Problem 52 - A Complete Resource Book in Chemistry for JEE Main

Question

The dissociation constant of a substituted benzoic acid at \(25^{\circ} \mathrm{C}\) is \(1 \times 10^{-4}\). The \(\mathrm{pH}\) of a \(0.01 \mathrm{~m}\) solution of its sodium salt is (a) 6 (b) 7 (c) 8 (d) 9

Step-by-Step Solution

Answer

The pH of the solution is 8.

Step 1: Understand the Problem

We have a substituted benzoic acid with a dissociation constant \( K_a = 1 \times 10^{-4} \). We need to find the pH of the sodium salt of this acid in a \( 0.01 \; \text{M} \) solution.

Step 2: Identify the Sodium Salt

The sodium salt of the benzoic acid dissociates completely in water to give benzoate ions and sodium ions. The benzoate ion is the conjugate base of the benzoic acid.

Step 3: Use the Relationship between \( K_w \), \( K_a \), and \( K_b \)

The equilibrium constant for the reaction of the conjugate base, \( K_b \), can be found using the relationship: \( K_w = K_a \times K_b \). Here, \( K_w = 1 \times 10^{-14} \) is the ion product constant of water.

Step 4: Calculate \( K_b \)

Rearrange the equation to solve for \( K_b \): \( K_b = \frac{K_w}{K_a} = \frac{1 \times 10^{-14}}{1 \times 10^{-4}} = 1 \times 10^{-10} \).

Step 5: Set Up the ICE Table for Benzoate Reaction

Set up an ICE table for the base dissociation of benzoate ion, \( \text{C}_6\text{H}_5\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{C}_6\text{H}_5\text{COOH} + \text{OH}^- \). Initially, the concentration of \( \text{C}_6\text{H}_5\text{COO}^- \) is \( 0.01 \; \text{M} \), and the concentrations of products are 0.

Step 6: Express \([\text{OH}^-]\) Using \( K_b \)

At equilibrium, \([\text{OH}^-] = x\). Using the formula for \( K_b \), \( K_b = \frac{x^2}{0.01 - x} \approx \frac{x^2}{0.01} \).

Step 7: Solve for \( x \)

Assuming \( x \ll 0.01 \), we have \( x^2 = 0.01 \times 1 \times 10^{-10} = 1 \times 10^{-12} \). This gives \( x = 1 \times 10^{-6} \; \text{M} \) for \([\text{OH}^-]\).

Step 8: Calculate the \(\text{pOH}\) and \(\text{pH}\)

The \( \text{pOH} = -\log_{10}(1 \times 10^{-6}) = 6\). Since \( \text{pH} = 14 - \text{pOH} \), the \( \text{pH} = 14 - 6 = 8 \).

Key Concepts

Substituted Benzoic Acid

Substituted benzoic acid refers to a derivative of benzoic acid. Benzoic acid itself is a simple aromatic carboxylic acid, represented as

• \( ext{C}_6 ext{H}_5 ext{COOH} \)

A substituted benzoic acid has one or more hydrogens of the benzene ring replaced by other atoms or groups, which can affect its chemical properties and reactivity.
These substitutions often play a significant role in determining the acid's strength, measured through its dissociation constant (

• \( K_a \)

). The value of \( K_a \) indicates how well the acid ionizes in solution. The larger the \( K_a \), the stronger the acid.
In the context of benzoic acids, substitutions can either increase or decrease the acidic strength by affecting the electron distribution around the carboxyl group. For instance, electron-withdrawing groups, like nitro groups, tend to increase acidity, while electron-donating groups, like alkyl groups, can decrease it.

Dissociation Constant

A dissociation constant (

• \( K_a \)

) measures the propensity of an acid to donate a proton to the solution, thereby forming its conjugate base. In the case of substituted benzoic acids, the expression for the dissociation constant is given by:

• \( K_a = \frac{[ ext{C}_6 ext{H}_5 ext{COO}^-][ ext{H}^+]}{[ ext{C}_6 ext{H}_5 ext{COOH}]} \)

This expression shows the concentrations of products and reactants at equilibrium. A larger \( K_a \) value corresponds to a stronger acid, as it signifies a greater degree of ionization.
In our exercise, the substituted benzoic acid has a \( K_a \) of \( 1 \times 10^{-4} \), suggesting it is a moderately strong acid compared to water's ionization scale. Understanding \( K_a \) values is crucial as they allow chemists to predict the behavior of acids in various environments and decide how different conditions might affect the acid-base balance.

Conjugate Base

The conjugate base is what remains after an acid has donated a proton. In this exercise, when the substituted benzoic acid loses a proton, it creates the benzoate ion (

• \( ext{C}_6 ext{H}_5 ext{COO}^- \)

).
This ion is a crucial species because it determines the solution's pH when the sodium salt is dissolved in water. Upon dissolution, the sodium salt dissociates completely into sodium (

• \( ext{Na}^+ \)

) and benzoate ions.
As a conjugate base, benzoate can then accept a proton, which demonstrates its role in buffer systems, especially when it reacts with hydroxide ions formed from water dissociation in the presence of basic solutions. The ability of the benzoate ion to accept protons makes it an important component in maintaining pH levels in various applications, including food preservation.

Equilibrium Constant

The equilibrium constant (

• \( K_b \) for bases

) is similar to the dissociation constant for acids but applies to the conjugate base's ability to accept protons. This measure is important for determining the basicity of substances in solution. In our example, the benzoate ion acts as a base, reacting with water to re-form benzoic acid and hydroxide ions (

• \( ext{C}_6 ext{H}_5 ext{COO}^- + ext{H}_2 ext{O} \rightleftharpoons ext{C}_6 ext{H}_5 ext{COOH} + ext{OH}^- \)

).
The equilibrium constant for this reaction—\( K_b \)—is calculated using

• \( K_w = K_a \times K_b \)

, where \( K_w \) is the ion product constant of water, equalling \( 1 \times 10^{-14} \).
Rearranging gives \( K_b = \frac{K_w}{K_a} \). This process helps us find how basic a conjugate base is—important for understanding how solutions behave at equilibrium. Knowing \( K_b \) allows us to calculate the amount of hydroxide ions in solution accurately, which is vital for subsequent pH calculation.