If equal volumes of AB_2 and XY (both are sal | vedclass

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(C) When equal volumes are mixed,the concentration of each species is halved. The reaction is $AB_2 + 2XY \rightarrow AY_2 + 2XB$. For precipitation t

Vedclass · Accepted Answer

$2.0 	imes 10^{-2} \ M \ AB_2, 2.0 	imes 10^{-2} \ M \ XY$

Vedclass · Answer

(C) When equal volumes are mixed,the concentration of each species is halved. The reaction is $AB_2 + 2XY ightarrow AY_2 + 2XB$. For precipitation to occur,the ionic product $Q_{sp}$ must be greater than $K_{sp}$. $Q_{sp} = [A^{2+}][Y^-]^2$. For option $C$: $[A^{2+}] = (2.0 	imes 10^{-2} / 2) = 1.0 	imes 10^{-2} \ M$ and $[Y^-] = (2.0 	imes 10^{-2} / 2) = 1.0 	imes 10^{-2} \ M$. $Q_{sp} = (1.0 	imes 10^{-2}) 	imes (1.0 	imes 10^{-2})^2 = 1.0 	imes 10^{-6}$. Since $1.0 	imes 10^{-6} > 5.2 	imes 10^{-7}$,a precipitate will form.

If equal volumes of $AB_2$ and $XY$ (both are salts) aqueous solutions are mixed,which of the following combinations will give a precipitate of $AY_2$ at $300 \ K$? (Given $K_{sp}$ (at $300 \ K$) for $AY_2 = 5.2 \times 10^{-7}$)

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