KCET · Chemistry · Ionic Equilibrium
\(\mathrm{H}_{2} \mathrm{~S}\) is passed into one \(\mathrm{nm}^{3}\) of a solution containing \(0.1\) mole of \(\mathrm{Zn}^{2+}\) and \(0.01\) mole of \(\mathrm{Cu}^{2+}\) till the sulphide ion concentration reaches to \(8.1 \times 10^{-19}\) moles. Which one of the following statements is true?
\(\left[K_{\mathrm{sp}}\right.\) of \(\mathrm{ZnS}\) and CuS are \(3 \times 10^{-22}\) and \(8 \times 10^{-36}\) respectively.]
- A Only \(\mathrm{ZnS}\) precipitates
- B Both CuS and \(\mathrm{ZnS}\) precipitate
- C Only CuS precipitates
- D No precipitation occurs
Answer & Solution
Correct Answer
(B) Both CuS and \(\mathrm{ZnS}\) precipitate
Step-by-step Solution
Detailed explanation
The ionic product for \(\mathrm{ZnS}\left(0.1 \times 8.1 \times 10^{-19}\right)\) and for CuS \(\left(0.01 \times 8.1 \times 10^{-19}\right)\) exceeds their \(K_{\mathrm{sp}}\) values. Hence, both get precipitated.
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