Question

An aqueous solution of an inorganic salt on treatment with $HCl$ gives a white precipitate. This solution contains
(A) $H{g_2}^{ + 2}$
(B) $H{g^{2 + }}$
(C) $Z{n^{2 + }}$
(D) $C{d^{2 + }}$

Answer 1

Hint: To check whether the salt will give white precipitate when treated with $HCl$ , we must know the general reaction of metal $M$ with $HCl$, which results in the formation of metal salt and liberation of ${H_2}$ gas hence, we will first check whether the given salt formed is of white precipitate then choose the suitable option from the given question.

Complete Step by Step Solution:
We know that whenever a metal reacts with hydrogen chloride, following reaction takes place:
$M(s) + nHCl(aq) \to {M^{n + }}C{l_n} + {H_2} \uparrow $
Now, among all the four options given, only $H{g_2}^{ + 2}$ is most prominent to react with $HCl$ due to very low value of the solubility product.
$H{g_2}^{ + 2} + 2HCl(aq) \to H{g_2}C{l_2} \downarrow $ (White ppt.) $ + {H_2} \uparrow $

In addition, $H{g^{2 + }}$ when react with $HCl$, will form a soluble complex with $C{l^ - }$ ion.
$Z{n^{2 + }}$ and $C{d^{2 + }}$ will also give soluble salts when react with $HCl$ as:
$Zn + 2HCl(aq) \to ZnC{l_2}(s) + {H_2} \uparrow $
$Cd + 2HCl(aq) \to CdC{l_2}(s) + {H_2} \uparrow $
Thus, an aqueous solution of an inorganic salt treated with $HCl$ , gives a white precipitate when the solution contains $H{g_2}^{ + 2}$ ions only.
Hence, the correct option is (A) $H{g_2}^{ + 2}$ .

Note: Since this is a conceptual-based problem hence, it is essential that the options given in the question are analyzed very carefully to give an accurate solution. Also, remember that Qualitative analysis of mercury $(Hg)$ uses this property of $H{g_2}^{ + 2}$to react with halogen ion as: $H{g_2}^{ + 2} + 2C{l^ - }(aq) \to H{g_2}C{l_2} \downarrow (s)$(white).