Question

The reagent(s) that can selectively precipitate $S^{2-}$ from a mixture of $S^{2-}$ and ${SO}_4^{2-}$ in aqueous solution is (are) :

• A. $Cu{Cl}_2$
• B. $Ba{Cl}_2$
• C. ${Pb\left(OOC{CH}_3\right)}_2$
• D. ${Na}_2\left[{Fe\left(CN\right)}_{5}NO\right]$

Answer 1

Answer: (A)

$Cu{Cl}_2$

A)
$CuC{l}_2$ exists as $C{u}^{2+}$ and $C{l}^{-1}$
Now $CuS{O}_4$ is highly soluble in water while, $CuS$ forms a black ppt.
B)
$BaC{l}_2$ exists as $B{a}^{2+}$ and $C{l}^{-1}$
Now $BaS$ is soluble in water while, $BaS{O}_4$ forms a white ppt.
C)
$P{b}^{2+}$ reacts with $S^{2-}$ and $S{O}_4^{2-}$ to form lead sulphide and lead sulphate.
But lead sulphate is more soluble than lead sulphide.
Hence PbS will form ppt.
D)

Here in the above complex, the reacting ion will be $N{a}^{+}$. The sodium ion will react with the sulphide and sulphate ions. However, both of them are soluble. Hence no ppt will be observed.

Hence, the correct option is A.