The solubility product of a sparingly soluble | vedclass

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(B) For the salt $A_2B$,the dissociation equilibrium is $A_2B \rightleftharpoons 2A^{+} + B^{2-}$.Let the solubility be $s \ mol \cdot L^{-1}$.The con

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$2 	imes 10^{-4}$

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(B) For the salt $A_2B$,the dissociation equilibrium is $A_2B ightleftharpoons 2A^{+} + B^{2-}$.Let the solubility be $s \ mol \cdot L^{-1}$.The concentrations of the ions are $[A^{+}] = 2s$ and $[B^{2-}] = s$.The solubility product expression is $K_{sp} = [A^{+}]^2 [B^{2-}] = (2s)^2(s) = 4s^3$.Given $K_{sp} = 3.2 	imes 10^{-11}$.Equating the two: $4s^3 = 3.2 	imes 10^{-11}$.$s^3 = \frac{3.2 	imes 10^{-11}}{4} = 0.8 	imes 10^{-11} = 8 	imes 10^{-12}$.Taking the cube root: $s = \sqrt[3]{8 	imes 10^{-12}} = 2 	imes 10^{-4} \ mol \cdot L^{-1}$.

Metal sulfide	Solubility product
$A$. $PbS$	$I$. $4.0 \times 10^{-53}$
$B$. $HgS$	$II$. $8.0 \times 10^{-28}$
$C$. $MnS$	$III$. $1.6 \times 10^{-24}$
$D$. $ZnS$	$IV$. $2.5 \times 10^{-13}$

The solubility product of a sparingly soluble salt $A_2B$ is $3.2 \times 10^{-11}$. Its solubility in $mol \cdot L^{-1}$ is

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