/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 a. What is solution equilibrium?... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

a. What is solution equilibrium? b. What factors determine the point at which a given solute-solvent combination reaches equilibrium?

Short Answer

Expert verified
Solution equilibrium is when the rates of dissolution and precipitation are equal. Factors include temperature, solute-solvent nature, and pressure (for gases).

Step by step solution

01

Define Solution Equilibrium

Solution equilibrium occurs when the rate of dissolution of a solute in a solvent equals the rate of precipitation of the solute from the solution. At this point, the concentration of dissolved solite remains constant over time.
02

Identify Factors Determining Equilibrium

The factors that determine the point at which a solute-solvent combination reaches equilibrium include temperature, the nature of the solute and solvent, and pressure (for gases). Temperature affects the kinetic energy of the molecules, while the nature of the solute and solvent determines the interaction strength between them. For gases, pressure influences how much gas dissolves in the solvent.

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 91Ó°ÊÓ!

Key Concepts

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

dissolution
Dissolution is the process where a solute dissolves in a solvent to form a homogeneous mixture known as a solution. When a solute dissolves, its molecules or ions separate and disperse uniformly throughout the solvent. This occurs due to the attractive forces between solute and solvent particles. For example, when you mix salt in water, the salt dissolves by breaking into individual ions that spread evenly throughout the water.

Several factors affect the rate and extent of dissolution:

  • Temperature: Higher temperatures usually increase the dissolution rate because the kinetic energy of molecules increases, causing solute particles to disperse faster.
  • Stirring: Stirring a solution helps distribute solute particles evenly and speeds up the dissolution process.
  • Surface area: Finely divided solutes dissolve quicker than larger chunks because more particles are exposed to the solvent.
solute-solvent interactions
Solute-solvent interactions play a crucial role in the formation of solutions. These interactions determine whether a solute will dissolve in a particular solvent and how well it will dissolve. There are three primary types of interactions:

  • Ionic Solvation: When ionic compounds like salt dissolve, the ions are surrounded by solvent molecules. For instance, in water, the positive areas of water molecules (hydrogen) are attracted to negatively charged ions, and the negative areas (oxygen) are attracted to positively charged ions.
  • Hydrogen Bonding: Solvents like water have strong hydrogen bonds, which can attract and dissolve solutes that are capable of forming hydrogen bonds, such as sugar.
  • Van der Waals forces: Nonpolar solutes dissolve in nonpolar solvents through Van der Waals forces. For instance, oil (nonpolar) dissolves in hexane (nonpolar) due to these weak intermolecular attractions.


The general principle is 'like dissolves like', meaning polar solvents like water dissolve polar solutes, and nonpolar solvents dissolve nonpolar solutes.
equilibrium factors
Solution equilibrium is reached when the rate of dissolution equals the rate of precipitation. Several factors influence when this equilibrium is achieved:

  • Temperature: Higher temperatures can either increase or decrease solubility, depending on the nature of the solute-solvent interaction. For most solid solutes in liquids, solubility increases with temperature.
  • Nature of the Solute and Solvent: Different chemicals interact differently. Stronger interactions favor easier and more extensive dissolution, impacting when equilibrium is reached.
  • Pressure (for Gases): For gaseous solutes, increased pressure increases solubility. This behavior is represented by Henry’s Law, which states that the solubility of a gas in a liquid is proportional to the pressure of the gas above the liquid.


Understanding these equilibrium factors helps in predicting and controlling the solubility of substances in various conditions.

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

Predicting Outcomes You have been investigating the nature of suspensions, colloids, and solutions and have collected the following observational data on four unknown samples. From the data, infer whether each sample is a solution, suspension, or colloid. $$ \begin{array}{l|l|l|l|l} \text { Sample } & \text { Color } & \begin{array}{l} \text { Clarity (clear } \\ \text { or cloudy) } \end{array} & \begin{array}{l} \text { Settle } \\ \text { out } \end{array} & \begin{array}{l} \text { Tyndall } \\ \text { effect } \end{array} \\ \hline 1 & \text { green } & \text { clear } & \text { no } & \text { no } \\ \hline 2 & \text { blue } & \text { cloudy } & \text { yes } & \text { no } \\\ \hline 3 & \text { colorless } & \text { clear } & \text { no } & \text { yes } \\ \hline 4 & \text { white } & \text { cloudy } & \text { no } & \text { yes } \end{array} $$ Based on your inferences in Data Table 1 , you decide to conduct one more test of the particles. You filter the samples and then reexamine the filtrate. You obtain the data found in Data Table 2 . Infer the classifications of the filtrate based on the data in Data Table 2 . $$ \begin{array}{|l|l|l|l|l|} \hline \text { Sample } & \text { Color } & \begin{array}{l} \text { Clarity } \\ \text { (clear or } \\ \text { cloudy) } \end{array} & \begin{array}{l} \text { On } \\ \text { filter } \\ \text { paper } \end{array} & \begin{array}{l} \text { Tyndall } \\ \text { effect } \end{array} \\ \hline 1 & \text { green } & \text { clear } & \text { nothing } & \text { no } \\ \hline 2 & \text { blue } & \text { cloudy } & \begin{array}{l} \text { gray } \\ \text { solid } \end{array} & \text { yes } \\ \hline 3 & \text { colorless } & \text { clear } & \text { none } & \text { yes } \\ \hline 4 & \text { colorless } & \text { clear } & \text { white } & \text{ no } \\ \hline \end{array} $$

Given an unknown mixture consisting of two or more substances, explain how you could determine whether that mixture is a true solution, a colloid, or a suspension.

Suppose you wanted to know how many grams of KCl would be left if 350 mL of a 2.0 M KCl solution were evaporated to dryness. a. What is the molar mass of KCl? b. How would heating the solution affect the mass of KCl remaining? c. How many grams of KCl would remain?

A solution is made by dissolving 26.42 g of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) in enough \(\mathrm{H}_{2} \mathrm{O}\) to make 50.00 \(\mathrm{mL}\) of solution. a. What is the molar mass of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) ? b. What is the molarity of this solution?

Explain why a suspension is considered a heterogeneous mixture.

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.