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(A) For a salt of type $AB$: $K_{sp} = s^2$,so $s = \sqrt{K_{sp}} = \sqrt{4.0 	imes 10^{-20}} = 2.0 	imes 10^{-10} \ M$. For a salt of type $A_2B$:

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$AB < AB_3 < A_2B$

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(A) For a salt of type $AB$: $K_{sp} = s^2$,so $s = \sqrt{K_{sp}} = \sqrt{4.0 	imes 10^{-20}} = 2.0 	imes 10^{-10} \ M$. For a salt of type $A_2B$: $K_{sp} = 4s^3$,so $s = \sqrt[3]{K_{sp}/4} = \sqrt[3]{3.2 	imes 10^{-11} / 4} = \sqrt[3]{8.0 	imes 10^{-12}} = 2.0 	imes 10^{-4} \ M$. For a salt of type $AB_3$: $K_{sp} = 27s^4$,so $s = \sqrt[4]{K_{sp}/27} = \sqrt[4]{2.7 	imes 10^{-31} / 27} = \sqrt[4]{1.0 	imes 10^{-32}} = 1.0 	imes 10^{-8} \ M$. Comparing the solubilities: $2.0 	imes 10^{-10} Thus,the increasing order is $AB < AB_3 < A_2B$.

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

The solubility products of three sparingly soluble salts $AB$,$A_2B$ and $AB_3$ are respectively $4.0 \times 10^{-20}$,$3.2 \times 10^{-11}$ and $2.7 \times 10^{-31}$. The increasing order of their solubility is

Similar Questions

The solubility products of three sparingly soluble salts are given below. What is the correct decreasing order of their molar solubility?
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$3$ $PQ_3$ $2.7 \times 10^{-35}$

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