\) (The Henry's law constant for \(\mathrm{O}_{2}\) at this temperature is \(1.71 \times 10^{-3} \mathrm{mol} / \mathrm{L}\) -atm.)

A supersaturated solution of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) is made by dissolving sucrose in hot water and slowly letting the solution cool to room temperature. After a long time, the excess sucrose crystallizes out of the solution. Indicate whether each of the following statements is true or false: (a) After the excess sucrose has crystallized out, the remaining solution is saturated. (b) Alter the excess sucrose has crystallized out, the system is now unstable and is not in equilibrium. (c) After the excess sucrose has crystallized out, the rate of sucrose molecules leaving the surface of the crystals to be hydrated by water is equal to the rate of sucrose molecules in water attaching to the surface of the crystals.

Which of the following in each pair is likely to be more soluble in water: (a) cyclohexane (C. \(\mathrm{H}_{12}\) ) or glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) (b) propionic acid \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right)\) or sodium propionate \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COONa}\right),(\mathbf{c}) \mathrm{HCl}\) or ethyl chloride \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}\right) ?\) Explain in each case.

Choose the best answer: A colloidal dispersion of one liquid in another is called (a) a gel, (b) an emulsion, (c) a foam, (d) an aerosol.

If you compare the solubilities of the noble gases in water, you find that solubility increases from smallest atomic weight to largest, 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 hydrogen-bond with water. (d) The heavier the gas, the more likely it is to make a saturated solution in water.

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