(A) For $(a)$ $MX$: $K_{sp} = s^2 = 4.0 \times 10^{-20} \implies s = \sqrt{4.0 \times 10^{-20}} = 2.0 \times 10^{-10} \ M$.For $(b)$ $P_2Q$: $K_{sp} =

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(A) For $(a)$ $MX$: $K_{sp} = s^2 = 4.0 \times 10^{-20} \implies s = \sqrt{4.0 \times 10^{-20}} = 2.0 \times 10^{-10} \ M$.For $(b)$ $P_2Q$: $K_{sp} = 4s^3 = 3.2 \times 10^{-11} \implies s^3 = 0.8 \times 10^{-11} = 8 \times 10^{-12} \implies s = 2 \times 10^{-4} \ M$.For $(c)$ $LY_3$: $K_{sp} = 27s^4 = 2.7 \times 10^{-31} \implies s^4 = 0.1 \times 10^{-31} = 1 \times 10^{-32} \implies s = 10^{-8} \ M$.Comparing the molar solubilities: $2.0 \times 10^{-10} Thus,the increasing order is $a < c < b$.

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

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The solubility products of some sparingly soluble electrolytes are given below. The increasing order of their molar solubility is:
$(a)$ Formula: $MX$,$K_{sp} = 4.0 \times 10^{-20}$
$(b)$ Formula: $P_2Q$,$K_{sp} = 3.2 \times 10^{-11}$
$(c)$ Formula: $LY_3$,$K_{sp} = 2.7 \times 10^{-31}$

A
$a < c < b$
B
$b < a < c$
C
$a < b < c$
D
$c < a < b$

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If the molar solubility (in $mol \cdot L^{-1}$) of a sparingly soluble salt $AB_4$ is $S$ and the corresponding solubility product is $K_{sp}$,then $S$ in terms of $K_{sp}$ is given by the relation:

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Derive the equation for the solubility and solubility product of a sparingly soluble salt $M_{x}X_{y}$.

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The solubility product of silver bromide is $5.0 \times 10^{-13}$. The quantity of potassium bromide (molar mass taken as $120 \ g \ mol^{-1}$) to be added to $1 \ L$ of $0.05 \ M$ solution of silver nitrate to start the precipitation of $AgBr$ is

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$A$ solution which is $10^{-3} \ M$ each in $Mn^{2+}, Fe^{2+}, Zn^{2+},$ and $Hg^{2+}$ is treated with $10^{-16} \ M$ sulphide ion. If $K_{sp}$ of $MnS, FeS, ZnS,$ and $HgS$ are $10^{-15}, 10^{-23}, 10^{-20},$ and $10^{-54}$ respectively,which one will precipitate first?

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Solubility product of $CaC_2O_4$ at a given temperature in pure water is $4 \times 10^{-9} \ (mol \ L^{-1})^2$. Solubility of $CaC_2O_4$ at the same temperature is

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The solubility products of some sparingly soluble electrolytes are given below. The increasing order of their molar solubility is:$(a)$ Formula: $MX$,$K_{sp} = 4.0 \times 10^{-20}$$(b)$ Formula: $P_2Q$,$K_{sp} = 3.2 \times 10^{-11}$$(c)$ Formula: $LY_3$,$K_{sp} = 2.7 \times 10^{-31}$

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If the molar solubility (in $mol \cdot L^{-1}$) of a sparingly soluble salt $AB_4$ is $S$ and the corresponding solubility product is $K_{sp}$,then $S$ in terms of $K_{sp}$ is given by the relation:

Derive the equation for the solubility and solubility product of a sparingly soluble salt $M_{x}X_{y}$.

The solubility product of silver bromide is $5.0 \times 10^{-13}$. The quantity of potassium bromide (molar mass taken as $120 \ g \ mol^{-1}$) to be added to $1 \ L$ of $0.05 \ M$ solution of silver nitrate to start the precipitation of $AgBr$ is

$A$ solution which is $10^{-3} \ M$ each in $Mn^{2+}, Fe^{2+}, Zn^{2+},$ and $Hg^{2+}$ is treated with $10^{-16} \ M$ sulphide ion. If $K_{sp}$ of $MnS, FeS, ZnS,$ and $HgS$ are $10^{-15}, 10^{-23}, 10^{-20},$ and $10^{-54}$ respectively,which one will precipitate first?

Solubility product of $CaC_2O_4$ at a given temperature in pure water is $4 \times 10^{-9} \ (mol \ L^{-1})^2$. Solubility of $CaC_2O_4$ at the same temperature is

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The solubility products of some sparingly soluble electrolytes are given below. The increasing order of their molar solubility is:
$(a)$ Formula: $MX$,$K_{sp} = 4.0 \times 10^{-20}$
$(b)$ Formula: $P_2Q$,$K_{sp} = 3.2 \times 10^{-11}$
$(c)$ Formula: $LY_3$,$K_{sp} = 2.7 \times 10^{-31}$