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) \(6.67\)

Short answer

The pH of the solution is 5.09, so the correct answer is (c) 5.09.

Step-by-step solution

Understanding the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation relates the pH of a solution to the pKa of the acid and the concentrations of the acid and its conjugate base. It's given by: \[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]where \([\text{A}^-]\) is the concentration of the acetate ion (sodium acetate) and \([\text{HA}]\) is the concentration of acetic acid.

Identify the values to substitute

From the problem, we have the pKa of acetic acid, \( \text{pKa} = 4.57 \), the concentration of sodium acetate \([\text{A}^-] = 0.10 \text{ M}\), and the concentration of acetic acid \([\text{HA}] = 0.03 \text{ M}\).

Substitute the values into the Henderson-Hasselbalch equation

Substitute the given values into the equation:\[ \text{pH} = 4.57 + \log \left( \frac{0.10}{0.03} \right) \]

Calculate the log term

Calculate \(\frac{0.10}{0.03} = 3.33\). Now calculate the logarithm: \[ \log(3.33) \approx 0.52 \]

Calculate the pH

Substitute the result from step 4 into the equation: \[ \text{pH} = 4.57 + 0.52 = 5.09 \]

Conclusion: Determine the correct option

The calculated pH of the solution is 5.09. Match this value with the given options to find that the correct answer is (c) 5.09.

Key concepts

pH Calculation

When we talk about pH calculation, we refer to the process of measuring how acidic or basic a solution is. pH is a scale used to specify the acidity or basicity of an aqueous solution. It generally ranges from 0 to 14. The lower the pH, the more acidic the solution is; the higher the pH, the more basic it is.
The pH is calculated using the equation:

• \[ ext{pH} = - ext{log}[ ext{H}^+] \]

where \([ ext{H}^+]\) is the concentration of hydrogen ions in the solution. However, for buffer solutions, a specific scenario, we use the Henderson-Hasselbalch equation.
This equation simplifies the process by relating pH to the pKa and the concentrations of an acid and its conjugate base. This is especially useful for weak acids, like acetic acid, and their salts, like sodium acetate. Understanding these relationships helps predict the behavior and characteristics of the solution's pH, providing insight into its capability to maintain stability against added acids or bases.

Acid-Base Equilibrium

Acid-base equilibrium deals with the balance between acids and bases in a solution. This balance is fundamental to many biochemical processes. It's crucial for understanding how solutions react to changes in pH levels.
The dissociation of acids and bases depends on their strengths, which are determined partly by their dissociation constants, termed \(K_a\) for acids or \(K_b\) for bases.
For acetic acid, a weak acid, the equilibrium can be represented as:

• \[ ext{CH}_3 ext{COOH} ightleftharpoons ext{CH}_3 ext{COO}^- + ext{H}^+ \]
• The equilibrium constant \(K_a\) for this reaction helps quantify the extent of dissociation.

In pretty straightforward terms, the equilibrium reflects how easily the acid donates protons.
In buffer solutions, the equilibrium concept supports the understanding that the solution resists drastic pH changes due to the presence of equal concentrations of the acid and its conjugate base, which compensates for the added acid or base.

Buffer Solution

A buffer solution is a special type of solution that resists changes in its pH when small amounts of acids or bases are added. This happens because of the presence of a weak acid and its conjugate base, or a weak base and its conjugate acid.
In our example, we have acetic acid (\[ ext{CH}_3 ext{COOH}\]) and its conjugate base, acetate (\[ ext{CH}_3 ext{COO}^-\]). These components work together to maintain a stable pH.

• The weak acid neutralizes added bases, while its conjugate base neutralizes added acids.
• This dual action helps mitigate drastic pH shifts.

A buffer's effectiveness depends on the concentration of its acid and conjugate base components. The buffer range is usually within one pH unit of the pKa, making the Henderson-Hasselbalch equation handy for determining the buffer's pH.
Buffer solutions are essential in biological systems where maintaining a constant pH is critical for proper enzyme activity and cellular functions.