Problem 258 - A Complete Resource Book in Chemistry for JEE Main

Question

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

Step-by-Step Solution

Answer

The \(pH\) of the solution is 6.

Step 1: Understanding the Relationship Between pH and pKa

The given dissociation constant, also known as the acid dissociation constant \(K_a\), is \(1.0 \times 10^{-4}\). To find the \(pH\) of the sodium salt of this acid, we first need to calculate the \(pK_a\) because \(pH\) and \(pK_a\) are related. Recall that \(pK_a = -\log_{10}(K_a)\).

Step 2: Calculating pKa

Calculate the \(pK_a\) using the formula: \[ pK_a = -\log_{10}(1.0 \times 10^{-4}) \]Therefore, \[ pK_a = 4 \].

Step 3: Using the Henderson-Hasselbalch Equation

For a sodium salt of a weak acid, the \(pH\) is generally higher than its \(pK_a\) because the salt acts as a conjugate base. Use the Henderson-Hasselbalch equation:\[ pH = pK_a + \log_{10}\left(\frac{[A^-]}{[HA]}\right) \]In this case, the sodium salt fully dissociates, so \([A^-] = [Na^+] = 0.01\, \text{M}\). Since \([HA] = 0\), we assume \([H^+]\) is minimally contributing to the \(pH\) of the solution.

Step 4: Solving for pH

Since the solution is a salt of a weak acid, the base form is predominant, simplifying the Henderson-Hasselbalch equation:\[ pH = pK_a + \log_{10}(0.01) \]This can be rewritten as:\[ pH = 4 + \log_{10}(10^{-2}) = 4 - 2 = 6 \].

Key Concepts

pKa calculation

The concept of calculating the pK\(_a\) is an important aspect when working with acids and bases. The pK\(_a\) value provides insight into the strength of an acid. In general, the pK\(_a\) is derived from the acid dissociation constant, which is denoted as \(K_a\). The formula used to calculate pK\(_a\) is:

• \( pK_a = -\log_{10}(K_a) \)

In our example, the dissociation constant \(K_a\) is given as \(1.0 \times 10^{-4}\). After applying this value to our formula, we find that:

• \( pK_a = -\log_{10}(1.0 \times 10^{-4}) = 4 \)

The pK\(_a\) value of 4 indicates that the acid is relatively weak. Knowing the pK\(_a\) helps predict how the acid will behave in different chemical environments.

dissociation constant

The dissociation constant, denoted as \(K_a\), is crucial as it quantifies the extent to which an acid ionizes in a solution. In simple terms, it measures the strength of an acid. A higher \(K_a\) indicates a stronger acid that dissociates more in solution, while a lower \(K_a\) value points to a weaker acid.

• The formula for the acid dissociation constant is: \( K_a = \frac{[H^+][A^-]}{[HA]} \)

This equation shows that \(K_a\) is the ratio of the concentration of dissociated ions to the concentration of the undissociated acid. When discussing acids, it's essential to consider whether you're dealing with a strong or weak acid. A dissociation constant of \(1.0 \times 10^{-4}\), like in our example, suggests that not all the acid molecules dissociate.
This establishes the acid as weak, setting the stage for how we handle it in further calculations like determining pH using its salt.

pH and pKa relationship

The relationship between pH and pK\(_a\) is pivotal in understanding acid-base chemistry, particularly through the Henderson-Hasselbalch equation. This equation elegantly ties pH, pK\(_a\), and the ratio of dissociated and undissociated forms:

• \( pH = pK_a + \log_{10}\left(\frac{[A^-]}{[HA]}\right) \)

For the sodium salt of a weak acid, which we have in this example, the salt dissociates fully, meaning the concentration of the conjugate base \(A^-\) equals the initial concentration of the salt. Under these conditions, the presence of the salt generally increases the pH compared to the pK\(_a\).
In our case, with a \(0.01\, \text{M}\) concentration of sodium salt and no undissociated acid present, we simplify our calculation to:

• \( pH = pK_a + \log_{10}(0.01) \)
• \( pH = 4 - 2 = 6 \)

Thus, the solution exhibits a pH greater than the pK\(_a\), reinforcing the concept that the conjugate base increases the solution's pH.