Problem 170 - A Complete Resource Book in Chemistry for JEE Main

Question

A buffer solution is prepared by mixing \(10 \mathrm{~mL}\) of \(1.0\) M acetic and \(20 \mathrm{~mL}\) of \(0.5 \mathrm{M}\) sodium acetate and then diluted to \(100 \mathrm{~mL}\) with distilled water. If the pKa of \(\mathrm{CH}_{3} \mathrm{COOH}\) is \(4.76\), what is the \(\mathrm{pH}\) of the buffer solution prepared? (a) \(5.21\) (b) \(4.76\) (c) \(4.34\) (d) \(5.35\)

Step-by-Step Solution

Answer

The pH of the buffer solution is 4.76.

Step 1: Calculate Moles of Each Component

First, calculate the moles of acetic acid (CH₃COOH) and sodium acetate (CH₃COONa) before dilution. For acetic acid: \[ \text{Moles of} \; \text{CH}_3\text{COOH} = 10\, \text{mL} \times \frac{1.0 \, \text{mol}}{1000 \, \text{mL}} = 0.01 \, \text{mol} \]For sodium acetate: \[ \text{Moles of} \; \text{CH}_3\text{COONa} = 20\, \text{mL} \times \frac{0.5 \, \text{mol}}{1000 \, \text{mL}} = 0.01 \, \text{mol} \]

Step 2: Calculate Concentrations After Dilution

Since the total volume after dilution is 100 mL:The concentration of acetic acid after dilution is:\[ \text{Concentration of CH}_3\text{COOH} = \frac{0.01 \, \text{mol}}{0.1 \, \text{L}} = 0.1 \, \text{M} \]The concentration of sodium acetate after dilution is:\[ \text{Concentration of CH}_3\text{COONa} = \frac{0.01 \, \text{mol}}{0.1 \, \text{L}} = 0.1 \, \text{M} \]

Step 3: Use Henderson-Hasselbalch Equation

Utilize the Henderson-Hasselbalch equation for buffers:\[\text{pH} = \text{pKa} + \log\left(\frac{[\text{CH}_3\text{COONa}]}{[\text{CH}_3\text{COOH}]}\right)\]Substitute the known values:\[\text{pH} = 4.76 + \log\left(\frac{0.1}{0.1}\right)\]\[\text{pH} = 4.76 + \log(1)\]

Step 4: Calculate the pH

Calculate the logarithm:\[\log(1) = 0\]Then the equation becomes:\[\text{pH} = 4.76 + 0 = 4.76\]

Key Concepts

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is essential for understanding buffer solutions. It's a formula used to calculate the pH of a buffer solution. A buffer is a mixture that resists changes in pH when small amounts of acid or base are added. The equation itself is derived from the expression of the acid dissociation constant (Ka), and it's written as:

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

Here, \( \text{pKa} \) is the negative logarithm of the acid dissociation constant \( \text{Ka} \), \( [\text{A}^-] \) is the concentration of the base form (the deprotonated species), and \( [\text{HA}] \) is the concentration of the acid form (the protonated species). It's used widely because it offers a straightforward approach to calculating pH, especially with weak acids and their conjugate bases present in the solution.

pH Calculation

Calculating the pH of a buffer solution involves understanding the balance between hydrogen ion concentration and the equilibrium established by the buffer components. The pH scale ranges from 0 to 14, where lower values are more acidic, higher values are more basic, and 7 is neutral. In the context of buffers, which are usually composed of a weak acid and its conjugate base, the Henderson-Hasselbalch equation reveals the pH by considering the ratio of these components. For example, when the concentrations of the deprotonated and protonated forms are equal, we find, as in our exercise, that the pH equals the pKa, which in this case is 4.76. Calculating pH provides insights into how well the buffer can neutralize added acids or bases.

Moles Calculation

Calculating moles is fundamental for determining the amounts of substances present in a solution, which is crucial for any stoichiometric or quantitative analysis. It involves using the relation between volume, concentration, and moles. The number of moles can be easily calculated with the formula:

• \( \text{Moles} = \text{volume in liters} \times \text{concentration in molarity} \)

In our example:

• For acetic acid (10 mL of 1.0 M solution): \( 0.01 \; \text{mol} \)
• For sodium acetate (20 mL of 0.5 M solution): \( 0.01 \; \text{mol} \)

These calculations serve as the foundation for later pH calculations in buffer solutions.

Acetic Acid

Acetic acid, with the chemical formula \( \text{CH}_3\text{COOH} \), is a weak organic acid, commonly known for its presence in vinegar. As a weak acid, it partially dissociates in water, making it effective for buffer solutions. Its dissociation is represented by:

• \( \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \)

The equilibrium established during this dissociation is crucial for the buffer action. When part of a buffer, its ability to donate protons (H⁺ ions) helps in maintaining a stable pH when acids or bases are introduced. Its pKa value, 4.76, is an important factor in calculating the pH of acetic acid-based buffers, as observed in the Henderson-Hasselbalch equation.

Sodium Acetate

Sodium acetate, represented as \( \text{CH}_3\text{COONa} \), acts as the conjugate base in acetic acid buffer solutions. In an aqueous solution, it dissociates fully to provide acetate ions:

• \( \text{CH}_3\text{COONa} \rightrightarrow \text{CH}_3\text{COO}^- + \text{Na}^+ \)

The release of acetate ions increases the concentration of the base form \( [\text{A}^-] \), which then interacts according to the Henderson-Hasselbalch equation. Sodium acetate strengthens the buffer's capacity to neutralize acids, ensuring the pH remains stable. Its presence in the acetic acid solution exemplifies how conjugate pairs work in tandem to resist pH changes.

Solution Dilution

Dilution affects the concentration of buffer components but maintains their ratio, crucial for the stability of the buffer. When solutions are diluted, their concentration decreases, but the moles of solute do not change. The relation for calculating the effect of dilution is:

• \( C_1 V_1 = C_2 V_2 \)

Where \( C_1 \) and \( V_1 \) are the initial concentration and volume, respectively, and \( C_2 \) and \( V_2 \) are the values after dilution. In our exercise, the final solution volume is 100 mL, making the concentrations of both acetic acid and sodium acetate 0.1 M after dilution. Despite the dilution, the ratio of concentrations stays constant, preserving the buffer's effectiveness in maintaining pH stability.