The solubility product of a sparingly soluble | vedclass

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(C) The dissociation of the salt is given by: $A_{2}X_{3(s)} \rightleftharpoons 2A^{3+}_{(aq)} + 3X^{2-}_{(aq)}$.Let the solubility be $s \ mol \ L^{-

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$3$

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(C) The dissociation of the salt is given by: $A_{2}X_{3(s)} ightleftharpoons 2A^{3+}_{(aq)} + 3X^{2-}_{(aq)}$.Let the solubility be $s \ mol \ L^{-1}$. Then $[A^{3+}] = 2s$ and $[X^{2-}] = 3s$.The solubility product $K_{sp} = (2s)^2(3s)^3 = 108s^5 = 1.1 	imes 10^{-23}$.$s^5 = \frac{1.1 	imes 10^{-23}}{108} \approx 1.018 	imes 10^{-25}$.$s \approx 10^{-5} \ mol \ L^{-1} = 10^{-5} \ mol \ (10^{-3} \ m^3)^{-1} = 0.01 \ mol \ m^{-3}$.For a sparingly soluble salt,the molar conductivity $\Lambda_m \approx \Lambda_m^{\infty} = \frac{\kappa}{C}$,where $\kappa$ is the specific conductance and $C$ is the concentration in $mol \ m^{-3}$.$\Lambda_m^{\infty} = \frac{3 	imes 10^{-5} \ S \ m^{-1}}{0.01 \ mol \ m^{-3}} = 3 	imes 10^{-3} \ S \ m^2 \ mol^{-1}$.Comparing this with $x 	imes 10^{-3} \ S \ m^2 \ mol^{-1}$,we get $x = 3$.

The solubility product of a sparingly soluble salt $A_{2}X_{3}$ is $1.1 \times 10^{-23}$. If specific conductance of the solution is $3 \times 10^{-5} \ S \ m^{-1}$,the limiting molar conductivity of the solution is $x \times 10^{-3} \ S \ m^{2} \ mol^{-1}$. The value of $x$ is ...

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