The conductivity of a saturated solution of B | vedclass

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(C) The relationship between equivalent conductance $(\Lambda_{eq})$,conductivity $(\kappa)$,and normality $(N)$ is given by: $\Lambda_{eq} = \frac{\k

Vedclass · Accepted Answer

$4 	imes 10^{-6} \ M^2$

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(C) The relationship between equivalent conductance $(\Lambda_{eq})$,conductivity $(\kappa)$,and normality $(N)$ is given by: $\Lambda_{eq} = \frac{\kappa 	imes 1000}{N}$.Given $\kappa = 3.06 	imes 10^{-6} \ \Omega^{-1} \ cm^{-1}$ and $\Lambda_{eq} = 1.53 \ \Omega^{-1} \ cm^2 \ eq^{-1}$.Substituting the values: $1.53 = \frac{3.06 	imes 10^{-6} 	imes 1000}{N}$.$N = \frac{3.06 	imes 10^{-3}}{1.53} = 2 	imes 10^{-3} \ N$.For $BaSO_4$,the molarity $(M)$ is equal to the normality $(N)$ because the n-factor is $2$ $(Ba^{2+} + SO_4^{2-})$,but the solubility $(S)$ is the molar concentration of the saturated solution.Since $BaSO_4 ightleftharpoons Ba^{2+} + SO_4^{2-}$,the solubility $S = M = 2 	imes 10^{-3} \ M$.The solubility product $K_{sp} = [Ba^{2+}][SO_4^{2-}] = S^2$.$K_{sp} = (2 	imes 10^{-3})^2 = 4 	imes 10^{-6} \ M^2$.

The conductivity of a saturated solution of $BaSO_4$ is $3.06 \times 10^{-6} \ \Omega^{-1} \ cm^{-1}$ and its equivalent conductance is $1.53 \ \Omega^{-1} \ cm^2 \ eq^{-1}$. Then the $K_{sp}$ for $BaSO_4$ will be:

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