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The best explanation for the solubility of $MnS$ in $dil.{\text{ }}HCl$ is that:A.solubility product of $MnC{I_2}$ is less than that of $MnS$B.concentration of $M{n^{2 + }}$ is lowered by the formation of complex ions with chloride ionsC.concentration of sulphide ions is lowered by oxidation to free sulphurD.concentration of sulphide ions is lowered by formation of the weak acid ${H_2}S$

Last updated date: 20th Jun 2024
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Hint: We have to write the chemical equation between $MnS$ and $HCl$ then apply the concept of common ion effect to find the solubility of $MnS$ in $dil.{\text{ }}HCl$. We know that the common ion effect is a concept in which a decrease in solubility of an ionic precipitate by the addition to the solution of a soluble compound with an ion in common with the precipitate.

For solving this question, let’s first write the equation showing how the reaction between the $MnS$ and $dil.{\text{ }}HCl$ would go, the reaction is given below:
$MnS{\text{ }} + {\text{ }}dil.{\text{ }}HCl{\text{ }} \to {\text{ }}MnC{l_2}{\text{ }} + {\text{ }}{H_2}S$
$MnS{\text{ }} + {\text{ }}2HCl{\text{ }} \to {\text{ }}MnC{l_2}{\text{ }} + {\text{ }}{H_2}S$
Now as we can see ${H_2}S$ in the product is being formed, but due to the presence of $HCl$ the dissociation of ${H_2}S$ will be decreased. The reason behind is the common ion effect. According to common ion effect at equilibrium if the presence of certain ions is more and that ion is common, then the dissociation of the solute will not happen. Similarly, in case of ${H_2}S$ and $HCl$ the common ion is ${H^ + }$ and due to its initial presence due to $HCl$ the dissociation of ${H_2}S$ is decreased.
Hence, the answer to this question will be option D. concentration of sulfide ions is lowered by the formation of the weak acid ${H_2}S$