(D) For $MX$: $K_{sp} = S^2 = 4.0 \times 10^{-8} \Rightarrow S = 2 \times 10^{-4} \text{ mol dm}^{-3}$.For $MX_2$: $K_{sp} = 4S^3 = 3.2 \times 10^{-14

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(D) For $MX$: $K_{sp} = S^2 = 4.0 \times 10^{-8} \Rightarrow S = 2 \times 10^{-4} \text{ mol dm}^{-3}$.For $MX_2$: $K_{sp} = 4S^3 = 3.2 \times 10^{-14}$ $\Rightarrow S^3 = 8 \times 10^{-15}$ $\Rightarrow S = 2 \times 10^{-5} \text{ mol dm}^{-3}$.For $M_3X$: $K_{sp} = 27S^4 = 2.7 \times 10^{-15}$ $\Rightarrow S^4 = 10^{-16}$ $\Rightarrow S = 1 \times 10^{-4} \text{ mol dm}^{-3}$.Comparing the solubilities: $2 \times 10^{-4} > 1 \times 10^{-4} > 2 \times 10^{-5}$.Thus,the order of solubility is $MX > M_3X > MX_2$.

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

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Solubility products $(K_{sp})$ of the salts of types $MX, MX_2$ and $M_3X$ at temperature $T$ are $4.0 \times 10^{-8}, 3.2 \times 10^{-14}$ and $2.7 \times 10^{-15}$ respectively. Solubilities (in $mol \text{ dm}^{-3}$) of the salts at temperature $T$ are in the order:

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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 products $(K_{sp})$ of the salts of types $MX, MX_2$ and $M_3X$ at temperature $T$ are $4.0 \times 10^{-8}, 3.2 \times 10^{-14}$ and $2.7 \times 10^{-15}$ respectively. Solubilities (in $mol \text{ dm}^{-3}$) of the salts at temperature $T$ are in the order:

Similar Questions

$A$ sparingly soluble salt having general formula $A_{x}B_{y}$ and molar solubility $S$ is in equilibrium with its saturated solution. Derive a relationship between solubility and the solubility product for such salt.

At $25^{\circ}C$,the solubility product $(K_{sp})$ of $Zn(OH)_2$ is $10^{-14}$. If $NH_4OH$ is $50\%$ dissociated,what will be the concentration of zinc in a $0.1 \ M$ $NH_4OH$ solution?

The solubility of $CaF_2$ in water is $1.7 \times 10^{-3} \ g/100 \ mL$ at $298 \ K$. Calculate the solubility product $(K_{sp})$ of $CaF_2$. (Molar mass of $CaF_2 = 78 \ g/mol$)

The molar solubility $(mol \ L^{-1})$ of a sparingly soluble salt $MX_4$ is $s$. The corresponding solubility product is $K_{sp}$. $s$ is given in terms of $K_{sp}$ by the relation:

For $AgCl$,$K_{sp} = 1.2 \times 10^{-10}$ and for $AgBr$,$K_{sp} = 3.5 \times 10^{-13}$. What is the relationship between their solubilities $(S)$?

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