(C) $AX_{2}$ is ionized as follows:$AX_{2} \rightleftharpoons A^{2+} + 2X^{-}$Solubility product of $AX_{2}$ is given by:$K_{sp} = [A^{2+}][X^{-}]^{2}

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(C) $AX_{2}$ is ionized as follows:$AX_{2} \rightleftharpoons A^{2+} + 2X^{-}$Solubility product of $AX_{2}$ is given by:$K_{sp} = [A^{2+}][X^{-}]^{2} = (S) \times (2S)^{2} = 4S^{3}$Given $K_{sp} = 3.2 \times 10^{-11}$$4S^{3} = 3.2 \times 10^{-11}$$S^{3} = 0.8 \times 10^{-11} = 8 \times 10^{-12}$$S = \sqrt[3]{8 \times 10^{-12}} = 2 \times 10^{-4} \, mol / L$Thus,the solubility is $2 \times 10^{-4} \, mol / L$.

Class 11 Chemistry 2 Equilibrium: Ionic Equilibrium Solubility product

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The solubility product of sparingly soluble salt $AX_{2}$ is $3.2 \times 10^{-11}$. Its solubility (in $mol / L$) is

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A
$5.6 \times 10^{-6}$
B
$3.1 \times 10^{-4}$
C
$2 \times 10^{-4}$
D
$4 \times 10^{-4}$

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2 Equilibrium: Ionic Equilibrium

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The correct order of solubility of the given salts in water at $298 K$ is:
Salt $K_{sp}$ at $298 K$
$AgBr$ $5.0 \times 10^{-13}$
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Salt	$K_{sp}$ at $298 K$
$AgBr$	$5.0 \times 10^{-13}$
$Zn(OH)_2$	$1.0 \times 10^{-15}$
$Hg_2Cl_2$	$1.3 \times 10^{-18}$

The solubility product of sparingly soluble salt $AX_{2}$ is $3.2 \times 10^{-11}$. Its solubility (in $mol / L$) is

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The solubility product of $AgBr$ is $4.9 \times 10^{-13}$ at a certain temperature. Calculate the solubility.

The solubility of $A_2X_3$ is $y \ mol \ m^{-3}$. Its solubility product is equal to .......

The $K_{sp}$ of $Mg(OH)_2$ is $1 \times 10^{-12}$. $0.01 \ M \ MgCl_2$ will show precipitation in a solution of $pH$ greater than:-

On the addition of a solution containing $CrO_4^{2-}$ ions to a solution containing $Ba^{2+}$,$Sr^{2+}$,and $Ca^{2+}$ ions,the precipitate obtained first will be of

The correct order of solubility of the given salts in water at $298 K$ is:Salt$K_{sp}$ at $298 K$$AgBr$$5.0 \times 10^{-13}$$Zn(OH)_2$$1.0 \times 10^{-15}$$Hg_2Cl_2$$1.3 \times 10^{-18}$

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The correct order of solubility of the given salts in water at $298 K$ is:
Salt $K_{sp}$ at $298 K$
$AgBr$ $5.0 \times 10^{-13}$
$Zn(OH)_2$ $1.0 \times 10^{-15}$
$Hg_2Cl_2$ $1.3 \times 10^{-18}$