/*! 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 12 When methanol, \(\mathrm{CH}_{3}... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 12

When methanol, \(\mathrm{CH}_{3} \mathrm{OH}\), is dissolved in water, a nonconducting solution results. When acetic acid, \(\mathrm{CH}_{3} \mathrm{COOH}\), dissolves in water, the solution is weakly conducting and acidic in nature. Describe what happens upon dissolution in the two cases, and account for the different results.

Short Answer

Expert verified
When methanol (CH鈧僌H) is dissolved in water, it forms hydrogen bonds with water molecules but does not release ions, resulting in a nonconducting solution. On the other hand, when acetic acid (CH鈧僀OOH) is dissolved in water, it undergoes weak ionization, donating protons to water molecules and forming hydronium ions (H鈧僌鈦�) and acetate ions (CH鈧僀OO鈦�). This produces a weakly conducting and acidic solution due to the presence of ions. The differences in conductivity and acidity can be attributed to the differences in ionization properties of methanol and acetic acid when dissolved in water.

Step by step solution

01

1. Dissolution of Methanol

When methanol (CH鈧僌H) is dissolved in water, it interacts with water molecules through hydrogen bonding. Since methanol is a covalent compound, there is no release of ions (i.e., charged particles) in the solution. As a result, the methanol-water mixture remains nonconducting, as there are no freely mobile ions present to carry an electric charge.
02

2. Dissolution of Acetic Acid

When acetic acid (CH鈧僀OOH) is dissolved in water, it undergoes a weak ionization process, in which a small fraction of acetic acid molecules donate a proton (H鈦�) to water molecules, forming hydronium ions (H鈧僌鈦�) and acetate ions (CH鈧僀OO鈦�). The presence of these ions allows the solution to be weakly conducting, as they are free to move and carry an electric charge. Additionally, since there is an excess of H鈧僌鈦� ions, the solution is acidic in nature, causing a decrease in pH.
03

3. Comparing the Dissolution Processes and Their Results

In summary, the main difference between the dissolution of methanol and acetic acid lies in the presence of ions in the solution. While both compounds form hydrogen bonds with water, methanol does not release any ions and its solution remains nonconducting. On the other hand, acetic acid undergoes partial ionization and forms a weakly conducting, acidic solution due to the presence of hydronium and acetate ions. Therefore, the observed differences in conductivity and acidity between the two solutions can be attributed to the difference in their ionization properties when dissolved in water.

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) How many grams of solute are present in \(50.0 \mathrm{~mL}\) of \(0.488 \mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} ?\) (b) If \(4.00 \mathrm{~g}\) of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) is dissolved in enough water to form \(400 \mathrm{~mL}\) of solution, what is the molarity of the solution? (c) How many milliliters of \(0.0250 \mathrm{M} \mathrm{CuSO}_{4}\) contain \(1.75 \mathrm{~g}\) of solute?

Determine the oxidation number for the indicated element in each of the following compounds: (a) Ti in \(\mathrm{TiO}_{2}\) (b) Sn in \(\mathrm{SnCl}_{3}^{-}\), (c) \(\mathrm{C}\) in \(\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\), (d) \(\mathrm{N}\) in \(\mathrm{N}_{2} \mathrm{H}_{4}\), (e) \(\mathrm{N}\) in \(\mathrm{HNO}_{2}\), (f) \(\mathrm{Cr}\) in \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\)

Tartaric acid, \(\mathrm{H}_{2} \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6}\), has two acidic hydrogens. The acid is often present in wines and precipitates from solution as the wine ages. A solution containing an unknown concentration of the acid is titrated with \(\mathrm{NaOH}\). It requires \(24.65 \mathrm{~mL}\) of \(0.2500 \mathrm{M} \mathrm{NaOH}\) solution to titrate both acidic protons in \(50.00 \mathrm{~mL}\) of the tartaric acid solution. Write a balanced net ionic equation for the

Name the spectator ions in any reactions that may be involved when each of the following pairs of solutions are mixed. (a) \(\mathrm{Na}_{2} \mathrm{CO}_{3}(a q)\) and \(\mathrm{Mg} \mathrm{SO}_{4}(a q)\) (b) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q)\) and \(\mathrm{Na}_{2} \mathrm{~S}(a q)\) (c) \(\left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4}(a q)\) and \(\mathrm{CaCl}_{2}(a q)\)

A 1.248-g sample of limestone rock is pulverized and then treated with \(30.00 \mathrm{~mL}\) of \(1.035 \mathrm{M} \mathrm{HCl}\) solution. The excess acid then requires \(11.56 \mathrm{~mL}\) of \(1.010 \mathrm{M} \mathrm{NaOH}\) for neutralization. Calculate the percent by mass of calcium carbonate in the rock, assuming that it is the only substance reacting with the \(\mathrm{HCl}\) solution.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Organic Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Nuclear Chemistry

Read Explanation

Kinetics

Read Explanation

Inorganic 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.