Question

A student in a laboratory diluted 150 mL of 0.2 M acetic acid with 100 mL of 0.1 N NaOH in Beaker A and with 100 mL of water in Beaker B. The pH of the diluted acetic acid in two beakers will be

• A. 4.46 in both the beakers
• B. 2.84 in both the beakers
• C. 4.46 in beaker A and 2.84 in beaker B
• D. 2.84 in beaker A and 4.46 in beaker B

Hint:

Henderson-Hasselbalch: pH = pKa + log([A-]/[HA]).
For weak acid alone: [H+] = sqrt(Ka × C).
Acetic acid pKa = 4.76.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
This question tests the application of the Henderson-Hasselbalch equation and weak acid pH calculations.

Step 2: Key Formula or Approach:
Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA])
For weak acid alone: [H+] = sqrt(Ka × C)

Step 3: Detailed Explanation:
Beaker A (Acetic acid + NaOH):
Moles of acetic acid = 0.15 × 0.2 = 0.03 mol.
Moles of NaOH = 0.1 × 0.1 = 0.01 mol.
NaOH reacts with acetic acid:
CH3COOH + NaOH → CH3COONa + H2O
After reaction:
- Moles of acetic acid left = 0.03 - 0.01 = 0.02 mol.
- Moles of acetate formed = 0.01 mol.
Total volume = 150 + 100 = 250 mL = 0.25 L.
\[ pH = 4.76 + \log \left( \frac{0.01/0.25}{0.02/0.25} \right) = 4.76 + \log(0.5) = 4.76 - 0.301 = 4.46 \] Beaker B (Acetic acid + water):
Moles of acetic acid = 0.03 mol.
Total volume = 150 + 100 = 250 mL = 0.25 L.
Concentration of acetic acid = 0.03 0.25 = 0.12 M.
\[ [H+] = \sqrt{Ka \times C} = \sqrt{(1.8 \times 10^{-5}) \times 0.12} = \sqrt{2.16 \times 10^{-6}} = 1.47 \times 10^{-3} \] pH = -log(1.47 × 10^{-3}) = 2.84.
Thus, pH in Beaker A is 4.46 and in Beaker B is 2.84.

Step 4: Final Answer:
Thus, the correct answer is 4.46 in beaker A and 2.84 in beaker B, which corresponds to option (C).
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