Ionic Equilibrium
The degree of dissociation of a 0.1 M weak monoprotic acid HA is 0.01. The dissociation constant Ka of the acid is:
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About This Question
- Subject
- chemistry
- Chapter
- equilibrium
- Topic
- ionic equilibrium
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
1.0 × 10⁻⁵
For a weak monoprotic acid HA ⇌ H⁺ + A⁻ with initial concentration c and degree of dissociation α, the equilibrium concentrations are: [HA] = c(1 − α), [H⁺] = cα, [A⁻] = cα. The dissociation constant Ka = [H⁺][A⁻]/[HA] = (cα)(cα)/c(1 − α) = cα²/(1 − α). Given c = 0.1 M and α = 0.01 (which is small, so 1 − α ≈ 1): Ka = 0.1 × (0.01)² / 1 = 0.1 × 10⁻⁴ = 1.0 × 10⁻⁵. The approximation 1 − α ≈ 1 is valid when α < 0.05 (5%). This result also confirms Ostwald's dilution law: Ka = cα², which holds for weak electrolytes with small α values.
This medium difficulty chemistry question is from the chapter equilibrium, covering the topic of ionic equilibrium. It appeared in the 2025 exam.
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