Acid–base Equilibria
A buffer is made by mixing 0.10 M acetic acid and 0.20 M sodium acetate. (Ka for acetic acid = 1.8×10⁻⁵). The pH of the buffer is (25°C):
Select the correct option:
Solution
5.04
- Buffer Type: Acidic buffer (Weak acid + Salt with strong base).
- Henderson–Hasselbalch Equation: pH=pKa+log[Acid][Salt].
- pKa Calculation:
- pKa=−log(1.8×10−5)
- pKa=5−log1.8=5−0.255=4.745.
- Substituting Values:
- [Salt]=0.20 M, [Acid]=0.10 M.
- pH=4.745+log(0.20/0.10)
- pH=4.745+log2
- pH=4.745+0.301=5.046.
- Result: The buffer pH is approximately 5.04.
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About This Question
- Subject
- chemistry
- Chapter
- equilibrium
- Topic
- acid–base equilibria
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
5.04
- Buffer Type: Acidic buffer (Weak acid + Salt with strong base).
- Henderson–Hasselbalch Equation: pH=pKa+log[Acid][Salt].
- pKa Calculation:
- pKa=−log(1.8×10−5)
- pKa=5−log1.8=5−0.255=4.745.
- Substituting Values:
- [Salt]=0.20 M, [Acid]=0.10 M.
- pH=4.745+log(0.20/0.10)
- pH=4.745+log2
- pH=4.745+0.301=5.046.
- Result: The buffer pH is approximately 5.04.
This medium difficulty chemistry question is from the chapter equilibrium, covering the topic of acid–base equilibria. It appeared in the 2025 exam.
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