The conductivity of a weak acid HA of concentration 0.001 mol L⁻¹ is 2.0 × 10⁻⁵ S cm⁻¹. If Λ ₘᵒ(HA) = 190 S cm² mol⁻¹, the ionization constant (Kₐ) of HA is equal to ______________ × 10⁻⁶. (Round off to the Nearest Integer)

Solution

Chemistry ma’am · 4 steps

Get Λₘ from conductivity, then α, and keep the 1-α in Ostwald's law.

1. Molar conductivity. Convert conductivity to molar conductivity: Λₘ = (κ × 1000)/c = (2.0 × 10⁻⁵ × 1000)/0.001 = 20 S cm² mol⁻¹.
2. Degree of dissociation. For a weak electrolyte, α = Λₘ/Λₘ° = 20/190 = 0.105. This is about 10%, too large to drop from 1 - α.

   Kₐ = cα²/(1-α) = (0.001 × (20/190)²)/(170/190) = (0.001 × 400)/(190 × 170)

3. Put in numbers. Kₐ = 0.4/32300 = 1.238 × 10⁻⁵ = 12.38 × 10⁻⁶.
4. So. Kₐ ≈ 12.38 × 10⁻⁶, which rounds to 12.

The common slip: 11. Using the approximation Kₐ = cα² and dropping 1 - α gives 11.08 × 10⁻⁶, which rounds to 11, not 12.

The idea to keep. Conductivity to Λₘ, then α = Λₘ/Λₘ°, then Ostwald's law. Use 1-α in the denominator unless α is really tiny.

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