Conductivity of a saturated solution of a sparingly soluble salt AB at 298 K is 1.85 × 10-5 S m-1. Solubility product of the salt AB at 298 K is Given ∧ m ° (AB) = 140 x 10-4 S m2 mol-1

Question

Conductivity of a saturated solution of a sparingly soluble salt AB at 298 K is 1.85 × 10-5 S m-1. Solubility product of the salt AB at 298 K is Given ∧ m ° (AB) = 140 x 10-4 S m2 mol-1 (A) 1.32 × 10-12 (B) 1.74 × 10-12 (C) 5.7 × 10-12 (D) 7.5 × 10-12

Tardigrade Admin · Accepted Answer

Lambdam°= (K × 1000/S) =(K × 1000/ Lambdam°)=(1.85 × 10-6 × 1000/140 × 10-4)=1.321 A B leftharpoons A++B- K sp =[A+][B-]=S2 =(1.321)2=1.74

Q. Conductivity of a saturated solution of a sparingly soluble salt $A B$ at $298 K$ is $1.85 \times 1 0^{- 5} S m^{- 1}$ . Solubility product of the salt $A B$ at $298 K$ is
Given $\land_{m}^{\circ}$ (AB) = 140 x $1 0^{- 4} S m^{2} m o l^{- 1}$

$1.32 \times 1 0^{- 12}$

$1.74 \times 1 0^{- 12}$

$5.7 \times 1 0^{- 12}$

$7.5 \times 1 0^{- 12}$

$Λ_{m}^{\circ} = \frac{K \times 1000}{S}$

$= \frac{K \times 1000}{Λ _{m}^{\circ}} = \frac{1.85 \times 1 0 ^{- 6} \times 1000}{140 \times 1 0 ^{- 4}} = 1.321$

$A B ⇌ A^{+} + B^{-}$

$K_{s p} = [A^{+}] [B^{-}] = S^{2}$

$= (1.321)^{2} = 1.74$

Q. Conductivity of a saturated solution of a sparingly soluble salt AB at 298K is 1.85×10−5Sm−1. Solubility product of the salt AB at 298K is Given ∧m∘​(AB) = 140 x 10−4Sm2mol−1

1.32×10−12

1.74×10−12

5.7×10−12

7.5×10−12