For a compound AB,the degree of ionization is | vedclass

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(D) The dissociation of $AB$ is given by: $AB \rightleftharpoons A^{+} + B^{-}$.Let the solubility be $s$. Since the degree of ionization is $80\%$,th

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$10^{-4}$

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(D) The dissociation of $AB$ is given by: $AB ightleftharpoons A^{+} + B^{-}$.Let the solubility be $s$. Since the degree of ionization is $80\%$,the concentration of ions formed is $0.8s$.Thus,$[A^{+}] = 0.8s$ and $[B^{-}] = 0.8s$.The solubility product expression is $K_{sp} = [A^{+}][B^{-}]$.Substituting the values: $(0.8s)(0.8s) = 6.4 	imes 10^{-9}$.$0.64 s^2 = 6.4 	imes 10^{-9}$.$s^2 = \frac{6.4 	imes 10^{-9}}{0.64} = 10 	imes 10^{-9} = 10^{-8}$.Therefore,$s = \sqrt{10^{-8}} = 10^{-4}$.

$S.No.$	$Formula$	$K_{sp}$
$1$	$PQ$	$4.0 \times 10^{-20}$
$2$	$PQ_2$	$3.2 \times 10^{-14}$
$3$	$PQ_3$	$2.7 \times 10^{-35}$

For a compound $AB$,the degree of ionization is $80\%$. What is the solubility $(s)$ of $AB$ if its solubility product $(K_{sp})$ is $6.4 \times 10^{-9}$?

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