Two salts A_2X and MX have the same value of | vedclass

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(B) For $A_2X$: The dissociation is $A_2X ightleftharpoons 2A^+ + X^{2-}$.$K_{sp} = [2S_1]^2 [S_1] = 4S_1^3$.$4.0 	imes 10^{-12} = 4S_1^3$ $\Righta

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(B) For $A_2X$: The dissociation is $A_2X ightleftharpoons 2A^+ + X^{2-}$.$K_{sp} = [2S_1]^2 [S_1] = 4S_1^3$.$4.0 	imes 10^{-12} = 4S_1^3$ $\Rightarrow S_1^3 = 10^{-12}$ $\Rightarrow S_1 = 10^{-4} 	ext{ mol L}^{-1}$.For $MX$: The dissociation is $MX ightleftharpoons M^{2+} + X^{2-}$.$K_{sp} = S_2^2$.$4.0 	imes 10^{-12} = S_2^2 \Rightarrow S_2 = 2.0 	imes 10^{-6} 	ext{ mol L}^{-1}$.Ratio $\frac{S(A_2X)}{S(MX)} = \frac{10^{-4}}{2.0 	imes 10^{-6}} = \frac{100}{2} = 50$.

Two salts $A_2X$ and $MX$ have the same value of solubility product of $4.0 \times 10^{-12}$. The ratio of their molar solubilities i.e. $\frac{S(A_2X)}{S(MX)} = \dots$ (Round off to the Nearest Integer).

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