(C) For $AgBrO_3$ (type $AB$): $K_{sp} = S^2$ $S = \sqrt{K_{sp}} = \sqrt{5.5 \times 10^{-5}} \approx 7.41 \times 10^{-3} \ M$. For $Ag_2SO_4$ (type $A

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(C) For $AgBrO_3$ (type $AB$): $K_{sp} = S^2$ $S = \sqrt{K_{sp}} = \sqrt{5.5 \times 10^{-5}} \approx 7.41 \times 10^{-3} \ M$. For $Ag_2SO_4$ (type $A_2B$): $K_{sp} = 4S^3$ $S = \sqrt[3]{\frac{K_{sp}}{4}} = \sqrt[3]{\frac{2 \times 10^{-5}}{4}} = \sqrt[3]{0.5 \times 10^{-5}} = \sqrt[3]{5 \times 10^{-6}} \approx 1.71 \times 10^{-2} \ M$. Comparing the two values,$S_{Ag_2SO_4} > S_{AgBrO_3}$ or $S_{AgBrO_3} < S_{Ag_2SO_4}$.

Class 11 Chemistry 6-2.Equilibrium-II (Ionic Equilibrium)Solubility product

Medium

If the solubility products of $AgBrO_3$ and $Ag_2SO_4$ are $5.5 \times 10^{-5}$ and $2 \times 10^{-5}$ respectively,what is the relationship between their solubilities?

A
$S_{AgBrO_3} > S_{Ag_2SO_4}$
B
$S_{AgBrO_3} = S_{Ag_2SO_4}$
C
$S_{AgBrO_3} < S_{Ag_2SO_4}$
D
None of the above

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6-2.Equilibrium-II (Ionic Equilibrium)

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Solubility product

The solubility of a sparingly soluble salt $A_3B_2$ (Molar mass $= M \ g/mol$) in water is $x \ g/L$. The ratio of the solubility product of the salt to the molar concentration of $B^{-3}$ ion is:

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If the solubility products of $AgCl$ and $AgBr$ are $1.0 \times 10^{-10}$ and $3.5 \times 10^{-13}$ respectively,then the relation between the solubilities (denoted by the symbol $S$) of these salts can correctly be represented as:

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At $25^{\circ} C$, the solubility product of a salt of $MX_{2}$ type is $3.2 \times 10^{-8}$ in water. The solubility (in $mol / L$) of $MX_{2}$ in water at the same temperature will be:

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At $25^{\circ}C$,the solubility product $(K_{sp})$ of $M(OH)_2$ is $32 \times 10^{-12}$. What is the $pOH$ of a saturated solution of $M(OH)_2$?

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How many grams of $CaC_2O_4$ (molecular weight $= 128$) on dissolving in $1 \ L$ of distilled water will give a saturated solution? $[K_{sp}(CaC_2O_4) = 2.5 \times 10^{-9} \ mol^2 L^{-2}]$

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If the solubility products of $AgBrO_3$ and $Ag_2SO_4$ are $5.5 \times 10^{-5}$ and $2 \times 10^{-5}$ respectively,what is the relationship between their solubilities?

Similar Questions

The solubility of a sparingly soluble salt $A_3B_2$ (Molar mass $= M \ g/mol$) in water is $x \ g/L$. The ratio of the solubility product of the salt to the molar concentration of $B^{-3}$ ion is:

If the solubility products of $AgCl$ and $AgBr$ are $1.0 \times 10^{-10}$ and $3.5 \times 10^{-13}$ respectively,then the relation between the solubilities (denoted by the symbol $S$) of these salts can correctly be represented as:

At $25^{\circ} C$, the solubility product of a salt of $MX_{2}$ type is $3.2 \times 10^{-8}$ in water. The solubility (in $mol / L$) of $MX_{2}$ in water at the same temperature will be:

At $25^{\circ}C$,the solubility product $(K_{sp})$ of $M(OH)_2$ is $32 \times 10^{-12}$. What is the $pOH$ of a saturated solution of $M(OH)_2$?

How many grams of $CaC_2O_4$ (molecular weight $= 128$) on dissolving in $1 \ L$ of distilled water will give a saturated solution? $[K_{sp}(CaC_2O_4) = 2.5 \times 10^{-9} \ mol^2 L^{-2}]$

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