Problem 140 - A Complete Resource Book in Chemistry for JEE Main

Question

The dissociation constant of acetic acid is \(1.6 \times 10^{-5}\). The degree of dissociation \((\alpha)\) of \(0.01 \mathrm{M}\) acetic acid in the presence of \(0.1 \mathrm{M} \mathrm{HCl}\) is equal to (a) \(0.4\) (b) \(0.026\) (c) \(1.6\) (d) \(0.016\)

Step-by-Step Solution

Answer

(d) 0.016

Step 1: Understand the Problem

We need to find the degree of dissociation (\(\alpha\)) for acetic acid \((\text{CH}_3\text{COOH})\) in the presence of \(0.1\,\text{M}\) \(\text{HCl}\). The dissociation constant \(K_a\) of acetic acid is given as \(1.6 \times 10^{-5}\). The concentration of acetic acid is \(0.01\, \text{M}\).

Step 2: Write the Equilibrium Expression

The dissociation of acetic acid can be represented by the equation: \[\text{CH}_3\text{COOH} \leftrightarrow \text{CH}_3\text{COO}^- + \text{H}^+\]. The equilibrium constant expression for this dissociation is \(K_a = \frac{[\text{CH}_3\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{COOH}]}\).

Step 3: Simplify the Equilibrium Calculation

In the presence of \(\text{HCl}\), which is a strong acid, it is assumed that \([\text{H}^+] \approx 0.1\,\text{M}\), thus overpowering the contribution from the acetic acid itself. Therefore, \(K_a = \frac{\alpha \cdot C \cdot 0.1 \cdot 0.01}{0.01(1-\alpha)}\), where \(C\) is the initial concentration of acetic acid, \(0.01\,\text{M}\), and \(\alpha\) is the degree of dissociation.

Step 4: Solve for Degree of Dissociation \(\alpha\)

Since \([\text{H}^+] \approx 0.1\,\text{M}\) is significantly larger than \([\text{CH}_3\text{COOH}]\), the term \((1 - \alpha)\) can be approximated to \(1\) due to a small \(\alpha\). So, \(1.6 \times 10^{-5} = (\alpha \cdot 0.01 \cdot 0.1)\). Solving \(\alpha = \frac{1.6 \times 10^{-5}}{0.01 \cdot 0.1}\), we get \(\alpha = 0.016\).

Key Concepts

Understanding Acetic Acid

Acetic acid, known chemically as \(\text{CH}_3\text{COOH}\), is a weak acid that is commonly found in household vinegar. In the chemical world, it’s an important organic compound due to its simple structure and versatile use.
Acetic acid, like many weak acids, partially dissociates in water. This means that only some of its molecules will release hydrogen ions \((\text{H}^+)\), while the rest remain intact in solution.
Understanding the behavior of acetic acid in water and other solutions is crucial for various chemical calculations, including those involving pH and buffering capacities. Here are some key points about acetic acid:

• Acetic acid has a distinctive sour taste and smell, akin to vinegar.
• In chemistry, it is often used as a solvent and reagent.
• Its dissociation in water is represented by the equilibrium: \(\text{CH}_3\text{COOH} \leftrightarrow \text{CH}_3\text{COO}^- + \text{H}^+\).
• The extent of this dissociation can be affected by the presence of other acids or bases in the solution, which can suppress or enhance the release of \(\text{H}^+\) ions.

Exploring the Dissociation Constant

The dissociation constant \(K_a\) is a pivotal factor in understanding how weak acids like acetic acid behave. It gives insight into the extent of dissociation, showing us how many of the acid's molecules release \(\text{H}^+\) ions into the solution.
In the context of acetic acid, the dissociation constant is given as \(1.6 \times 10^{-5}\). This small value indicates that only a tiny fraction of acetic acid molecules dissociate at equilibrium.
Why do we care about \(K_a\)? It helps chemists predict the behavior of acids under different conditions. For acetic acid, the equation is:

• \(K_a = \frac{[\text{CH}_3\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{COOH}]}\)

This expression is crucial for calculating the concentrations of ions in solutions containing acetic acid. The concentration expressed by \(K_a\) is temperature-dependent and also affected by the ionic strength of the solution. In practice, knowing \(K_a\) allows us to determine the degree of dissociation \((\alpha)\) in various settings, enhancing our ability to manipulate chemical reactions as needed.

Understanding the Equilibrium Expression

Equilibrium expressions are fundamental in chemistry, especially when describing reactions where products and reactants are present simultaneously. For acetic acid, the equilibrium expression reflects the balance between its dissociated and non-dissociated forms.
When we write the equilibrium expression for acetic acid dissociation:

• \(\text{CH}_3\text{COOH} \leftrightarrow \text{CH}_3\text{COO}^- + \text{H}^+\)

This equilibrium is dynamic, meaning the forward and reverse reactions occur at the same rate, achieving a stable state where no net change ensues.
The expression to mathematically represent this balance is:

• \(K_a = \frac{[\text{CH}_3\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{COOH}]}\)

In practical applications, knowing this expression helps us navigate questions of reaction direction, predict concentrations, and gauge the impact of changes in conditions, such as the introduction of a strong acid like \(\text{HCl}\).
Through such expressions, chemists can excel in maintaining and manipulating reaction conditions to suit specific scientific or industrial goals.