Problem 176 - A Complete Resource Book in Chemistry for JEE Main

Question

A buffer solution is prepared by mixing \(20 \mathrm{ml}\) of \(0.1 \mathrm{M}\) \(\mathrm{CH}_{3} \mathrm{COOH}\) and \(40 \mathrm{ml}\) of \(0.5 \mathrm{M} \mathrm{CH}_{3} \mathrm{COONa}\) and then diluted by adding \(100 \mathrm{ml}\) of distilled water. The \(\mathrm{pH}\) of resulting buffer solution is (Given \(\mathrm{pKa} \mathrm{CH}_{3} \mathrm{COOH}=4.76\) ) (a) \(5.76\) (b) \(4.67\) (c) \(3.48\) (d) \(5.9\)

Step-by-Step Solution

Answer

The pH of the buffer solution is 5.76 (option a).

Step 1: Calculate Initial Moles of Acetic Acid

First, find the moles of acetic acid (\( \mathrm{CH}_3\mathrm{COOH} \)). Use the formula: \( \text{moles} = M \times V \). For 0.1 M \( \mathrm{CH}_3\mathrm{COOH} \), in 20 ml, the moles are \( 0.1 \times 0.02 = 0.002 \) moles.

Step 2: Calculate Initial Moles of Sodium Acetate

Find the moles of sodium acetate (\( \mathrm{CH}_3\mathrm{COONa} \)). For 0.5 M \( \mathrm{CH}_3\mathrm{COONa} \), in 40 ml, the moles are \( 0.5 \times 0.04 = 0.02 \) moles.

Step 3: Calculate Total Volume after Dilution

The buffer solution is diluted by adding 100 ml of water. So, the total volume becomes \( 20 + 40 + 100 = 160 \) ml or 0.16 L.

Step 4: Calculate Concentrations in the Diluted Solution

After dilution, the concentration of acetic acid \( [\mathrm{CH}_3\mathrm{COOH}] = \frac{0.002}{0.16} = 0.0125 \) M and the concentration of sodium acetate \( [\mathrm{CH}_3\mathrm{COONa}] = \frac{0.02}{0.16} = 0.125 \) M.

Step 5: Use Henderson-Hasselbalch Equation

To find the pH, use the Henderson-Hasselbalch equation: \[ \mathrm{pH} = \mathrm{pKa} + \log\left(\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}\right) \]Substitute \( \mathrm{pKa} = 4.76 \), \([\mathrm{A^-}] = 0.125\), \([\mathrm{HA}] = 0.0125\) into the equation to get:\[ \mathrm{pH} = 4.76 + \log\left(\frac{0.125}{0.0125}\right) = 4.76 + 1 = 5.76 \]

Step 6: Conclusion

The pH of the resulting buffer solution is 5.76. The correct answer is option (a).

Key Concepts

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a fundamental formula used to estimate the pH of buffer solutions. It provides a quick way to calculate pH when you know the concentrations of the acid and its conjugate base in a solution. The equation is expressed as:

• \[ \text{pH} = \text{pKa} + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \]

Here:

• \([\text{A}^-]\) is the concentration of the conjugate base.
• \([\text{HA}]\) is the concentration of the weak acid.
• \(\text{pKa}\) is the negative logarithm of the acid dissociation constant, \(K_a\), and is a measure of the strength of the acid.

This equation simplifies the process of pH calculation by replacing the need to calculate hydrogen ion concentration directly. By using the ratio of the base to acid concentrations, one can quickly find the pH provided the \(\text{pKa}\) is known. This is particularly useful for buffer solutions made from weak acids and their salts.

Acetic Acid

Acetic acid, also known by its chemical formula \(\text{CH}_3\text{COOH}\), is a common weak acid found in vinegar. It partially dissociates in water to produce hydrogen ions \(\text{H}^+\) and acetate ions \(\text{CH}_3\text{COO}^-\). The pKa of acetic acid is approximately 4.76, indicating that it is weakly acidic.

In buffer solutions, acetic acid acts as the weak acid component and pairs with its conjugate base, acetate, to resist changes in pH. This unique property makes acetic acid a popular choice in laboratory and industrial buffer solutions.

Understanding acetic acid's role is crucial in the context of buffer solutions, as it provides the necessary equilibrium with sodium acetate, allowing the solution to stabilize the pH within a certain range.

Sodium Acetate

Sodium acetate, with the formula \(\text{CH}_3\text{COONa}\), is the sodium salt of acetic acid. In solution, it dissociates completely into sodium ions \(\text{Na}^+\) and acetate ions \(\text{CH}_3\text{COO}^-\), which is the conjugate base of acetic acid.

When mixed with acetic acid in solution, sodium acetate contributes the acetate ions that are part of the buffer system. This means there is a reservoir of acetate ions that can react with any added hydrogen ions \(\text{H}^+\), or hydroxide ions \(\text{OH}^-\), helping to maintain a stable pH condition.

Sodium acetate's role in a buffer is vital as it determines the capacity of the buffer to resist pH changes. The more sodium acetate present, the greater the buffer capacity, allowing the solution to neutralize added acids or bases effectively without significant pH shifts.

pH Calculation

Calculating the pH of a buffer solution can seem complex but is made simple using the Henderson-Hasselbalch equation. The equation relies on the concentrations of the weak acid and its conjugate base in the solution, as well as the known \(\text{pKa}\) value.

To calculate the pH:

• Determine the molar concentrations of the acid and base after any dilution.
• Insert these values into the Henderson-Hasselbalch equation.
• Solve for pH using the equation: \[ \text{pH} = \text{pKa} + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \]

In the given exercise, the pH was calculated to be 5.76, which is a straightforward application of the equation provided the correct concentrations are used.

Understanding the steps to calculate pH is essential for mastering buffer solutions and is vital for anyone studying chemistry at an advanced level.

Dilution and Concentration

In buffer solutions, dilution and concentration are key concepts that influence the final properties of the buffer.

Dilution involves adding a solvent, such as water, to the solution, which decreases the concentration of solutes (acid and base in this context). However, this doesn't change the ratio of acid to base, which is why buffer solutions remain effective at resisting changes in pH.

To find concentrations after dilution:

• Calculate the initial number of moles for each component.
• Determine the new total volume of the solution after dilution.
• Divide the initial moles by the new volume to find the new concentrations.

For the exercise given, after diluting with 100 ml of water, the concentrations of acetic acid and sodium acetate were calculated as 0.0125 M and 0.125 M respectively.

These adjusted concentrations were then used in the Henderson-Hasselbalch equation to deduce the final pH.