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The best explanation for the solubility of \text{MnS} in \text{dil. HCl} is that

Option: 1

Solubility product of \mathrm{MnCl}_{2} is less than that of \mathrm{MnS}


Option: 2

Concentration of \mathrm{Mn}^{2+} is lowered by the formation of complex ions with chloride ions


Option: 3

Concentration of sulphide ions is lowered by oxidation to free sulphur


Option: 4

Concentration of sulphide ions is lowered due to formation of the weak acid \mathrm{H}_{2}\text{S}


Answers (1)

best_answer

It is a characteristic property of sulphides of Group IV basic radicals. Concentration of sulphide ions is lowered by formation of the weak acid \mathrm{H}_{2} \mathrm{S} in the presence of \mathrm{HCl} which leads to dissolution of \mathrm{MnS}

\mathrm{\mathrm{MnS}(s)+2 \mathrm{HCl} \rightleftharpoons \mathrm{MnCl}_{2}(a q)+\mathrm{H}_{2} S}

Hence, the correct answer is Option (4)

Posted by

Ritika Harsh

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