Problem 53 - Analytical Chemistry

Question

Use the Henderson-Hasselbalch equation to find the value of \(\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right] /\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-}\right]\) in a solution at (a) \(\mathrm{pH} 3.00,\) and (b) \(\mathrm{pH}\) 5.00. For \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}, \mathrm{pK}_{a}\) is 4.20 .

Step-by-Step Solution

Answer

(a) The ratio is approximately 15.8 at pH 3.00; (b) The ratio is approximately 0.16 at pH 5.00.

Step 1: Understand the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is useful for estimating the pH of a buffer solution. It is given by: \[ \mathrm{pH} = \mathrm{pK}_a + \log \left( \frac{\left[\text{A}^-\right]}{\left[\text{HA}\right]} \right) \] where \(\text{HA}\) is the weak acid and \(\text{A}^-\) is its conjugate base. For this problem, \( \text{HA} = \text{C}_6\text{H}_5\text{COOH} \) and \( \text{A}^- = \text{C}_6\text{H}_5\text{COO}^- \).

Step 2: Solve for part (a) - Determining ratio at pH 3.00

For a solution of \( \text{C}_6\text{H}_5\text{COOH} \) at pH 3.00, use the Henderson-Hasselbalch equation: \[ 3.00 = 4.20 + \log \left( \frac{\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{COO}^-\right]}{\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}\right]} \right) \] Rearrange to find the ratio: \[ \log \left( \frac{\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{COO}^-\right]}{\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}\right]} \right) = 3.00 - 4.20 = -1.20 \] Take the antilog: \[ \frac{\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{COO}^-\right]}{\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}\right]} = 10^{-1.20} \approx 0.063 \] So, \[ \frac{\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}\right]}{\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{COO}^-\right]} \approx 15.8 \]

Step 3: Solve for part (b) - Determining ratio at pH 5.00

For the pH 5.00 solution, use the same method: \[ 5.00 = 4.20 + \log \left( \frac{\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{COO}^-\right]}{\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}\right]} \right) \] Rearrange to find the ratio: \[ \log \left( \frac{\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{COO}^-\right]}{\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}\right]} \right) = 5.00 - 4.20 = 0.80 \] Take the antilog: \[ \frac{\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{COO}^-\right]}{\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}\right]} = 10^{0.80} \approx 6.31 \] Therefore, \[ \frac{\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}\right]}{\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{COO}^-\right]} \approx 0.16 \]

Key Concepts

Buffer Solutions

Buffer solutions are chemical mixtures that can resist changes in pH when small amounts of an acid or a base are added to them. They are significant in many biological and chemical processes, maintaining stability in the pH of the environment.
A buffer solution typically contains a weak acid and its conjugate base, or a weak base and its conjugate acid. These pairs effectively work by neutralizing added acids or bases, thereby maintaining a steady pH level.
To understand better, consider a solution containing acetic acid (\( ext{CH}_3 ext{COOH} \)) and its conjugate base acetate (\( ext{CH}_3 ext{COO}^- \)). Added hydrogen ions are picked up by the acetate ions, and added hydroxide ions by the acetic acid, keeping the pH relatively constant.
There are numerous real-world applications for buffer solutions:

• Blood is a natural buffer that maintains a pH around 7.4, essential for physiological processes.
• Buffers are used in laboratories to maintain the pH of experiments that could be sensitive to pH fluctuations.
• Additionally, they're crucial in manufacturing to ensure products maintain the correct pH level for safety and quality.

pKa of Benzoic Acid

The pKa value is a fundamental concept to understand when dealing with acids and bases. It represents the acid dissociation constant, which is a measure of the strength of an acid in solution.
For benzoic acid (\( ext{C}_6 ext{H}_5 ext{COOH} \)), the pKa is 4.20. This value determines the pH at which half the acid is dissociated. A lower pKa value means a stronger acid, because it dissociates more in solution.
In the context of buffer solutions, knowing the pKa is crucial as it helps to calculate the pH of the buffer using the Henderson-Hasselbalch equation. This equation relates pH, pKa, and the ratio of concentrations of the conjugate acid-base pair: \[ ext{pH} = ext{pKa} + \log \left( \frac{[ ext{A}^-]}{[ ext{HA}]} \right) \]
Benzoic acid is widely used for understanding acidity and buffering due to its relatively stable pKa. It is often included in educational settings to practice calculations related to chemical equilibria and buffer formulations. Understanding the pKa gives insights into how different substances react under various pH conditions and how to manipulate conditions to maintain desired acidity levels.

Acid-Base Equilibrium

In chemistry, acid-base equilibrium refers to the balance between acidic and basic (alkaline) species in a solution. It is a key concept in understanding how solutions behave when acids or bases are added.
The equilibrium is based on the reversible proton transfer that occurs between an acid and a base. According to the Bronsted-Lowry theory, an acid donates protons, while a base accepts protons. This defines the relationship between conjugate acid-base pairs.
When we focus on benzoic acid and its conjugate base, benzoate (\( ext{C}_6 ext{H}_5 ext{COOH} \) and \( ext{C}_6 ext{H}_5 ext{COO}^- \) respectively), the equilibrium can be represented as:\[ ext{C}_6 ext{H}_5 ext{COOH} \leftrightarrow ext{C}_6 ext{H}_5 ext{COO}^- + ext{H}^+ \]
The position of equilibrium depends on the pH of the solution compared to the pKa of the acid. At pH values lower than the pKa, the free acidic form dominates. Conversely, at higher pH levels, the conjugate base form is more prevalent.

• This balance is vital for reactions and processes where specific pH levels are necessary to ensure stability and efficiency.
• Acid-base equilibrium calculations help determine how a solution will behave when components are added or conditions are altered, such as in buffer solutions.

Understanding these dynamics allows chemists to predict and manipulate how acids and bases will interact in different chemical environments.