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Question: Answered & Verified by Expert
On passing $\mathrm{H}_{2} \mathrm{~S}$ into a solution containing both $\mathrm{Zn}^{2+}$ and $\mathrm{Cu}^{2+}$ in acidic medium, only CuS gets precipitate(d) This is because
ChemistryIonic EquilibriumCOMEDKCOMEDK 2012
Options:
  • A $K_{\mathrm{sp}}$ of $\mathrm{CuS} < K_{\mathrm{sp}}$ of $\mathrm{ZnS}$
  • B $K_{\mathrm{sp}}$ of $\mathrm{CuS}=K_{\mathrm{sp}}$ of $\mathrm{ZnS}$
  • C $K_{s p}$ of CuS $>K_{\mathrm{sp}}$ of $\mathrm{ZnS}$
  • D CuS is more stable than $\mathrm{ZnS}$
Solution:
1258 Upvotes Verified Answer
The correct answer is: $K_{\mathrm{sp}}$ of $\mathrm{CuS} < K_{\mathrm{sp}}$ of $\mathrm{ZnS}$
On passing $\mathrm{H}_{2} \mathrm{~S}$ into a solution containing $\mathrm{Zn}^{2+}$ and $\mathrm{Cu}^{2+}$ ions, $\mathrm{ZnS}$ and $\mathrm{CuS}$ are formed.
The solubility product of CuS is less than that of $\mathrm{ZnS}$.
$\therefore$ CuS get precipitated.

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