Conductivity of a saturated solution of a sparingly soluble salt ( A B ) at ( 298 K ) is ( 1.85 10⁻⁵ Sm ⁻¹ ). Solubility product of the salt AB at ( 298 ~K ) is Given ( _ m ^ (A B)=140 10⁻⁴ ~S ~m ² ~mol ⁻¹ )

Step-by-step solution

Given, ( _ m ^ =140 10⁻⁴ Sm ² ~mol ⁻¹ ) ( K=1.85 10⁻⁵ Sm ⁻¹ ) Now, molar conductivity of the saturated salt is, [ array l . _ m ^ = K 1000 C = 1.85 10⁻⁵ 1000 S [Concentration ( C )= Solubility (S) ] 140 10⁻⁴= 1.85 10⁻⁵ 1000 S S= 1.85 10⁻⁵ 1000 140 10⁻⁴ S=1.32 10⁻⁶ array ] For the saturated salt [ array l A B A⁺+B⁻ K_ s p = [A⁺ ] [B⁻ ]=S² K_ s p [A⁺ ] [B⁻ ]=S² K_ s p =S² = (1.32 10⁻⁶ )² =1.74 10⁻¹² array ]