Question

20 ml of CH\(_3\)COOH solution is neutralised by 28.08 ml of NaOH. In the 20 ml of same solution 14.04 ml of same NaOH solution is poured, then pH of resultant solution will be: (pKa of CH\(_3\)COOH = 4.75)

• A. \(4.75 \)
• B. \(7 \)
• C. \(3.5 \)
• D. \(4.5 \)

Answer 1

The Correct Option is A

Concept: When a weak acid is partially neutralised by a strong base, a buffer solution is formed. The pH of such a solution is given by the Henderson–Hasselbalch equation: \[ \text{pH} = \text{pKa} + \log \left( \frac{\text{Salt}}{\text{Acid}} \right) \]
Step 1: Understand the neutralisation reaction.
\[ \text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COO}^- \text{Na}^+ + \text{H}_2\text{O} \]
Step 2: Identify equivalence and half-equivalence point.
Given: • 28.08 ml NaOH is required for complete neutralisation. • 14.04 ml NaOH = half of 28.08 ml. Thus, the system is at half-equivalence point.
Step 3: Apply buffer concept.
At half-equivalence point: \[ [\text{Salt}] = [\text{Acid}] \] So, \[ \log \left( \frac{\text{Salt}}{\text{Acid}} \right) = \log(1) = 0 \]
Step 4: Calculate pH.
\[ \text{pH} = \text{pKa} + 0 = 4.75 \]