Conductivity of a saturated solution of a spa | vedclass

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(B) Given,$\Lambda_{m}^{\circ} = 140 	imes 10^{-4} \ S \ m^{2} \ mol^{-1}$,$K = 1.85 	imes 10^{-5} \ S \ m^{-1}$.For a sparingly soluble salt,the mo

Vedclass · Accepted Answer

$1.74 	imes 10^{-12}$

Vedclass · Answer

(B) Given,$\Lambda_{m}^{\circ} = 140 	imes 10^{-4} \ S \ m^{2} \ mol^{-1}$,$K = 1.85 	imes 10^{-5} \ S \ m^{-1}$.For a sparingly soluble salt,the molar conductivity $\Lambda_{m}^{\circ}$ is related to solubility $S$ (in $mol \ m^{-3}$) by the formula $\Lambda_{m}^{\circ} = \frac{K}{S}$.$S = \frac{K}{\Lambda_{m}^{\circ}} = \frac{1.85 	imes 10^{-5} \ S \ m^{-1}}{140 	imes 10^{-4} \ S \ m^{2} \ mol^{-1}} = 1.32 	imes 10^{-3} \ mol \ m^{-3}$.Since $1 \ m^{3} = 1000 \ L$,the solubility in $mol \ L^{-1}$ is $S = \frac{1.32 	imes 10^{-3}}{1000} = 1.32 	imes 10^{-6} \ mol \ L^{-1}$.For the salt $AB$,$K_{sp} = S^{2} = (1.32 	imes 10^{-6})^{2} = 1.74 	imes 10^{-12}$.

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 $\Lambda_{m}^{\circ}(AB) = 140 \times 10^{-4} \ S \ m^{2} \ mol^{-1}$.

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