The specific conductance of 0.0025 M acetic acid is 5 × 10⁻⁵ S cm⁻¹ at a certain temperature. The dissociation constant of acetic acid is __________ × 10⁻⁷ (Nearest integer)
Consider limiting molar conductivity of CH₃ COOH as 400 S cm² mol⁻¹

Solution

Chemistry ma’am · 4 steps

Turn conductivity into molar conductivity, compare it with the limit to get α, then use Ostwald's law.

1. Molar conductivity. With κ in S cm⁻¹ and C in mol L⁻¹, Λₘ = (1000 κ)/C = (1000 × 5 × 10⁻⁵)/0.0025 = 20 S cm² mol⁻¹.
2. Degree of dissociation. For a weak electrolyte, α = Λₘ/Λₘ° = 20/400 = 0.05.

   Kₐ = Cα²/(1-α) = (0.0025 × (0.05)²)/0.95 = (6.25 × 10⁻⁶)/0.95

3. Work it out. Kₐ = 6.58 × 10⁻⁶ = 65.8 × 10⁻⁷. Here α is 5%, so keeping the 1-α term changes the nearest integer.
4. So. Kₐ ≈ 65.8 × 10⁻⁷, which rounds to 66.

The common slip: 63. Using the shortcut Kₐ = Cα² and ignoring the 1-α term gives 62.5 × 10⁻⁷, about 63, not 66.

The idea to keep. For a weak electrolyte, go from κ to Λₘ to α = Λₘ/Λₘ°, then use K = Cα²/(1-α). Drop 1-α only when the answer does not need that precision.

Answer from the NTA final answer key. Solution by PrepWiser.