/*! 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 13 The common ion effect for ionic ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 16: Problem 13

The common ion effect for ionic solids (salts) is to significantly decrease the solubility of the ionic compound in water. Explain the common ion effect.

Short Answer

Expert verified
The common ion effect significantly decreases the solubility of ionic solids in water due to the presence of a common ion in the solution. As a result, the equilibrium shifts towards precipitation, reducing the solubility of the ionic compound. This can be explained using Le Ch芒telier's principle and the solubility product constant (Ksp). For example, when adding sodium fluoride (NaF) to a solution containing calcium fluoride (CaF2), the increase in fluoride ions (F-) causes the system to shift towards the solid side to reduce the F- concentration, ultimately leading to a decrease in the solubility of CaF2.

Step by step solution

01

Define Solubility and the Common Ion Effect

Solubility refers to the maximum amount of solute that can dissolve in a solvent at a given temperature. When an ionic solid (salt) dissolves in water, it dissociates into its constituent ions. The common ion effect occurs when two (or more) solutes, which have the same ions, are mixed in a solution. The presence of a common ion in the solution suppresses the ionization of the acid (or base) and affects the solubility of the ionic solids present.
02

Discuss the Role of Solubility Product Constant (Ksp)

The solubility product constant (Ksp) is used to describe the equilibrium between a solid compound and its dissolved ions in a saturated solution. The larger the Ksp, the more soluble the compound is. The common ion effect impacts the solubility of the ionic solid, and in most cases, it reduces the solubility of the compound. This is because, at equilibrium, the concentration of the common ion increases, which makes the system shift towards the precipitation side (according to Le Ch芒telier's principle), resulting in the decrease of solubility.
03

Provide an Example to Illustrate the Common Ion Effect

Let's take the example of calcium fluoride (CaF2) dissolving in water: CaF2(s) 鈬� Ca2+(aq) + 2F鈭�(aq) The solubility product constant (Ksp) for this reaction is given by: Ksp = [Ca2+][F鈭抅^2 Now, consider adding a common ion, such as sodium fluoride (NaF), to the solution: NaF(s) 鈫� Na+(aq) + F鈭�(aq) As there's an increase in the concentration of fluoride ions (F-) in the solution, the equilibrium will shift towards the solid side (formation of more CaF2) to reduce the concentration of F- ions, according to Le Ch芒telier's principle. This shift will lead to a decrease in the solubility of calcium fluoride (CaF2). In conclusion, the common ion effect for ionic solids (salts) significantly reduces the solubility of the ionic compound in water due to the presence of a common ion in the solution, which causes a shift in the equilibrium towards precipitation, as per Le Ch芒telier's principle.

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 91影视!

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

The concentration of \(\mathrm{Pb}^{2+}\) in a solution saturated with \(\mathrm{PbBr}_{2}(s)\) is \(2.14 \times 10^{-2} M .\) Calculate \(K_{\text {sp }}\) for \(\mathrm{PbBr}_{2}\).

Aluminum ions react with the hydroxide ion to form the precipitate \(\mathrm{Al}(\mathrm{OH})_{3}(s)\), but can also react to form the soluble complex ion \(\mathrm{Al}(\mathrm{OH})_{4}^{-} .\) In terms of solubility, \(\mathrm{Al}(\mathrm{OH})_{3}(s)\) will be more soluble in very acidic solutions as well as more soluble in very basic solutions. a. Write equations for the reactions that occur to increase the solubility of \(\mathrm{Al}(\mathrm{OH})_{3}(s)\) in very acidic solutions and in very basic solutions. b. Let's study the \(\mathrm{pH}\) dependence of the solubility of \(\mathrm{Al}(\mathrm{OH})_{3}(s)\) in more detail. Show that the solubility of \(\mathrm{Al}(\mathrm{OH})_{3}\), as a function of \(\left[\mathrm{H}^{+}\right]\), obeys the equation $$S=\left[\mathrm{H}^{+}\right]^{3} K_{\mathrm{sp}} / K_{\mathrm{w}}^{3}+K K_{\mathrm{w}} /\left[\mathrm{H}^{+}\right]$$ where \(S=\) solubility \(=\left[\mathrm{Al}^{3+}\right]+\left[\mathrm{Al}(\mathrm{OH})_{4}^{-}\right]\) and \(K\) is the equilibrium constant for $$\mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{Al}(\mathrm{OH})_{4}^{-}(a q)$$ c. The value of \(K\) is \(40.0\) and \(K_{\mathrm{se}}\) for \(\mathrm{Al}(\mathrm{OH})_{3}\) is \(2 \times 10^{-32}\). Plot the solubility of \(\mathrm{Al}(\mathrm{OH})_{3}\) in the \(\mathrm{pH}\) range \(4-12\).

A solution is prepared by adding \(0.10 \mathrm{~mol} \mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6} \mathrm{Cl}_{2}\) to \(0.50 \mathrm{~L}\) of \(3.0 \mathrm{M} \mathrm{NH}_{3} .\) Calculate \(\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\right]\) and \(\left[\mathrm{Ni}^{2+}\right]\) in this solution. \(K_{\text {owerall }}\) for \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\) is \(5.5 \times 10^{8} .\) That is, $$5.5 \times 10^{\mathrm{x}}=\frac{\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\right]}{\left[\mathrm{Ni}^{2+}\right]\left[\mathrm{NH}_{3}\right]^{6}}$$ for the overall reaction $$\mathrm{Ni}^{2+}(a q)+6 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}(a q)$$

Describe how you could separate the ions in each of the following groups by selective precipitation. a. \(\mathrm{Ag}^{+}, \mathrm{Mg}^{2+}, \mathrm{Cu}^{2+}\) c. \(\mathrm{Pb}^{2+}, \mathrm{Bi}^{3+}\) b. \(\mathrm{Pb}^{2+}, \mathrm{Ca}^{2+}, \mathrm{Fe}^{2+}\)

You have two salts, \(\mathrm{AgX}\) and \(\mathrm{Ag} \mathrm{Y}\), with very similar \(K_{\mathrm{sp}}\) values. You know that \(\mathrm{HX}\) is a strong acid and \(\mathrm{HY}\) is a weak acid. Which salt is more soluble in an acidic solution? Explain.
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Organic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Nuclear Chemistry

Read Explanation

Making Measurements

Read Explanation

Physical Chemistry

Read Explanation
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.