/*! 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 129 The conductivity of \(0.01 \math... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 129

The conductivity of \(0.01 \mathrm{~mol} / \mathrm{dm}^{3}\) aqueous acetic acid at \(300 \mathrm{~K}\) is \(19.5 \times 10^{-5} \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}\) and limiting molar conductivity of acetic acid at the same temperature is \(390 \mathrm{ohm}^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1} .\) The degree of dissociation of acetic acid is (a) \(0.05\) (b) \(0.5 \times 10^{-2}\) (c) \(5 \times 10^{-7}\) (d) \(5 \times 10^{-3}\)

Short Answer

Expert verified
The degree of dissociation of acetic acid is (d) \(5 \times 10^{-3}\).

Step by step solution

01

Identify Given Values

The problem provides us with the following information:- Molarity of acetic acid solution, \( C = 0.01 \text{ mol/dm}^3 \).- Conductivity of the solution, \( \kappa = 19.5 \times 10^{-5} \text{ ohm}^{-1} \text{ cm}^{-1} \).- Limiting molar conductivity, \( \Lambda^0 = 390 \text{ ohm}^{-1} \text{ cm}^2 \text{ mol}^{-1} \).
02

Compute Molar Conductivity

Molar conductivity \( \Lambda \) is calculated using the formula:\[\Lambda = \frac{\kappa \times 1000}{C}\]Substituting the values:\[\Lambda = \frac{19.5 \times 10^{-5} \times 1000}{0.01} = 1.95 \text{ ohm}^{-1} \text{ cm}^2 \text{ mol}^{-1}\]
03

Determine Degree of Dissociation

The degree of dissociation \( \alpha \) is given by:\[\alpha = \frac{\Lambda}{\Lambda^0}\]Substitute the values:\[\alpha = \frac{1.95}{390} = 0.005\]Thus, \( \alpha = 5 \times 10^{-3} \).
04

Choose the Correct Answer

From the options provided, \( \alpha = 5 \times 10^{-3} \) corresponds to option (d).

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!

Key Concepts

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

Degree of Dissociation
Understanding the degree of dissociation is key to grasping how much of a weak acid like acetic acid separates into ions in a solution. A weak acid will only partially dissociate, meaning that not all of the acid's molecules will break into ions. This concept is crucial because it gives us insight into the acid's behavior in solution.
  • The degree of dissociation, denoted as \( \alpha \), tells us the fraction of the original substance that has dissociated into ions.
  • Calculated by taking the molar conductivity of a solution and dividing it by its limiting molar conductivity (\( \alpha = \frac{\Lambda}{\Lambda^0} \)).
  • A larger value of \( \alpha \) indicates more dissociation, while a smaller value suggests less dissociation.
Knowing \( \alpha \) helps us understand the solution's composition and predict its reactivity and conductivity.
Molar Conductivity
Molar conductivity \( \Lambda \) represents how well a solution can conduct electricity, taking into account the concentration of ions present. It's a measure that combines both the number of ions and their mobility in the solution.
  • Calculated using the formula \( \Lambda = \frac{\kappa \times 1000}{C} \), where \( \kappa \) is the conductivity and \( C \) is the concentration.
  • It provides insight into how conductive a solution will be as the concentration changes.
  • Molar conductivity typically increases as the concentration of ions decreases, because ions move more freely in less concentrated solutions.
This measure is pivotal for understanding how well solutions like acetic acid can conduct electricity under different conditions.
Limiting Molar Conductivity
Limiting molar conductivity is the value of molar conductivity when a solution is infinitely dilute, meaning ions are completely free to move without any hindrance from nearby ions.
  • Symbolized as \( \Lambda^0 \), it represents the upper limit of conductivity for a solution.
  • This measure is determined under the assumption that there's no interaction between ions, which only occurs in very dilute solutions.
  • It allows comparison between different electrolytes, as it reflects the inherent ability of ions to conduct electricity.
Knowing \( \Lambda^0 \) is essential for calculating the degree of dissociation and for understanding the fundamental properties of ionic solutions.
Solution Conductivity
Solution conductivity, denoted by \( \kappa \), is a measure of a solution's ability to conduct electric current. It depends on the concentration and mobility of the ions present in the solution.
  • Measured in ohm\(^{-1}\) cm\(^{-1}\), it reflects the ease with which electric current can pass through a solution.
  • Higher conductivity indicates more available charge carriers in the solution, such as ions from dissociated molecules.
  • The conductivity of a solution can change with temperature and concentration, affecting the overall behavior and efficiency of the solution in conducting electricity.
Understanding solution conductivity is vital for practical applications like electrolysis and other processes involving ionic solutions.
Ionic Dissociation
Ionic dissociation involves the splitting of molecules into ions when dissolved in water or another solvent. For weak acids like acetic acid, dissociation is partial, meaning not all molecules become ions.
  • Only a fraction of the acid's molecules will dissociate to form hydronium ions \( H_3O^+ \) and acetate ions \( CH_3COO^- \).
  • Partial dissociation is characterized by a specific equilibrium condition where the forward and reverse reactions occur at equal rates.
  • The extent of dissociation is influenced by factors such as temperature and concentration.
This concept helps us to understand the behavior and characteristics of weak acids and other electrolytes in solution.
Chemical Equilibrium
Chemical equilibrium refers to the state in a chemical reaction where the rates of the forward and reverse reactions are equal. For acetic acid in solution, this concept is linked to its partial dissociation.
  • At equilibrium, the concentration of reactants and products remains constant over time, though the reaction continues to occur both forward and backward.
  • For acetic acid, equilibrium involves its dissociation into ions and the recombination of those ions back into acetic acid molecules.
  • Le Chatelier's principle explains how changes in concentration, temperature, or pressure affect chemical equilibrium, allowing predictions of the system's response to different conditions.
Understanding chemical equilibrium is essential for predicting the behavior of chemical reactions and preparing solutions with desired properties.

