(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 6-2.Equilibrium-II (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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6-2.Equilibrium-II (Ionic Equilibrium)

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Solubility product

The solubility product of $PbS$ is $3.4 \times 10^{-28}$. If $[Pb^{2+}] = 1 \times 10^{-2} \ mol/L$,what concentration of $[S^{2-}]$ is required to precipitate $PbS$?

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If the solubility of $AgCl$ (molar mass $= 143 \, g/mol$) in water at $25 \, ^\circ C$ is $1.43 \times 10^{-4} \, g/100 \, mL$ of solution,then the value of $K_{sp}$ is?

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The low solubility of $BaSO_4$ in water can be attributed to

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The solubility product of $BaSO_4$ is $1.5 \times 10^{-9}$. The precipitation in a $0.01 \ M \ Ba^{2+}$ solution will start,on adding $H_2SO_4$ of concentration:

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The $pH$ of a saturated solution of $Ca(OH)_2$ is $12.25$. Calculate its solubility product $(K_{sp})$.

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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 product of $PbS$ is $3.4 \times 10^{-28}$. If $[Pb^{2+}] = 1 \times 10^{-2} \ mol/L$,what concentration of $[S^{2-}]$ is required to precipitate $PbS$?

If the solubility of $AgCl$ (molar mass $= 143 \, g/mol$) in water at $25 \, ^\circ C$ is $1.43 \times 10^{-4} \, g/100 \, mL$ of solution,then the value of $K_{sp}$ is?

The low solubility of $BaSO_4$ in water can be attributed to

The solubility product of $BaSO_4$ is $1.5 \times 10^{-9}$. The precipitation in a $0.01 \ M \ Ba^{2+}$ solution will start,on adding $H_2SO_4$ of concentration:

The $pH$ of a saturated solution of $Ca(OH)_2$ is $12.25$. Calculate its solubility product $(K_{sp})$.

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