Ionic Equilibrium - Weak Acids
A 0.1 M acetic acid solution has a degree of dissociation of 1.34 \times 10^{-2} at 25°C. What is the dissociation constant K_a of acetic acid?
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About This Question
- Subject
- chemistry
- Chapter
- equilibrium
- Topic
- ionic equilibrium - weak acids
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
1.80×10−5
The acid dissociation constant K_a is related to the degree of dissociation \alpha and the initial concentration C by the approximation K_a \approx C\alpha^2 when \alpha << 1. Given C = 0.1 M and \alpha = 1.34 \times 10^{-2}: K_a = 0.1 \times (1.34 \times 10^{-2})^2 = 0.1 \times 1.7956 \times 10^{-4} = 1.796 \times 10^{-5} \approx 1.80 \times 10^{-5}. Since \alpha = 0.0134 << 1, the approximation (1-\alpha) \approx 1 is valid and introduces negligible error. Option 1.34 \times 10^{-3} is incorrect; it comes from K_a = C\alpha rather than C\alpha^2, forgetting to square the degree of dissociation. Option 1.34 \times 10^{-2} simply equates K_a with \alpha, which is dimensionally and conceptually wrong. Option 2.69 \times 10^{-4} doubles the correct answer, possibly from an error where \alpha is multiplied by 2 instead of squared. This calculation pattern is directly from NCERT Equilibrium and is a standard numerical in JEE Main. Plausibility check: K_a of acetic acid is known to be ~1.8 \times 10^{-5}, confirming our answer is correct in both magnitude and sign.
This medium difficulty chemistry question is from the chapter equilibrium, covering the topic of ionic equilibrium - weak acids. It appeared in the 2025 exam.
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