Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 16: Problem 15

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 is a phenomenon in which the solubility of an ionic compound decreases when a common ion is present in the solution. This occurs due to the Le Chatelier's principle, stating that a system at equilibrium will adjust to counteract any changes. When a common ion is introduced to a solubility equilibrium (e.g., \( AB_{(s)} <=> A^{+}_{(aq)} + B^{-}_{(aq)} \)), the equilibrium shifts to oppose the increased ion concentration, favoring precipitation. For example, adding sodium fluoride (NaF) to a solution of calcium fluoride (CaF2) will result in the increased concentration of F-, shifting the equilibrium left and decreasing the solubility of CaF2.

Step by step solution

01

Define the Common Ion Effect

The common ion effect is a phenomenon that occurs when an ionic solute (like a salt) is dissolved in a solution containing a common ion. In this context, the common ion refers to an ion that is common between the ionic solute and the solution. The presence of the common ion in the solution results in a decrease in the solubility of the ionic solute.
02

Understand the Le Chatelier's Principle

To explain the common ion effect, we need to understand the Le Chatelier's principle. This principle states that if a chemical system at equilibrium experiences a change in concentration, temperature, or total pressure, the equilibrium will shift in the direction that counteracts the change. In the context of the common ion effect, the presence of the common ion disturbs the equilibrium which results in the adjustment of the system to reestablish the equilibrium.
03

Relate the Common Ion Effect to Solubility Equilibria

When an ionic solid dissolves in water, it reaches a solubility equilibrium where the rate of dissolution is equal to the rate of precipitation. For example, for a salt AB, the solubility equilibrium can be represented as: \[ AB_{(s)} <=> A^{+}_{(aq)} + B^{-}_{(aq)} \] If we now add a solution containing common ions (A+ or B-), it leads to an increase in the concentration of these ions in the solution. According to Le Chatelier's principle, the equilibria will shift to oppose the change. In this case, the equilibrium will shift left, which means more AB salt precipitates and ultimately the solubility of the salt decreases.
04

Give an Example

Consider a solution of calcium fluoride (CaF2) and an added source of the common ion F-, such as sodium fluoride (NaF). The solubility equilibrium in this system can be represented as: \[ CaF2_{(s)} <=> Ca^{2+}_{(aq)} + 2F^{-}_{(aq)} \] Upon adding NaF, the concentration of F- increases, and according to Le Chatelier's principle, the equilibrium will shift to the left. This causes more CaF2 to precipitate out of the solution, and thus the solubility of CaF2 is significantly reduced. In summary, the common ion effect causes a decrease in the solubility of ionic compounds in water due to the presence of a common ion that shifts the solubility equilibrium to favor precipitation.

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.

Start your free trial

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

Calculate the solubility of each of the following compounds in moles per liter. Ignore any acid–base properties. a. \(A g_{3} P O_{4}, K_{s p}=1.8 \times 10^{-18}\) b. \(\mathrm{CaCO}_{3}, K_{\mathrm{sp}}=8.7 \times 10^{-9}\) c. $\mathrm{Hg}_{2} \mathrm{Cl}_{2}, K_{\mathrm{sp}}=1.1 \times 10^{-18}\left(\mathrm{Hg}_{2}^{2+} \right.$ is the cation in is the cation in solution.\()\)

Calculate the solubility of solid $\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\left(K_{\mathrm{sp}}=1 \times 10^{-54}\right)\( in a \)0.10-M \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}$ solution.

\(\mathrm{Ag}_{2} \mathrm{S}(s)\) has a larger molar solubility than CuS even though \(\mathrm{Ag}_{2} \mathrm{S}\) has the smaller \(K_{\mathrm{sp}}\) value. Explain how this is possible.

A solution contains \(1.0 \times 10^{-5} M \mathrm{Na}_{3} \mathrm{PO}_{4} .\) What concentrations of \(\mathrm{A} \mathrm{g} \mathrm{NO}_{3}\) will cause precipitation of solid \(\mathrm{Ag}_{3} \mathrm{PO}_{4}\) \(\left(K_{\mathrm{sp}}=1.8 \times 10^{-18}\right) ?\)

The copper(l) ion forms a complex ion with \(\mathrm{CN}^{-}\) according to the following equation: $$\mathrm{Cu}^{+}(a q)+3 \mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{Cu}(\mathrm{CN})_{3}^{2-}(a q) \quad K=1.0 \times 10^{11}$$ a. Calculate the solubility of $\mathrm{CuBr}(s)\left(K_{\mathrm{sp}}=1.0 \times 10^{-5}\right)\( in \)1.0 \mathrm{L}\( of \)1.0 \mathrm{M} \mathrm{NaCN}$ . b. Calculate the concentration of \(\mathrm{Br}^{-}\) at equilibrium. c. Calculate the concentration of \(\mathrm{CN}^{-}\) at equilibrium.
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Chemical Analysis

Read Explanation

Nuclear Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Kinetics

Read Explanation

Chemical Reactions

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.

Sign-up for free