Calculate degree of ionization:

Problem:

The resistance of a 0.02 mol dm^-3 solution of acetic acid in a cell (cell constant =0.2063 cm^-1) was found to be 888 ohm. What is the degree of ionization of the acid at this concentration? (Given Λm for acetic acid = 387.9 ×10^-4 ohm^-1 mol^-1 m²)

Answer:

K = 1/R (Cell constant)

= 1/888 ohm (0.2063 cm^-1)

= 2.324 × 10^-4 ohm^-1 cm^-1

Λ_m= K/c = 2.324 × 10_-4 ohm_-1 cm_-1/ 0.02 mol dm_-3

=1000× 2.324× 10_-4 ohm_-1 cm_-1/0.02molcm_-3

= 11.62  ohm_-1  mol_-1 cm₂

Λ ^∞ = 387.9 × 10^-4 ohm^-1 mol^-1 m² = 387.9 ohm^-1  mol^-1 cm²

α = Λ_m/Λ ^∞  = 11.62 ohm^-1  mol^-1 cm²/387.9 ohm^-1  mol^-1 cm²

= 2.99 × 10^-2

= ~0.03