Problem 85 - A Complete Resource Book in Chemistry for JEE Main

Question

At \(25^{\circ} \mathrm{C}\) the \(\mathrm{pH}\) of solution containing \(0.10 \mathrm{M}\) sodium acetate and \(0.03 \mathrm{M}\) acetic acid is [pKa value of \(\left.\mathrm{CH}_{3} \mathrm{COOH}=4.57\right]\) (a) \(3.24\) (b) \(4.59\) (c) \(5.09\) (d) \(6.67\)

Step-by-Step Solution

Answer

The pH of the solution is approximately 5.09, choice (c).

Step 1: Identify the Use of the Henderson-Hasselbalch Equation

To find the pH of the buffer solution containing sodium acetate and acetic acid, we use the Henderson-Hasselbalch equation: \[ \text{pH} = \text{pKa} + \log\left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \] where \([\text{A}^-]\) is the concentration of the acetate ion and \([\text{HA}]\) is the concentration of acetic acid.

Step 2: Insert Known Values into the Equation

First, identify and insert the known values:- The concentration of the acetate ion (\([\text{A}^-]\)) is equal to the concentration of sodium acetate, which is 0.10 M.- The concentration of acetic acid (\([\text{HA}]\)) is given as 0.03 M.- The pKa of acetic acid is 4.57.

Step 3: Calculate the Logarithmic Term

Calculate the ratio inside the logarithm:\[ \frac{[\text{A}^-]}{[\text{HA}]} = \frac{0.10}{0.03} = \frac{10}{3} \approx 3.33 \]Then, compute the logarithm:\[ \log (3.33) \approx 0.52 \]

Step 4: Compute the pH

Using the Henderson-Hasselbalch equation, solve for pH:\[ \text{pH} = 4.57 + 0.52 = 5.09 \]

Key Concepts

Understanding Buffer Solutions

A buffer solution is a special type of solution that resists changes in pH, even when small amounts of acids or bases are added. This remarkable ability is crucial in various biological and chemical processes.
The two primary components of a buffer are:

• Weak acid (like acetic acid)
• Conjugate base (like sodium acetate in our case)

These components work together to maintain a stable pH. When an acid is added, the conjugate base neutralizes it. Similarly, when a base is added, the weak acid component takes action. This ensures that the pH remains relatively constant, making buffer solutions essential in maintaining the pH balance in systems like the human body.

The Role of Henderson-Hasselbalch Equation in pH Calculation

To determine the pH of a buffer solution, the Henderson-Hasselbalch equation is immensely valuable. It provides a simple relationship between the pH, the pKa of the acid, and the concentration ratio of the conjugate base to acid. The equation is:\[\text{pH} = \text{pKa} + \log\left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]Here,

• \(\text{pH}\) is what we're trying to find.
• \(\text{pKa}\) represents the acid's dissociation constant, a measure of its strength.
• \([\text{A}^-]\) is the concentration of the conjugate base.
• \([\text{HA}]\) is the concentration of the acid.

In situations where you know the concentrations and pKa, this equation makes it straightforward to solve for pH, as done in the original exercise.

Exploring Acetic Acid in Buffers

Acetic acid is a weak acid, commonly represented by the formula \(\text{CH}_3\text{COOH}\). It partially dissociates in water to release hydrogen ions \(\text{H}^+\), contributing to the acidity of the solution.
When used in a buffer, such as the one in the exercise, acetic acid works to neutralize added bases. This ability stems from its dissociation equilibrium:\[\text{CH}_3\text{COOH} \leftrightarrow \text{CH}_3\text{COO}^- + \text{H}^+\]The pKa of acetic acid is crucial in buffer calculations as it represents the equilibrium point between dissociated and undissociated forms of the acid. In our exercise, the pKa value is 4.57.
This value plays a central role in the Henderson-Hasselbalch equation, helping us determine the pH as seen in the original solution.

The Function of Sodium Acetate

Sodium acetate, depicted as \(\text{CH}_3\text{COONa}\), is the salt form of acetic acid's conjugate base, \(\text{CH}_3\text{COO}^-\). It dissolves in water to provide acetate ions, contributing to the buffering capability.
In the buffer solution, sodium acetate serves multiple roles:

• It furnishes acetate ions that react with any added protons \(\text{H}^+\), thereby counteracting acidity increases.
• By doing so, it ensures that the buffer remains effective, maintaining a stable pH environment.

This setup of a weak acid together with its conjugate base is what gives the solution its buffering properties, effectively stabilizing pH even when disturbances occur. In our calculation, sodium acetate's concentration is crucial as it determines the ratio term in the Henderson-Hasselbalch equation, helping us pinpoint the pH accurately.