Solubility of a salt M_2X_3 is y mol dm^{-3 | vedclass

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(D) The dissociation of the salt $M_2X_3$ is given by: $M_2X_3(s) \rightleftharpoons 2M^{3+}(aq) + 3X^{2-}(aq)$ If the solubility is $y \ mol \ dm^{-3

Vedclass · Accepted Answer

$108y^5$

Vedclass · Answer

(D) The dissociation of the salt $M_2X_3$ is given by: $M_2X_3(s) ightleftharpoons 2M^{3+}(aq) + 3X^{2-}(aq)$ If the solubility is $y \ mol \ dm^{-3}$,then the concentration of $M^{3+}$ is $2y \ mol \ dm^{-3}$ and the concentration of $X^{2-}$ is $3y \ mol \ dm^{-3}$. The solubility product $K_{sp}$ is defined as: $K_{sp} = [M^{3+}]^2 [X^{2-}]^3$ Substituting the values: $K_{sp} = (2y)^2 	imes (3y)^3$ $K_{sp} = (4y^2) 	imes (27y^3)$ $K_{sp} = 108y^5 \ mol^5 \ dm^{-15}$ Therefore,the correct option is $(D)$.

Solubility of a salt $M_2X_3$ is $y \ mol \ dm^{-3}$. The solubility product of the salt will be

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