The stoichiometry and solubility product of a | vedclass

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(C) From the graph,at $[X] = 1 	imes 10^{-3} \ M$,the corresponding $[Y] = 2 	imes 10^{-3} \ M$.Since the concentration of $Y$ is twice that of $X$,

Vedclass · Accepted Answer

$XY_{2}, 4 	imes 10^{-9} \ M^{3}$

Vedclass · Answer

(C) From the graph,at $[X] = 1 	imes 10^{-3} \ M$,the corresponding $[Y] = 2 	imes 10^{-3} \ M$.Since the concentration of $Y$ is twice that of $X$,the stoichiometry of the salt is $XY_{2}$.The dissociation equilibrium is: $XY_{2(s)} ightleftharpoons X_{(aq)}^{2+} + 2Y_{(aq)}^{-}$.The solubility product constant $K_{sp}$ is given by: $K_{sp} = [X^{2+}][Y^{-}]^{2}$.Substituting the values: $K_{sp} = (1 	imes 10^{-3}) 	imes (2 	imes 10^{-3})^{2}$.$K_{sp} = (10^{-3}) 	imes (4 	imes 10^{-6}) = 4 	imes 10^{-9} \ M^{3}$.

The stoichiometry and solubility product of a salt with the solubility curve given below is,respectively:

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