Determine the degree of dissociation:

Problem:

Determine the degree of dissociation, and the ionic product constant of water at 298 K. Given density (at 298 K) = 0.9971 g cm^-3, conductivity at 298 K = 5.8 × 10^-8 ohm^-1 cm^-1.

Answer:

H₂O = H⁺  + OH ^-

Λ_m = κV

V is the volume in cm3 containing 1 g mole of the electrolyte

V = Molar mass/density = 18.016/0.9971 = 18.06 cm³

c for H₂O = 1000/18.06 = 55.3

Λ_m =κV = 5.8 × 10^-8 × 18.06 = 1.05 × 10^-6 ohm^-1 cm²

Λ ^∞ = Λ ^∞ (H⁺) + Λ ^∞ (OH ^- )

= 349.8 + 198.0 = 547.8 ohm^-1 cm².

α = Λ_m/ Λ ^∞ = 1.05 × 10^-6 ohm^-1 cm²/547.8 ohm^-1  cm² = 1.9 × 10^-9

H⁺ = OH ^- = cα = 55.3 × 1.9 × 10^-9    = 1.05 × 10^-7 M

H⁺ × OH ^- = Kw  = (1.05 × 10^-7)2 = 1.1 × 10^-14 (moles/dm³)²

K_w = 1.1 × 10^-14 (moles/dm³)²