Determine the solubility product:

Problem:

A saturated solution of silver chloride at 293 K has a conductivity of 3.41 × 10^-6 ohm^-1 cm^-1. The conductivity of water used was 1.60 × 10^-6 ohm^-1 cm^-1.  Determine the solubility product of silver chloride. Given that Λ ∝ (Ag⁺) = 61.92 ohm-1 cm² mol^-1 and Λ∝ (Cl^-)= 76.34 ohm^-1   cm² mol^-1

Answer:

K_AgCl = (3.41 - 1.60) × 10^-6 ohm^-1 cm^-1

= 1.81 × 10^-6 ohm^-1 cm^-1

Λ_AgCl  = Λ ∞ Ag⁺ + Λ ∞ Cl ^-  = (61.92 + 76.34) ohm^-1 cm² mol^-1

= 138.26 ohm^-1 cm² mol^-1

Λ_AgCl = κ/c, therefore,

c= κ/Λ _AgCl = 1.81 × 10^-6 ohm^-1 cm^-1/ 138.26 ohm^-1 cm² mol^-1

=0.01309×10^-6  mol cm^-3

= 1.31 × 10^-5   mol dm^-3

Ksp = [Ag⁺] [Cl ^- ]

= (1.31 × 10^-5] [1.31 × 10^-5]

= 1.71 × 10^-10     (mol dm^-3)²