The solubility product of As_2S_3 is 2.8 time | vedclass

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Question

(A) For the salt $As_2S_3$,the dissociation equilibrium is: $As_2S_3(s) ightleftharpoons 2As^{3+}(aq) + 3S^{2-}(aq)$.Let the solubility be $S \ 	ex

Vedclass · Accepted Answer

$1.09 	imes 10^{-15} \ 	ext{mol/L}$

Vedclass · Answer

(A) For the salt $As_2S_3$,the dissociation equilibrium is: $As_2S_3(s) ightleftharpoons 2As^{3+}(aq) + 3S^{2-}(aq)$.Let the solubility be $S \ 	ext{mol/L}$. Then $[As^{3+}] = 2S$ and $[S^{2-}] = 3S$.The solubility product expression is $K_{sp} = [As^{3+}]^2 [S^{2-}]^3 = (2S)^2 (3S)^3 = 4S^2 	imes 27S^3 = 108S^5$.Given $K_{sp} = 2.8 	imes 10^{-72}$,we have $108S^5 = 2.8 	imes 10^{-72}$.$S^5 = \frac{2.8 	imes 10^{-72}}{108} \approx 0.0259 	imes 10^{-72} = 2.59 	imes 10^{-74}$.Taking the fifth root,$S = (2.59 	imes 10^{-74})^{1/5} \approx 1.09 	imes 10^{-15} \ 	ext{mol/L}$.

The solubility product of $As_2S_3$ is $2.8 \times 10^{-72}$. What is the solubility of $As_2S_3$?

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