 An aqueous solution of a substance gives a white precipitate on treatment with dilute hydrochloric acid, which dissolves on heating. When hydrogen sulphide is passed through the hot acidic solution, a black precipitate is obtained. The substance is a/an:[A] $H{{g}_{2}}^{2+}salt$[B] $C{{u}^{2+}}salt$[C] $A{{g}^{+}}salt$[D] $P{{b}^{2+}}salt$ Verified
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Hint: The correct option here is the salt which is placed in group (I) as well as group (II). Its chloride is not fully precipitated due to its partial solubility in water. The remaining ions of the metal in the acidic solution from its sulphide when hydrogen sulphide is passed through it.

When a salt is treated with an acid, we get other salt with evolution of a gas like carbon dioxide or hydrogen gas and water.
The substance given to us in the question reacts with dilute hydrochloric acid, giving us a white precipitate which dissolves on heating.
In the first option we have $H{{g}_{2}}^{2+}salt$. When we treat it with dilute HCl, we get a white precipitate of mercurous chloride which is also known as calomel. The reaction is-

$H{{g}_{2}}^{2+}+2HCl(dil.)\to H{{g}_{2}}C{{l}_{2}}\downarrow$

Although we get a white precipitate of mercurous chloride, upon heating it does not dissolve which makes it different from other group I chlorides. Therefore, this cannot be the correct answer.
In the next option, we have $C{{u}^{2+}}salt$. When we treat it with dilute hydrochloric acid, it shows no reaction.
In the next option, we have $A{{g}^{+}}salt$. When it is treated with dil. HCl, we get a precipitate of silver chloride. We can write the reaction as-

$A{{g}^{+}}+HCl(dil.)\to AgCl\downarrow$

Although we get the precipitate here too but upon heating, it converts back to silver and the solution turns greyish- black. Therefore, this cannot be the correct option either.
In the last option, we have$P{{b}^{2+}}salt$. When it is treated with dilute HCl, it gives us lead chloride in the form of a white precipitate but it is slightly soluble in water therefore, complete precipitation does not take place. The reaction is-

$P{{b}^{2+}}+2HCl(dil.)\to PbC{{l}_{2}}\downarrow$

The precipitate is not completely soluble in cold water but upon heating, it dissolves completely and we get a colourless solution.
When hydrogen sulphide is passed through this acidic solution, we get a black coloured precipitate of lead (II) sulphide. Lead chloride is not fully precipitated and hence the remaining lead ions form lead sulphide when hydrogen sulphide is passed through it. We can write the reaction as-

$P{{b}^{2+}}+{{H}_{2}}S\to PbS\downarrow +{{H}_{2}}(g)$

So, the correct answer is “Option D”.

Note: Cations are put in 6 different groups and each group has a common reagent which is used to separate the cations from the solution. Lead is put in group (I) as well as (II) as it shows positive results to the reagents in both the groups. On reaction with dilute hydrochloric acid, it forms a partially soluble chloride salt which puts it in group (I) and on passing hydrogen sulphide through it, it forms an insoluble sulphide, which puts it in group (II).