Question

The pH of the solution containing 50 mL each of 0.10 M sodium acetate and 0.01 M acetic acid is
[Given pKa of CH3COOH = 4.57]

• A. 5.57
• B. 3.57
• C. 4.57
• D. 2.57

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
A mixture of a weak acid (\(CH_3COOH\)) and its salt with a strong base (\(CH_3COONa\)) constitutes an acidic buffer.
Key Formula or Approach:
The pH of an acidic buffer is calculated using the Henderson-Hasselbalch equation:
\[ pH = pK_a + \log \frac{[\text{Salt}]}{[\text{Acid}]} \]
Note: Since volumes are equal (50 mL each), the final concentration ratio remains the same as the initial ratio.
Step 2: Detailed Explanation:
Given:
\(pK_a = 4.57\)
Concentration of Salt (\(CH_3COONa\)) = 0.10 M
Concentration of Acid (\(CH_3COOH\)) = 0.01 M

Substitute these values into the formula:
\[ pH = 4.57 + \log \left( \frac{0.10}{0.01} \right) \]
Simplify the fraction:
\[ \frac{0.10}{0.01} = 10 \]
Now, calculate the logarithm:
\[ pH = 4.57 + \log_{10}(10) \]
Since \(\log_{10}(10) = 1\):
\[ pH = 4.57 + 1 = 5.57 \]
Step 3: Final Answer:
The pH of the solution is 5.57.