(D) Let the solubility of all salts be $s$. For $M_2X$: $M_2X \rightleftharpoons 2M^+ + X^{2-}$. $K_{sp} = (2s)^2(s) = 4s^3$. For $QY_2$: $QY_2 \right

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$K_{sp}(M_2X) = K_{sp}(QY_2) = K_{sp}(PZ_2)$

(D) Let the solubility of all salts be $s$. For $M_2X$: $M_2X \rightleftharpoons 2M^+ + X^{2-}$. $K_{sp} = (2s)^2(s) = 4s^3$. For $QY_2$: $QY_2 \rightleftharpoons Q^{2+} + 2Y^-$. $K_{sp} = (s)(2s)^2 = 4s^3$. For $PZ_2$: $PZ_2 \rightleftharpoons P^{2+} + 2Z^-$. $K_{sp} = (s)(2s)^2 = 4s^3$. Since all salts have the same solubility $s$,their $K_{sp}$ values are equal: $K_{sp}(M_2X) = K_{sp}(QY_2) = K_{sp}(PZ_2)$.

Class 11 Chemistry 2 Equilibrium: Ionic Equilibrium Solubility product

Medium

If the solubility of salts $M_2X$,$QY_2$,and $PZ_2$ is the same,what is the relationship between their $K_{sp}$ values?

A
$K_{sp}(M_2X) > K_{sp}(QY_2) > K_{sp}(PZ_2)$
B
$K_{sp}(M_2X) = K_{sp}(QY_2) < K_{sp}(PZ_2)$
C
$K_{sp}(M_2X) > K_{sp}(QY_2) = K_{sp}(PZ_2)$
D
$K_{sp}(M_2X) = K_{sp}(QY_2) = K_{sp}(PZ_2)$

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If the ${K_{sp}}$ for $HgSO_4$ is $6.4 \times 10^{-5}$,then the solubility of the salt is:

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The solubility product of a sparingly soluble salt $AX_2$ is $3.2 \times 10^{-11}$. Its solubility (in moles/litres) is

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The solubility of a sparingly soluble salt $AB_2$ is $1 \times 10^{-6} \text{ mol/dm}^3$. Calculate its 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: $LY_3$,$K_{sp} = 2.7 \times 10^{-31}$

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Solubility of $Th_3(PO_4)_4$ $(M_w = m)$ is $w \ g / 100 \ mL$ at $25 \ ^oC$. The $K_{sp}$ is:

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If the solubility of salts $M_2X$,$QY_2$,and $PZ_2$ is the same,what is the relationship between their $K_{sp}$ values?

Similar Questions

If the ${K_{sp}}$ for $HgSO_4$ is $6.4 \times 10^{-5}$,then the solubility of the salt is:

The solubility product of a sparingly soluble salt $AX_2$ is $3.2 \times 10^{-11}$. Its solubility (in moles/litres) is

The solubility of a sparingly soluble salt $AB_2$ is $1 \times 10^{-6} \text{ mol/dm}^3$. Calculate its solubility product.

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: $LY_3$,$K_{sp} = 2.7 \times 10^{-31}$

Solubility of $Th_3(PO_4)_4$ $(M_w = m)$ is $w \ g / 100 \ mL$ at $25 \ ^oC$. The $K_{sp}$ 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: $LY_3$,$K_{sp} = 2.7 \times 10^{-31}$