Problem 172 - A Complete Resource Book in Chemistry for JEE Main

Question

Calculate the ratio of \(\mathrm{pH}\) of a solution containing 1 mole of \(\mathrm{CH}_{3} \mathrm{COONa}+1\) mole of HCl per litre to that of a solution containing 1 mole of \(\mathrm{CH}_{3} \mathrm{COONa}+1\) mole of \(\mathrm{CH}_{3} \mathrm{COOH}\) per litre. (a) \(\frac{2}{1}\) (b) \(\frac{1}{2}\) (c) \(\frac{2}{3}\) (d) \(\frac{3}{2}\)

Step-by-Step Solution

Answer

Assuming a simplification in analysis, the \\( \frac{\text{pH}_1}{\text{pH}_2} \\) ratio leans toward a higher denominator impact if matching general numerical choices; otherwise detail driven deduction points to (theoretically) approximate strong acid dominance computed consideration.

Step 1: Understand the Buffer Solutions

Identify the type of solutions given. The first solution is a strong acid (HCl) with its conjugate base (CH3COONa), which will not form a buffer as HCl is strong. The second solution is a weak acid (CH3COOH) and its conjugate base (CH3COONa), which forms a buffer solution.

Step 2: Calculate pH of the First Solution

In the first solution, HCl is a strong acid that will completely dissociate, neutralizing the acetate ion from CH3COONa, leading to a non-buffered acidic solution. As all acetate ions are consumed, the pH is determined by the strong acid, and can be considered very low. For calculation purposes, assume the use of strong acid pH determination, pH ≈ 0.

Step 3: Use Henderson-Hasselbalch Equation for the Second Solution

For the second solution, use the Henderson-Hasselbalch equation for buffer solutions: \( \text{pH} = \text{pKa} + \log \left( \frac{[\text{Salt}]}{[\text{Acid}]} \right) \). Here, \([\text{Salt}] = 1\) M and \([\text{Acid}] = 1\) M, thus: \( \text{pH} = \text{pKa} \).

Step 4: Calculate the Ratio of pH Values

Since the first solution has a pH approximately equal to 0, and the second solution’s pH equals the pKa of acetic acid (approximately 4.76), the ratio of pH values is practically infinite as dividing anything by 0 leads to an undefined or infinite answer. However, if considering simplified theoretical values for illustration without complete dissociation consideration, take the pH ratios for general grading selection.

Key Concepts

Buffer Solutions

Buffer solutions are crucial in maintaining the pH of a system. They resist abrupt changes in pH when small amounts of acid or base are added. This stability is due to the presence of weak acids or bases and their conjugate pairs in the solution. For example, a mixture of acetic acid, (CH₃COOH), and its conjugate base, the acetate ion (CH₃COO⁻, often found as sodium acetate, CH₃COONa), forms a buffer solution.

A buffer works by reacting with any added acid or base to moderate changes in pH:

• If a strong acid is added to the buffer, the additional hydrogen ions (H⁺) are neutralized by the conjugate base (CH₃COO⁻), forming the weak acid (CH₃COOH) and water, thus minimizing the increase in H⁺ in the solution.
• Conversely, if a strong base is added, the OH⁻ ions are neutralized by the weak acid, forming water and the conjugate base, thus stabilizing the pH.

However, not every solution with an acid and a salt is a buffer. For instance, a solution of HCl (a strong acid) and CH₃COONa will not act as a buffer because HCl dissociates completely, overwhelming the buffering action.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a formula commonly used to calculate the pH of buffer solutions. It provides insight into the pH based on the concentration ratio of the buffer's components: the weak acid and its conjugate base. The equation is expressed as:

\[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{Salt}]}{[\text{Acid}]} \right) \]

In this equation:

• \(\text{pH}\) stands for the acidity or basicity of the buffer solution.
• \(\text{pKa}\) is the negative logarithm of the acid dissociation constant (Ka) of the weak acid, reflecting its tendency to donate protons. It is a crucial factor in determining the pH of the buffer.
• \([\text{Salt}]\) refers to the concentration of the conjugate base in the solution.
• \([\text{Acid}]\) refers to the concentration of the weak acid.

The Henderson-Hasselbalch equation is particularly useful when the concentrations of the acid and its conjugate base are known. It shows that at equal concentrations of acid and salt, the pH equals the pKa of the acid, demonstrating the power of buffers to maintain a stable pH close to the pKa value.

pH Calculation

Calculating the pH of a solution is fundamental in chemistry to understand the solution's acidity or basicity. The pH scale ranges from 0 to 14:

• A pH below 7 indicates an acidic solution.
• A pH of 7 is considered neutral.
• A pH above 7 signifies a basic (alkaline) solution.

In the exercise, one solution consisted of HCl, a strong acid, and CH₃COONa. Since HCl fully dissociates, it releases a large concentration of H⁺ ions, resulting in a very low pH, often close to 0 for concentrated solutions.

For the second solution, consisting of CH₃COOH and CH₃COONa as a buffer, the pH is calculated using the Henderson-Hasselbalch equation. Since the concentrations of CH₃COOH and CH₃COONa are equal, the equation simplifies to: \[ \text{pH} = \text{pKa} \]

This results in a pH value approximately equal to the pKa of acetic acid, which is about 4.76. Understanding these calculations helps predict how a solution behaves when exposed to additional acids or bases, an essential skill for managing chemical reactions and processes effectively.