Problem 110 - A Complete Resource Book in Chemistry for JEE Main

Question

A weak monobasic acid is half neutralized by a strong base. If the \(\mathrm{pH}\) of the solution is \(5.4\), its pKa is (a) \(6.8\) (b) \(2.7\) (c) \(5.4\) (d) \(10.8\)

Step-by-Step Solution

Answer

The pKa of the acid is 5.4, thus the answer is (c).

Step 1: Understanding the Concept

When a weak monobasic acid is half neutralized, it means that half of the acid has reacted with a base. At this point, the concentrations of the acid and its conjugate base are equal, i.e., \([HA]=[A^-]\). This is when \( ext{pH} = ext{pK}_a\) according to the Henderson-Hasselbalch equation.

Step 2: Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is given by \( ext{pH} = ext{pK}_a + ext{log}rac{[A^-]}{[HA]}\). When the acid is half neutralized, \([A^-]=[HA]\), so the term \( ext{log}rac{[A^-]}{[HA]} = 0\). Therefore, \( ext{pH} = ext{pK}_a\).

Step 3: Substitution and Conclusion

Given that \( ext{pH} = 5.4\) at half-neutralization, and knowing \( ext{pH} = ext{pK}_a\), it follows that \( ext{pK}_a = 5.4\). Thus, the correct answer is option (c) \(5.4\).

Key Concepts

Understanding pKa

The term "pKa" is a key concept in chemistry, especially when dealing with acids and bases. It stands for the negative logarithm of the acid dissociation constant, \(K_a\). In simpler terms, \(pK_a\) helps us understand how easily an acid donates its proton to a base.
To calculate \(pK_a\), we use the formula: \[pK_a = -\log(K_a).\] A smaller \(pK_a\) value indicates a stronger acid because it means the acid dissociates more in water, donating more protons.
For weak acids, \(pK_a\) values are higher, reflecting their lower tendency to release protons compared to strong acids.

• This explains why weak acids have \(pK_a\) values often greater than 3 or 4.
• Knowing the \(pK_a\) allows chemists to predict the behavior of an acid in different pH environments, which is crucial for many applications like buffer solutions and titrations.

Characteristics of a Weak Monobasic Acid

A weak monobasic acid is an acid that can donate only one proton (H⁺) per molecule in a solution. "Monobasic" indicates this single proton transfer.
What makes it "weak" is its limited ability to donate protons compared to strong acids. This is due to its relatively high \(pK_a\) value, which suggests less dissociation in water.

• Weak acids do not completely ionize in solution, so their equilibrium lies significantly with the undissolved acid.
• Examples of weak monobasic acids include acetic acid (CH₃COOH) and formic acid (HCOOH).

Understanding these acids is important as they play a central role in buffer solutions. They help stabilize pH by reacting with added bases or acids, thereby maintaining a relatively constant pH in solutions.

The Concept of Half-Neutralization

Half-neutralization is a critical concept, particularly in titrations involving weak acids and strong bases. It refers to the point where half of the original acid has been neutralized by the base.
At this stage, the concentration of the acid, \( [HA] \), equals the concentration of its conjugate base, \( [A^-] \). As a result, the term \( \log \frac{[A^-]}{[HA]} \) in the Henderson-Hasselbalch equation becomes zero.

• This simplifies the equation to \( pH = pK_a \), providing a direct link between \(pH\) and \(pK_a\).
• At half-neutralization, knowing the pH directly informs us of the \(pK_a\), highlighting the acid's strength under equilibrium conditions.

This specific point is valuable in determining the \(pK_a\) experimentally, as shown by the example problem where at the half-neutralization point, \(pH\) is equal to the given \(pK_a\) value, simplifying analysis.