COMEDK · Chemistry · 8. Ionic Equilibrium
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
- 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 \(\mathrm{ZnS}\)
Answer & Solution
Correct Answer
(A) \(K_{\mathrm{sp}}\) of \(\mathrm{CuS} < K_{\mathrm{sp}}\) of \(\mathrm{ZnS}\)
Step-by-step Solution
Detailed explanation
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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