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

Of the following metals, those that cannot be obtained by the electrolysis of the aqueous solution of their salts are (a) \(\mathrm{Ag}\) and \(\mathrm{Mg}\) (b) \(\mathrm{Ag}\) and \(\mathrm{Al}\) (c) \(\mathrm{Mg}\) and \(\mathrm{Al}\) (d) \(\mathrm{Cu}\) and \(\mathrm{Cr}\)

When a copper wire is placed in a solution of \(\mathrm{AgNO}_{3}\), the solution acquires blue colour. This is due to the formation of (a) \(\mathrm{Cu}^{+}\)ions (b) \(\mathrm{Cu}^{2+}\) ions (c) soluble complex of copper with \(\mathrm{AgNO}_{3}\) (d) \(\mathrm{Cu}\) ion by the reduction of \(\mathrm{Cu}\)

In electrolysis of dilute \(\mathrm{H}_{2} \mathrm{SO}_{4}\), what is liberated at anode? (a) \(\mathrm{H}_{2}\) (b) \(\mathrm{SO}_{4}^{2-}\) (c) \(\mathrm{SO}_{2}\) (d) \(\mathrm{O}_{2}\)

One ampere of current is passed for 9650 seconds through molten \(\mathrm{AlCl}_{3}\). What is the weight in grams of \(\mathrm{Al}\) deposited at cathode? (Atomic weight of \(\mathrm{Al}=27\) ) (a) \(0.9\) (b) \(9.0\) (c) \(0.18\) (d) \(18.0\)

Match the following Column-I (a) \(\mathrm{Pt}\left|\mathrm{H}_{2}(1 \mathrm{~atm})\right| \mathrm{H}^{+}(1 \mathrm{M})\) (b) \(\mathrm{Pt}\left|\overrightarrow{\mathrm{Cl}}^{-}\right| \mathrm{Cl}_{2}\) (c) \(\mathrm{Pt}\left[\mathrm{H}_{2}(1 \mathrm{~atm}) \mid \mathrm{H}^{+}\left(\mathrm{C}_{1} \| \mathrm{H}^{+}\left(\mathrm{C}_{2}\right) \mathrm{H}_{2}(1 \mathrm{~atm}) \mid \mathrm{Pt}\right.\right.\) (d) \(\mathrm{Pt} \mathrm{H}_{2}\left(\mathrm{p}_{1} \mathrm{~atm}\right)\left[\mathrm{H}^{+}(1 \mathrm{M}) \| \mathrm{H}^{+}(1 \mathrm{M})\left|\mathrm{H}_{2}\left(\mathrm{p}_{2} \mathrm{~atm}\right)\right| \mathrm{Pt}\right.\) Column-II (p) Wrong representation (q) Concentration cell (r) Standard hydrogen electrode (s) \(\mathrm{E}^{\circ}=0\) (t) The cell may be commercially feasibly
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Making Measurements

Read Explanation

The Earths Atmosphere

Read Explanation

Inorganic Chemistry

Read Explanation

Kinetics

Read Explanation

Ionic and Molecular Compounds

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.