Question

A solution which is ${10^{ - 3}}M$ each in $M{n^{2 + }},F{e^{2 + }},Z{n^{2 + }}$_and $H{g^{2 + }}$ is treated with ${10^{ - 16}}M$ sulphide ion. If ${K_{sp}}$ of $MnS,FeS,ZnS$and $HgS$are ${10^{ - 15}},{10^{ - 25}},{10^{ - 20}}$ and ${10^{ - 54}}$ respectively, which one will precipitate first?
A. $FeS$
B. $MnS$
C. $HgS$
D. $ZnS$

Answer 1

Hint: In this question here many types of different ions are reacted with sulphide ion after which we have to say which product will precipitate first out of all the given products. For which we have to justify them on the basis of their ${K_{sp}}$ because we know that whose ${K_{sp}}$ is lowest will precipitate faster. So, let’s start with the complete solution.

Complete step by step solution:
As firstly, we just calculate the Ionic product of the solution by which we can compare the ${K_{sp}}$ so, by calculation of ionic product of the solution we get,
Ionic product $ = {10^{ - 3}} \times {10^{ - 16}}$
As we know that, when the base of the multiplying variables are same then the powers are got added from which the Ionic product will be,
Ionic product $ = {10^{ - 19}}$
We also know that, In the event that the ionic product is greater than the ${K_{sp}}$ , the metal sulphide with the lowest solubility will precipitate first. In this case, every salt is the same valence type.
As from which we get that, the salt having the lowest ${K_{sp}}$ value will provide the precipitate first.
According to the question, the lowest valued ${K_{sp}}$is ${10^{ - 54}}$ and this ${K_{sp}}$ belongs to $HgS$ .
Therefore, the correct answer is $HgS$ , because it forms the precipitate first.

Additional Information :
Solubility is defined as the property of a substance known as solute to get dissolved in a solvent for the formation of a solution. The solubility of ionic compounds that dissociate and form cations and anions in water varies to a great extent. Some compounds are highly soluble and can even absorb moisture from the atmosphere whereas, on the other hand, the others are highly insoluble.

Hence, the correct option is (C)

Note: Here in this question a term is defined which is ${K_{sp}}$ . So, let’s see what this is and what this stands for. The equilibrium constant for a solid material dissolving in an aqueous solution is the solubility product constant, ${K_{sp}}$ . It stands for the degree of solute dissolution in solution. A substance's ${K_{sp}}$ value increases with how soluble it is.