An aqueous solution of substance gives a white precipitate on treatmen...
Explanation:
When an aqueous solution of a substance gives a white precipitate on treatment with dilute hydrochloric acid (HCl) and the precipitate dissolves on heating, it indicates the presence of a metal cation that forms an insoluble chloride. This helps to narrow down the options.
Step 1: White Precipitate Formation
The formation of a white precipitate upon treatment with dilute HCl indicates the presence of a chloride ion (Cl-) in the solution. The chloride ion reacts with the metal cation to form an insoluble chloride salt. Let's examine each option:
a) Hg22+ salt: The mercury(I) ion does form an insoluble chloride (Hg2Cl2), but it does not dissolve on heating. Therefore, option A can be eliminated.
b) Cu2+ salt: The copper(II) ion does form an insoluble chloride (CuCl2), but it does not dissolve on heating. Therefore, option B can be eliminated.
c) Ag+ salt: The silver ion (Ag+) forms an insoluble chloride (AgCl), which dissolves in ammonia (NH3) solution but not on heating. Therefore, option C can be eliminated.
d) Pb2+ salt: The lead(II) ion (Pb2+) forms an insoluble chloride (PbCl2), which dissolves on heating. This fits the given observations and is consistent with the experimental results. Therefore, option D is the correct answer.
Step 2: Confirmation with H2S
To further confirm the presence of lead(II) ion (Pb2+), hydrogen sulfide (H2S) is passed through the hot acidic solution. Hydrogen sulfide reacts with the lead(II) ion to form a black precipitate of lead sulfide (PbS). This confirms that the substance is a lead(II) salt.
In conclusion, the substance in the aqueous solution is a lead(II) salt (Pb2+ salt) based on the observations of a white precipitate with dilute HCl that dissolves on heating, and the formation of a black precipitate with H2S.
An aqueous solution of substance gives a white precipitate on treatmen...
PbCl2 white precipitate
PbS black ppt