Problem 129 - A Complete Resource Book in Chemistry for JEE Main

Question

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) \(5 \times 10^{-3}\)

Step-by-Step Solution

Answer

The degree of dissociation of acetic acid is (a) 0.05.

Step 1: Understanding the Concepts

The problem involves calculating the degree of dissociation of acetic acid given its conductivity in a solution. The degree of dissociation \( \alpha \) is found using the formula \( \alpha = \frac{\Lambda_c}{\Lambda_0} \), where \( \Lambda_c \) is the molar conductivity of the solution, and \( \Lambda_0 \) is the limiting molar conductivity.

Step 2: Calculate Molar Conductivity

The molar conductivity \( \Lambda_c \) of the solution can be calculated with the formula \( \Lambda_c = \frac{\text{Conductivity}}{c} \), where conductivity is given in \( \text{ohm}^{-1} \text{cm}^{-1} \) and \( c \) is the concentration in \( \text{mol/dm}^3 \). Here, \( \text{Conductivity} = 19.5 \times 10^{-5} \text{ ohm}^{-1} \text{ cm}^{-1} \) and \( c = 0.01 \text{ mol/dm}^3 \).

Step 3: Compute Molar Conductivity of the Solution

Substitute the given values into the formula to find \( \Lambda_c \): \[ \Lambda_c = \frac{19.5 \times 10^{-5}}{0.01} = 19.5 \text{ ohm}^{-1} \text{ cm}^2 \text{ mol}^{-1} \]

Step 4: Calculate the Degree of Dissociation

Now, using \( \Lambda_0 = 390 \text{ ohm}^{-1} \text{ cm}^{2} \text{ mol}^{-1} \), find \( \alpha \) as follows: \[ \alpha = \frac{19.5}{390} = 0.05 \]

Step 5: Identify Correct Answer

The calculated degree of dissociation \( \alpha = 0.05 \) corresponds to option (a).

Key Concepts

Conductivity

Conductivity is a measure of a solution's ability to conduct electricity. It tells us how well ions can travel through the solution under an electric field.
- Measured in units like \(\text{ohm}^{-1} \text{cm}^{-1}\), it relies on the nature of the dissolved ions and their concentration.
In a solution, ions carry electrical charges, so more ions or more mobile ions lead to higher conductivity.
For example, in acetic acid, which only partially dissociates in water, the number of ions is lower compared to a fully dissociated electrolyte, resulting in lower conductivity.
Understanding conductivity allows us to deduce information about ion concentration in the solution.

Molar Conductivity

Molar conductivity (\(\Lambda_c\)) is the conductivity of the solution divided by its concentration. It provides a way to compare how well different solutions conduct electricity, regardless of their concentration:

• Formula: \(\Lambda_c = \frac{\text{Conductivity}}{c} \)
• Units: \(\text{ohm}^{-1} \text{cm}^2 \text{mol}^{-1}\)

A higher molar conductivity indicates that each ion in the solution has a higher ability to conduct electricity.
In the case of acetic acid, the molar conductivity tells us about the effectiveness of its ions in moving through the solution.
This concept helps us determine the degree of dissociation of weak electrolytes like acetic acid by comparing it to its theoretical maximum value.

Limiting Molar Conductivity

Limiting molar conductivity (\(\Lambda_0\)) is the molar conductivity at infinite dilution, meaning when the concentration approaches zero.
At this point, the ions are far apart and do not hinder each other's movement, yielding the maximum possible molar conductivity:

• Provides benchmark for comparing actual molar conductivity \(\Lambda_c \)
• Particularly useful for calculating the degree of dissociation of weak acids and bases

For acetic acid, \(\Lambda_0\) tells us the ideal scenario of its ion conduction.
By comparing \(\Lambda_c\) with \(\Lambda_0\), we can assess how dissociated the acetic acid is by calculating its degree of dissociation using \(\alpha = \frac{\Lambda_c}{\Lambda_0} \).

Acetic Acid

Acetic acid is a common weak acid with the formula \( ext{CH}_3 ext{COOH}\).
It is commonly known as the main component of vinegar.
In water, acetic acid partially dissociates into ions:

• \(\text{CH}_3 ext{COOH} \rightleftharpoons \text{CH}_3 ext{COO}^- + \text{H}^+\)
• Only some molecules form ions, so conductivity is lower than a strong acid

The degree of dissociation reflects the fraction of acetic acid molecules that release ions in water.
This property is crucial for understanding its behavior in solutions and calculating related parameters like molar conductivity.
Studying acetic acid's dissociation helps us comprehend its acidity and its electrolytic properties in aqueous solutions.