Question

Conductivity of a saturated solution of a sparingly soluble salt $AB$ at $298\, K$ is $1.85 \times 10^{-5} S m^{-1}$. Solubility product of the salt $AB$ at $298\, K$ is
Given $\wedge _m ^{\circ} $(AB) = 140 x $10^{-4} S m^2 mol^{-1}$

• A. $1.32 \times 10^{-12}$
• B. $1.74 \times 10^{-12}$
• C. $5.7 \times 10^{-12}$
• D. $7.5 \times 10^{-12}$

Answer 1

Answer: (B)

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

$=\frac{K \times 1000}{\Lambda_{m}^{\circ}}=\frac{1.85 \times 10^{-6} \times 1000}{140 \times 10^{-4}}=1.321$

$A B \rightleftharpoons A^{+}+B^{-}$

$K_{ sp }=\left[A^{+}\right]\left[B^{-}\right]=S^{2}$

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