Calculate the solubility:

Problem:

The conductivity of a saturated aqueous solution of barium sulphate at 298 K is 1.84 × 10^-3 ohm^-1 m^-1 and that of water is 1.60 × 10^-4 ohm m^-1. The ionic conductivities at infinite dilution of Ba²⁺ and SO₄^2- ions at 298 K are 127.2 ×10^-4 ohm-1 m² mol^-1 and 160.0 × 10^-4   ohm^-1 cm² mol^-1 respectively. Calculate the solubility and solubility product of barium sulphate at 298 K.

Answer:

Lets c be the solubility of the salt

c = K/Λ_{BaSO 4}   = K/ ½  (Λ^∞ Ba²⁺ + Λ ^∞ SO₄^2- )

K = conductivity of the salt

= (1.84 × 10^-3 - 1.60 × 10^-4) ohm^-1 m^-1.

= 1.68 × 10^-3 ohm^-1 m^-1.

c = 1.68 × 10^-3 ohm^-1 m^-1 / ½  (127.2 × 10^-4 + 160.0 × 10^-4) ohm^-1  m² mol^-1

= 0.0117 mol m^-3

= 1.17× 10^-4 mol dm^-3

Therefore, the solubility product = [1.17 × 10^-4] [1.17 × 10^-4]

= 1.370 × 10^-8 (mol dm^-3)