1 × 10 - 5   M   AgNO 3 is added to 1   L of saturated solution of AgBr . The conductivity of this solution at 298   K is _ _ _ _ _ _ _ × 10 - 8   S   m - 1 . [Given : K sp ( AgBr ) = 4 . 9 × 10 - 13 at 298   K λ Ag + 0 = 6 × 10 - 3 Sm 2   mol - 1 λ Br - 0 = 8 × 10 - 3 Sm 2   mol - 1 λ NO 3 - 0 = 7 × 10 - 3 Sm 2   mol - 1

Step-by-step solution

The concentrations of the ions can be calculated as follows, Ag + = 10 - 5 M NO 3 - = 10 - 5 M Br - = Ksp Ag + = 4 . 9 × 10 - 8 M Λ m = k 1000 × M = Specific   conductance 1000 × Molarity For Ag + 6 × 10 - 3 = k Ag + 1000 × 10 - 5 k Ag + = 6 × 10 - 5 Sm - 1 ⇒ 6000 × 10 - 8 Sm - 1 for Br - 8 × 10 - 3 = k Br - 1000 × 4 . 9 × 10 - 8 k Br - = 39 . 2 × 10 - 8 Sm - 1 for NO 3 - 7 × 10 - 3 = k NO 3 - 1000 × 10 - 5 k NO 3 - = 7 × 10 - 5 S