(D) For $MX$: $K_{sp} = S^2 = 4 \times 10^{-8} \implies S = 2 \times 10^{-4} \, M$.For $MX_2$: $K_{sp} = 4S^3 = 3.2 \times 10^{-14} \implies S^3 = 0.8

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(D) For $MX$: $K_{sp} = S^2 = 4 \times 10^{-8} \implies S = 2 \times 10^{-4} \, M$.For $MX_2$: $K_{sp} = 4S^3 = 3.2 \times 10^{-14} \implies S^3 = 0.8 \times 10^{-14} = 8 \times 10^{-15} \implies S = 2 \times 10^{-5} \, M$.For $M_3X$: $K_{sp} = 27S^4 = 2.7 \times 10^{-15} \implies S^4 = 10^{-16} \implies S = 10^{-4} \, M$.Comparing the solubilities: $2 \times 10^{-4} > 10^{-4} > 2 \times 10^{-5}$.Thus,the order is $MX > M_3X > MX_2$.

Class 11 Chemistry 2 Equilibrium: Ionic Equilibrium Solubility product

Medium

Solubility product $(K_{sp})$ of salts of type $MX$,$MX_2$,and $M_3X$ at temperature $T$ are $4 \times 10^{-8}$,$3.2 \times 10^{-14}$,and $2.7 \times 10^{-15}$ respectively. The solubility of the salts at temperature $T$ are in the order:

A
$MX > MX_2 > M_3X$
B
$M_3X > MX_2 > MX$
C
$MX_2 > M_3X > MX$
D
$MX > M_3X > MX_2$

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The solubility of a salt $A_2B_3$ is $10^{-3} \ M$. What will be its solubility product?

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The solubility of $CaCl_2$ is $1.11 \times 10^{-8} \ g/100 \ mL$. If the molar mass of $CaCl_2$ is $111 \ g/mol$,find the $K_{sp}$.

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The required amount of $KBr$ (molar mass $= 119 \ g/mol$) in grams to start the precipitation of $AgBr$ in $500 \ mL$ solution of $0.05 \ M \ AgNO_3$ will be :- ($K_{SP}$ of $AgBr = 5 \times 10^{-13}$)

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$Mg(OH)_2$ is precipitated when $NaOH$ is added to a solution of $Mg^{2+}$. If the final concentration of $Mg^{2+}$ is $10^{-10} \ M$,the concentration of $OH^{-} \ (M)$ in the solution is (Given: Solubility product for $Mg(OH)_2 = 5.6 \times 10^{-12}$)

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The molar solubility of $Zn(OH)_{2}$ in $0.1 \, M \, NaOH$ solution is $x \times 10^{-18} \, M$. The value of $x$ is ...... . (Nearest integer)
(Given : The solubility product of $Zn(OH)_{2}$ is $2 \times 10^{-20}$)

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Solubility product $(K_{sp})$ of salts of type $MX$,$MX_2$,and $M_3X$ at temperature $T$ are $4 \times 10^{-8}$,$3.2 \times 10^{-14}$,and $2.7 \times 10^{-15}$ respectively. The solubility of the salts at temperature $T$ are in the order:

Similar Questions

The solubility of a salt $A_2B_3$ is $10^{-3} \ M$. What will be its solubility product?

The solubility of $CaCl_2$ is $1.11 \times 10^{-8} \ g/100 \ mL$. If the molar mass of $CaCl_2$ is $111 \ g/mol$,find the $K_{sp}$.

The required amount of $KBr$ (molar mass $= 119 \ g/mol$) in grams to start the precipitation of $AgBr$ in $500 \ mL$ solution of $0.05 \ M \ AgNO_3$ will be :- ($K_{SP}$ of $AgBr = 5 \times 10^{-13}$)

$Mg(OH)_2$ is precipitated when $NaOH$ is added to a solution of $Mg^{2+}$. If the final concentration of $Mg^{2+}$ is $10^{-10} \ M$,the concentration of $OH^{-} \ (M)$ in the solution is (Given: Solubility product for $Mg(OH)_2 = 5.6 \times 10^{-12}$)

The molar solubility of $Zn(OH)_{2}$ in $0.1 \, M \, NaOH$ solution is $x \times 10^{-18} \, M$. The value of $x$ is ...... . (Nearest integer)(Given : The solubility product of $Zn(OH)_{2}$ is $2 \times 10^{-20}$)

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The molar solubility of $Zn(OH)_{2}$ in $0.1 \, M \, NaOH$ solution is $x \times 10^{-18} \, M$. The value of $x$ is ...... . (Nearest integer)
(Given : The solubility product of $Zn(OH)_{2}$ is $2 \times 10^{-20}$)