Problem 49 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The dissociation constant of a substituted benzoic acid at \(25^{\circ} \mathrm{C}\) is \(1.0\) \(\times 10^{-4}\). The \(\mathrm{pH}\) of a \(0.01 \mathrm{M}\) solution of its sodium salt is

Step-by-Step Solution

Answer

The pH is 8.

Step 1: Understanding the Problem

We need to determine the pH of a 0.01 M solution of the sodium salt of a substituted benzoic acid whose dissociation constant \( K_a \) is given as \( 1.0 \times 10^{-4} \). This involves finding the pH of a solution where the benzoate ion acts as a base.

Step 2: Applying the Hydrolysis Equation

The benzoate ion (the conjugate base of benzoic acid) will hydrolyze in water according to the equation: \[ C_6H_5COO^- + H_2O \rightleftharpoons C_6H_5COOH + OH^- \]. We need to find the equilibrium concentration of \( OH^- \) to determine the pH.

Step 3: Calculating the Base Ionization Constant

We use the relationship \( K_w = K_a \times K_b \), where \( K_w \) is the ion-product constant of water \( (1.0 \times 10^{-14}) \) to find \( K_b \). Hence, \[ K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-4}} = 1.0 \times 10^{-10} \].

Step 4: Setting Up the Equilibrium Expression

For the hydrolysis of the benzoate ion, the equilibrium expression is: \( K_b = \frac{[C_6H_5COOH][OH^-]}{[C_6H_5COO^-]} \). Assuming an initial concentration of 0.01 M for \([C_6H_5COO^-]\) and \( x \) as the change due to hydrolysis, the expression becomes \( K_b = \frac{x^2}{0.01-x} \approx \frac{x^2}{0.01} \) since \( x \) is very small compared to 0.01 M.

Step 5: Solving for Hydroxide Ion Concentration

Substitute \( K_b \) into the expression: \[ 1.0 \times 10^{-10} = \frac{x^2}{0.01} \]. Solving for \( x \), we get \[ x = \sqrt{1.0 \times 10^{-12}} = 1.0 \times 10^{-6} \]. Therefore, \([OH^-] = 1.0 \times 10^{-6} \text{ M} \).

Step 6: Calculating the pH from pOH

The pOH is calculated as \( ext{pOH} = - ext{log}([OH^-]) = - ext{log}(1.0 \times 10^{-6}) = 6 \). Use the relationship \( ext{pH} + ext{pOH} = 14 \) to find the pH: \( ext{pH} = 14 - 6 = 8 \).

Key Concepts

Dissociation Constant

In chemistry, the dissociation constant, often represented as \( K_a \), is a critical parameter used to describe the extent to which an acid dissociates into its ions in a solution. The dissociation of an acid is essentially its ability to donate protons (\( H^+ \)) to the solution.
The higher the dissociation constant, the stronger the acid, as it more readily donates protons. Conversely, a lower \( K_a \) implies a weaker acid, meaning it donates protons less easily. For our exercise, the substituted benzoic acid has a \( K_a = 1.0 \times 10^{-4} \), indicating it's a weak acid.

To calculate the pH of a solution containing this acid, knowing \( K_a \) helps us to understand how much the acid dissociates. In this exercise, understanding the dissociation constant is essential for finding the base ionization constant \( K_b \), which further aids in determining the pH of the solution.

Benzoic Acid

Benzoic acid, \( C_6H_5COOH \), is a simple aromatic carboxylic acid. It appears as a white crystalline solid and is known for being a preservative in food. Structurally, it contains a benzene ring bonded to a carboxyl group \( -COOH \).
A substituted benzoic acid features additional atoms or groups replacing hydrogen atoms on this ring. This substitution can influence the acid's properties, especially its dissociation behavior.

• Substitutions can increase or decrease the acid strength.
• It affects the acid's ability to donate protons.

Understanding benzoic acid and its substitutions helps in comprehending how the substituted benzoic acid behaves in a solution and how its ions interact in the process of hydrolysis to determine the solution's pH.

Base Ionization Constant

The base ionization constant, \( K_b \), measures the extent to which a base can accept protons and thus generate hydroxide ions \( OH^- \) in a solution. It is comparable to \( K_a \) but in the context of bases.
In the scenario with the benzoate ion, which forms from the dissociation of benzoic acid, we treat it as a weak base in a solution through its hydrolysis. To find \( K_b \), we use the relationship \( K_w = K_a \times K_b \). Here, \( K_w \) is the ion-product constant for water, equal to \( 1.0 \times 10^{-14} \).

Therefore, calculating \( K_b \) for a given conjugate base helps us understand the degree of hydrolysis and further aids in pH calculations. Given \( K_a = 1.0 \times 10^{-4} \) for the substituted benzoic acid, we determined \( K_b = 1.0 \times 10^{-10} \). This value indicates a weak base.

Hydrolysis

Hydrolysis involves the reaction of a salt with water, leading to the formation of the corresponding weak acid or base and resulting in a pH change in the solution. In the context of our exercise, the benzoate ion \( C_6H_5COO^- \) acts as the conjugate base.
When benzoate ions undergo hydrolysis, they react with water to form benzoic acid \( C_6H_5COOH \) and hydroxide ions \( OH^- \):
\[ C_6H_5COO^- + H_2O \rightleftharpoons C_6H_5COOH + OH^- \]
This reaction results in the production of \( OH^- \) ions, which make the solution basic, thereby increasing its pH.
By setting up the equilibrium expression and solving for \( OH^- \) concentration, as explained in the solution steps, we can accurately find the pH of the solution derived from the salt of benzoic acid.

Conjugate Base

A conjugate base is formed when an acid donates a proton (\( H^+ \)) and undergoes a dissociation process. For an acid \( HA \), the conjugate base is \( A^- \). In this article's context, the benzoic acid \( C_6H_5COOH \) donates a proton to form its conjugate base, the benzoate ion \( C_6H_5COO^- \).
The concept of conjugate bases is pivotal for acid-base reactions because they are integral in base ionization and hydrolysis processes. When dissolved in water, a conjugate base can gain a proton, reverting to its acidic form, or interact with water, altering the solution's pH.

• The conjugate base's strength is inversely related to the acid's strength.
• The stronger the acid, the weaker its conjugate base, and vice versa.

Understanding how benzoate acts as a conjugate base allows us to predict its behavior in solution and how it affects the pH through the hydrolysis mechanism.