/*! 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 We have seen that ions in aqueou... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 6

We have seen that ions in aqueous solution are stabilized by the attractions between the ions and the water molecules. Why then do some pairs of ions in solution form precipitates? [Section 4.2]

Short Answer

Expert verified
In aqueous solutions, ions are stabilized by their interactions with water molecules through solvation. However, some pairs of ions in solution form precipitates because their attractive forces with each other are stronger than the stabilizing interactions with water molecules. These precipitates are formed when the solubility equilibrium is reached, and the ion product exceeds the solubility product constant (Ksp) for the substance involved. This indicates that the interactions between the ions overcome the stabilizing effect of water molecules.

Step by step solution

01

Introduction to solvation and stabilization

In aqueous solutions, ions (either positively charged cations or negatively charged anions) are surrounded by water molecules. These water molecules are attracted to the ions due to their polar nature. This interaction is called "solvation" and it stabilizes ions in the solution.
02

Formation of precipitates

Despite the stabilization imparted by water molecules, some pairs of ions in the solution might form precipitates. This is because the ions in the solution can interact with each other as well, forming products with very low solubility. Precipitation happens when the attractive forces between the ions in the pair are stronger than the stabilizing interaction with the water molecules.
03

Examples of precipitation reactions

An example of this interaction causing precipitation is the reaction between a soluble silver salt (e.g., silver nitrate, AgNO3) and a soluble chloride salt (e.g., sodium chloride, NaCl), leading to the formation of silver chloride (AgCl), a highly insoluble precipitate: \[ AgNO_3 + NaCl \rightarrow AgCl_{(s)} + NaNO_3 \] In this case, the strong attraction between the Ag+ and Cl- ions is stronger than the interactions between these ions and the water molecules, which leads to the formation of AgCl precipitate.
04

An understanding of solubility equilibrium

It is important to remember that the solubilities of different compounds vary, and this plays a significant role in the formation of the precipitate. Precipitation reactions are governed by the solubility equilibrium and the solubility product constant (Ksp), which is specific for each substance. If the ion product of a solution exceeds the Ksp value for the substance, the excess ions will form insoluble precipitates. This indicates that the attractive forces between the ions in the solution overcome the stabilizing interactions with water molecules.

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

A 0.5895-g sample of impure magnesium hydroxide is dissolved in \(100.0 \mathrm{~mL}\) of \(0.2050 \mathrm{M} \mathrm{HCl}\) solution. The excess acid then needs \(19.85 \mathrm{~mL}\) of \(0.1020 \mathrm{M} \mathrm{NaOH}\) for neutralization. Calculate the percentage by mass of magnesium hydroxide in the sample, assuming that it is the only substance reacting with the \(\mathrm{HCl}\) solution.

Separate samples of a solution of an unknown ionic compound are treated with dilute \(\mathrm{AgNO}_{3}, \mathrm{~Pb}\left(\mathrm{NO}_{3}\right)_{2}\), and \(\mathrm{BaCl}_{2}\). Precipitates form in all three cases. Which of the following could be the anion of the unknown salt: \(\mathrm{Br}^{-}, \mathrm{CO}_{3}{ }^{2-}, \mathrm{NO}_{3}{ }^{-}\)?

You are presented with a white solid and told that due to careless labeling it is not clear if the substance is barium chloride, lead chloride, or zinc chloride. When you transfer the solid to a beaker and add water, the solid dissolves to give a clear solution. Next a \(\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)\) solution is added and a white precipitate forms. What is the identity of the unknown white solid? [Section 4.2]

(a) How many milliliters of \(0.120 \mathrm{M} \mathrm{HCl}\) are needed to completely neutralize \(50.0 \mathrm{~mL}\) of \(0.101 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) solution? (b) How many milliliters of \(0.125 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) are needed to neutralize \(0.200 \mathrm{~g}\) of \(\mathrm{NaOH}\) ? (c) If \(55.8 \mathrm{~mL}\) of a \(\mathrm{BaCl}_{2}\) solution is needed to precipitate all the sulfate ion in a \(752-\mathrm{mg}\) sample of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\), what is the molarity of the \(\mathrm{BaCl}_{2}\) solution? (d) If \(42.7 \mathrm{~mL}\) of \(0.208\) \(\mathrm{M} \mathrm{HCl}\) solution is needed to neutralize a solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\), how many grams of \(\mathrm{Ca}(\mathrm{OH})_{2}\) must be in the solution?

The U.S. standard for arsenate in drinking water requires that public water supplies must contain no greater than 10 parts per billion ( \(\mathrm{ppb})\) arsenic. If this arsenic is present as arsenate, \(\mathrm{AsO}_{4}{ }^{3-}\), what mass of sodium arsenate would be present in a \(1.00\)-L sample of drinking water that just meets the standard? Parts per billion is defined on a mass basis as $$ \mathrm{ppb}=\frac{\mathrm{g} \text { solute }}{\text { g solution }} \times 10^{9} $$
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

The Earths Atmosphere

Read Explanation

Nuclear Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Ionic and Molecular Compounds

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.