Question

20 mL of a solution of acetic acid required 28.4 mL of 0.1 M NaOH for its neutralization. A solution (X) was prepared by mixing 20 mL of the above acetic acid and 14.2 mL of 0.1 M NaOH solution. What is the pH of the solution (X)? (\( pK_a \) value of acetic acid is 4.75).

• A. 7.0
• B. 4.75
• C. 3.5
• D. 4.82

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Neutralization of a weak acid with a strong base forms a salt. If only half of the acid is neutralized, a buffer solution containing the weak acid and its salt is formed.

Step 2: Key Formula or Approach:
Henderson-Hasselbalch equation: \[ pH = pK_a + \log \frac{[\text{salt}]}{[\text{acid}]} \]

Step 3: Detailed Explanation:
1. Find moles of acid: At neutralization, moles acid = moles base.
Moles acid in 20 mL = \( 28.4 \text{ mL} \times 0.1 \text{ M} = 2.84 \text{ mmol} \).
2. Prepare solution X: Mix 2.84 mmol acid with \( 14.2 \text{ mL} \times 0.1 \text{ M} = 1.42 \text{ mmol} \) NaOH.
3. Reaction: \( CH_3COOH + NaOH \to CH_3COONa + H_2O \).
Initial: acid = 2.84, base = 1.42.
Reacted: acid = -1.42, base = -1.42, salt = +1.42.
Final: acid = 1.42 mmol, salt = 1.42 mmol.
4. Since [acid] = [salt], the log term in HH equation becomes \( \log(1) = 0 \).
\( pH = pK_a = 4.75 \).

Step 4: Final Answer:
The pH of the solution is 4.75.