Problem 247 - A Complete Resource Book in Chemistry for JEE Main

Question

Find the \(\mathrm{pH}\) of a \(10^{-2} \mathrm{M}\) solution of sodium salt of substituted benzoic acid if the dissociation constant of substituted benzoic acid is \(1 \times 10^{-6}\) at \(298 \mathrm{~K}\)

Step-by-Step Solution

Answer

The pH of the solution is 9.

Step 1: Understanding the Problem

First, let's analyze the problem. We need to find the pH of a sodium salt solution derived from a substituted benzoic acid. We know the acid dissociation constant \(K_a = 1 \times 10^{-6}\). This indicates that the acid is weak. Since sodium salts are fully dissociated in solution, the benzoate ion (the conjugate base of the acid) will be in solution at \(10^{-2} \text{ M}\).

Step 2: Identify the Equilibrium Reaction

When benzoate is dissolved in water, it acts as a base. The equilibrium reaction is:\[ ext{C}_7 ext{H}_5 ext{O}_2^- + ext{H}_2 ext{O} ightleftharpoons ext{C}_7 ext{H}_6 ext{O}_2 + ext{OH}^- \]We are provided the \(K_a\), but we need to find \(K_b\) to analyze this reaction.

Step 3: Calculate the Base Dissociation Constant

We can find \(K_b\) using the relationship between \(K_a\) and \(K_b\). This is given by:\[ K_w = K_a imes K_b \] Where \(K_w\) at \(298 ext{ K}\) is \(1 \times 10^{-14}\). Therefore, \[ K_b = \frac{1 \times 10^{-14}}{1 \times 10^{-6}} = 1 \times 10^{-8} \]

Step 4: Establish the Base Equilibrium Expression

The expression for \(K_b\) is:\[ K_b = \frac{[ ext{C}_7 ext{H}_6 ext{O}_2][ ext{OH}^-]}{[ ext{C}_7 ext{H}_5 ext{O}_2^-]} \]Initially, the solution has \(10^{-2} \text{ M}\) \( ext{C}_7 ext{H}_5 ext{O}_2^-\), and at equilibrium, let change in concentration of \( ext{OH}^-\) be \(x\). Then, \[ K_b = \frac{x^2}{10^{-2}-x} \approx \frac{x^2}{10^{-2}} \] Assuming \(x\) is much smaller than \(10^{-2}\).

Step 5: Solve for Hydroxide Ion Concentration \( [\text{OH}^-] \)

Using the approximate expression from the previous step:\[ 1 \times 10^{-8} = \frac{x^2}{10^{-2}} \]Solving for \(x\):\[ x^2 = 1 \times 10^{-8} \times 10^{-2} = 1 \times 10^{-10} \]\[ x = 1 \times 10^{-5} \]Thus, \([\text{OH}^-] = 1 \times 10^{-5} \text{ M}\).

Step 6: Calculate \( \text{pOH} \) and \( \text{pH} \)

The pOH is calculated using:\[ \text{pOH} = -\log_{10}(10^{-5}) = 5 \]Finally, calculate the pH using the relationship:\[ \text{pH} + \text{pOH} = 14 \]\[ \text{pH} = 14 - 5 = 9 \]

Step 7: Final Conclusion

Thus, the pH of the solution is 9, indicating that it is slightly basic.

Key Concepts

Sodium Salt of Acid

Sodium salts are commonly formed when an acid reacts with sodium hydroxide (NaOH). This chemical process leads to the formation of a salt and water. In the context of benzoic acid, its sodium salt can be represented as sodium benzoate. This compound is highly soluble in water, where it dissociates completely into its constituent ions.
When dissolved, sodium ions (Na⁺) and benzoate ions (C₇H₅O₂⁻) are formed. The benzoate ion acts as the conjugate base of benzoic acid. This characteristic plays a significant role in pH calculations as it indicates that the solution might be basic.
Understanding the behavior of ionic compounds like the sodium salt of an acid in solution is crucial in grasping subsequent concepts such as ion equilibrium and pH impact.

Acid Dissociation Constant

The acid dissociation constant, represented as \(K_a\), measures the strength of an acid in solution. For weak acids, such as benzoic acid, this value is relatively low. This signifies that the acid doesn't fully dissociate in water, leaving a significant amount of the acid in its molecular form.
Mathematically, the acid dissociation constant is expressed through the equilibrium concentration of the products and reactants. Given by the formula:

• \( K_a = \frac{[H^+][A^-]}{[HA]} \)

A higher \(K_a\) value indicates a stronger acid, capable of donating more protons ( H⁺) into the solution. However, our example's \(K_a = 1 \times 10^{-6}\) denotes a relatively weak acid. Such acids have significant relevance when calculating the pH of solutions derived from their salts, as they affect the equilibrium dynamics.

Equilibrium Reaction

In chemical solutions, equilibrium reactions depict a state where the forward and backward reactions occur at the same rate. This concept is fundamental when dealing with weak acids and their conjugate bases.
For the sodium salt of benzoic acid, the equilibrium reaction involves its dissociation in water, forming benzoate ions and hydroxide ions ( OH⁻). The equation for this is:

• \( \text{C}_7\text{H}_5\text{O}_2^- + \text{H}_2\text{O} \rightleftharpoons \text{C}_7\text{H}_6\text{O}_2 + \text{OH}^- \)

In equilibrium reactions, the concentrations of reactants and products remain stable, allowing the use of constants like \(K_b\) to understand the reaction's behavior. Equilibrium provides insights into calculating pH by showing how far a reaction has proceeded and how much of a product, such as hydroxide ions, is present.

Base Dissociation Constant

The base dissociation constant \(K_b\) is a crucial factor when dealing with the conjugate bases of weak acids. In our example, the benzoate ion serves as the conjugate base and interacts with water to form hydroxide ions, making the solution basic.
\(K_b\) is calculated using the formula:

• \( K_b = \frac{[OH^-][HA]}{[A^-]} \)

The relationship between \(K_a\) and \(K_b\) is essential for understanding these dynamics. They are related through the ion product constant for water \(K_w\), given by:

• \( K_w = K_a \times K_b = 1 \times 10^{-14} \)

This relationship allows us to find \(K_b\) if \(K_a\) is known, using the equation:

• \( K_b = \frac{K_w}{K_a} \)

Understanding \(K_b\) helps in determining how strongly the base will dissociate, directly influencing the pH of the solution.