Why only As^{3+} gets precipitated as As_2S_3 | vedclass

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Question

(A) The precipitation of metal sulfides depends on the solubility product $(K_{sp})$ and the concentration of sulfide ions $(S^{2-})$.In an acidic med

Vedclass · Accepted Answer

Solubility product of $As_2S_3$ is less than that of $ZnS$

Vedclass · Answer

(A) The precipitation of metal sulfides depends on the solubility product $(K_{sp})$ and the concentration of sulfide ions $(S^{2-})$.In an acidic medium,the dissociation of $H_2S$ is suppressed due to the common ion effect of $H^+$ ions,resulting in a very low concentration of $S^{2-}$ ions.Since the $K_{sp}$ of $As_2S_3$ is extremely low,even this low concentration of $S^{2-}$ is sufficient to exceed the ionic product of $As_2S_3$,causing it to precipitate.Conversely,the $K_{sp}$ of $ZnS$ is relatively higher,so the low concentration of $S^{2-}$ in the acidic medium is insufficient to exceed the $K_{sp}$ of $ZnS$,preventing its precipitation.

Why only $As^{3+}$ gets precipitated as $As_2S_3$ and not $Zn^{2+}$ as $ZnS$ when $H_2S$ is passed through an acidic solution containing $As^{3+}$ and $Zn^{2+}$?

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