Question

Which of the following pairs of cations can be separated by passing $H_{2}S$ gas through the mixture in the presence of $0.2M$ $HCl$?

• A. $P{b}^{2+},C{u}^{2+}$
• B. $A{g}^{+},C{u}^{2+}$
• C. $C{d}^{2+},B{i}^{3+}$
• D. $C{u}^{2+},Z{n}^{2+}$

Answer 1

The correct option is D $C{u}^{2+},Z{n}^{2+}$

We can separate $C{u}^{2+}$ and $Z{n}^{2+}$ by passing $H_{2}S$ gas through their mixture in the presence of $0.2M$ $HCl$ as $C{u}^{2+}$ belongs to group II and $Z{n}^{2+}$ belong to group IV.

The $K_{sp}$ value of group II sulphide is quite low in comparison to the $K_{sp}$ values of group IV sulphides. Hence, group II ion gets precipitated while group IV ion does not get precipitate in acidic medium, where $S^{2-}$ ion concentration is quite low.