Question

The \(pK_a\) of acetic acid is \(4.74\). The concentration of \( \mathrm{CH_3COOH} \) is \(0.01\, M\). The pH of \( \mathrm{CH_3COOH} \) is _ _ _ _ _ .

Hint:

For weak acids: \(\text{pH} = \frac{1}{2}(pK_a - \log C)\) — memorize this shortcut!

Answer 1

Correct Answer: 3.37

Step 1: Understanding the Concept:
For a weak monobasic acid like acetic acid, the \(\text{pH}\) depends on the dissociation constant and initial concentration.
Step 2: Key Formula or Approach:
For a weak acid:
\[ \text{pH} = \frac{1}{2} (\text{p}K_{a} - \log C) \]
Step 3: Detailed Explanation:
Given:
\(\text{p}K_{a} = 4.74\).
Concentration \(C = 0.01\text{ M} = 10^{-2}\text{ M}\).
\[ \log C = \log(10^{-2}) = -2 \]
Substitute into the formula:
\[ \text{pH} = \frac{1}{2} (4.74 - (-2)) \]
\[ \text{pH} = \frac{1}{2} (4.74 + 2) = \frac{6.74}{2} \]
\[ \text{pH} = 3.37 \]
Step 4: Final Answer:
The \(\text{pH}\) of the solution is 3.37.