Question

The ion that cannot be precipitated by both $HCl$ and ${H_2}S$ is :
 $P{b^{2 + }}$
 $C{u^{2 + }}$
 $A{g^ + }$
 $S{n^{2 + }}$

Answer 1

Hint: Chemical precipitation is the process of turning a liquid into a solid by making the liquid either super-saturated or rendering the material not soluble. In the given question we need to tell the ion that cannot be precipitated by both the given reagents. We need to use the laws of solubility. Using them we get that $S{n^{2 + }}$ is precipitated only in case of ${H_2}S$ .

Complete Step by Step Solution:
Before we start answering the question let us first learn the meaning of a few terms used here.

Precipitate is a substance that has been physically or chemically isolated from a solution or suspension. It usually takes the form of an insoluble crystalline or amorphous solid.
The degree to which a substance dissolves in a solvent to form a solution is known as solubility (usually expressed as grams of solute per litre of solvent). One fluid's (liquid or gas) solubility in another might be absolute (completely miscible; for example, methanol and water) or partial (oil and water dissolve only slightly).

The largest product of an electrolyte's ionic concentrations or activities that can still maintain equilibrium with the undissolved phase at a given temperature is known as the solubility product.

Now we will use solubility principles along with their exceptions to answer our question.
Using laws of solubility we see that $S{n^{2 + }}$ is the ion that $HCl$ cannot precipitate only ${H_2}S$ can . As a chocolate-colored precipitate of $SnS$ that is soluble in yellow ammonium sulphide, $S{n^{2 + }}$ is not precipitated by $HCl$ but is instead precipitated by ${H_2}S$ . $S{n^{2 + }}$ combines with ${H_2}S$ to produce a brown precipitate of insoluble stannous sulphide, whereas it reacts with $HCl$ to produce soluble stannous chloride.

The product of the reaction between $S{n^{2 + }}$ and $HCl$ is soluble molecule $SnC{l_2}$.
Additionally, it produces the compound $SnS$ , a brown precipitate, when it combines with ${H_2}S$ .
 $C{u^{2 + }}$ dissolves in concentrated $HCl$ but crystallises upon dilution. ${H_2}S$ also causes it to precipitate.
Both $HCl$ and ${H_2}S$ can precipitate $P{b^{2 + }}$ and $A{g^ + }$ .
Insoluble chlorides and sulphides of $P{b^{2 + }}$ , $C{u^{2 + }}$ , and $A{g^ + }$ can be precipitated by $HCl$ and ${H_2}S$ , respectively.
Hence, we can say that $S{n^{2 + }}$ does not precipitate with both $HCl$ and ${H_2}S$ . It only precipitates out when reacted with ${H_2}S$ .
Hence, option D. is the answer.

Note: It should always be kept in mind that the exceptions involved in the principles of solubility are as important as the laws themselves. Also, these questions can always be given by changing the ion or the reagent and can be asked in different ways. There is not just one way of asking this question.