A) \[P{{b}^{2+}}\]
B) \[F{{e}^{3+}}\]
C) \[Z{{n}^{2+}}\]
D) \[C{{u}^{2+}}\]
Correct Answer: A
Solution :
\[P{{b}^{2+}}\] can be precipitated both by \[HCl\] and \[{{H}_{2}}S\]. The reason is that the solubility products of \[PbC{{l}_{2}}\] and \[PbS\] are quite less.
You need to login to perform this action.
You will be redirected in
3 sec
