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Question: The value of \({{K}_{sp}}\) for CuS, \(A{{g}_{2}}S\) and HgS are \({{10}^{-31}},{{10}^{-42}}\text{ a...

The value of Ksp{{K}_{sp}} for CuS, Ag2SA{{g}_{2}}S and HgS are 10−31,10−42 and 10−54{{10}^{-31}},{{10}^{-42}}\text{ and 1}{{\text{0}}^{-54}} respectively. The correct order of their solubility in water is:
[A] Ag2S>HgS>CuSA{{g}_{2}}S>HgS>CuS
[B] HgS>CuS>Ag2SHgS>CuS>A{{g}_{2}}S
[C] HgS>Ag2S>CuSHgS>A{{g}_{2}}S>CuS
[D] Ag2S>CuS>HgSA{{g}_{2}}S>CuS>HgS

Explanation

Solution

To solve this, find out the relation between the solubility product and the solubility. Use the values of the solubility product given in the question and find out the solubility. Compound with the highest value of solubility will be the most soluble one in water.

Complete step by step answer:
To solve this, firstly let us find out the relation between the solubility product and solubility.
We know that Ksp{{K}_{sp}} is the solubility product of a reaction. It is the measure for the degree of dissociation of the ions in a solution. It is written in terms of concentration of the ions-
Ksp=[cation]×[anion]Ksp=\left[ cation \right]\times \left[ anion \right]
I.e. Ksp{{K}_{sp}} is the product of concentration of anions and cations present in the solution.

If we take a salt MxXy{{M}_{x}}{{X}_{y}} and write its dissociation reaction as: MxXy⇌xMy++yXx−{{M}_{x}}{{X}_{y}}\rightleftharpoons x{{M}^{y+}}+y{{X}^{x-}}
The, it’s Ksp{{K}_{sp}} will be Ksp=[s]+×[s]−Ksp={{\left[ s \right]}^{+}}\times {{\left[ s \right]}^{-}} where S stands for solubility.

Now, let’s use this relation for the given question.
Firstly we have CuS. We can write its dissociation reaction as-
CuS⇌Cu2++S2−CuS\rightleftharpoons C{{u}^{2+}}+{{S}^{2-}}
Therefore, its Ksp{{K}_{sp}} will be: Ksp=s×s=s2{{K}_{sp}}=s\times s={{s}^{2}}

The value of the solubility product is given to us as 10−31{{10}^{-31}}. Therefore, we can write that:
s2=10−31 or,s=10−31=3.16×10−16 \begin{aligned} & {{s}^{2}}={{10}^{-31}} \\\ & or,s=\sqrt{{{10}^{-31}}}=3.16\times {{10}^{-16}} \\\ \end{aligned}
Then we have Ag2SA{{g}_{2}}S.
Ag2S⇌2Ag++S2−A{{g}_{2}}S\rightleftharpoons 2A{{g}^{+}}+{{S}^{2-}}
Therefore, Ksp=(2s)2×s=4s3{{K}_{sp}}={{\left( 2s \right)}^{2}}\times s=4{{s}^{3}}

Or, 4s3=10−42 or,s=10−424=6.23×10−15 \begin{aligned} & 4{{s}^{3}}={{10}^{-42}} \\\ & or,s=\sqrt{\frac{{{10}^{-42}}}{4}}=6.23\times {{10}^{-15}} \\\ \end{aligned}
And lastly we have HgS. Dissociation equation will be:
HgS⇌Hg2++S2−HgS\rightleftharpoons H{{g}^{2+}}+{{S}^{2-}}
Therefore, Ksp=s×s=s2{{K}_{sp}}=s\times s={{s}^{2}}

Or, s2=10−54 or,s=10−54=1×10−27 \begin{aligned} & {{s}^{2}}={{10}^{-54}} \\\ & or,s=\sqrt{{{10}^{-54}}}=1\times {{10}^{-27}} \\\ \end{aligned}
We can see from the above calculations that the value of solubility is lowest for HgS and highest for Ag2SA{{g}_{2}}S.
So, the correct answer is “Option D”.

Note: The solubility product can be calculated by using the concentration of the ions of the dissociated salt. It is dependent on other surrounding factors like temperature, pressure and nature of the electrolyte. These factors affect the concentration of the ions and thus affect the solubility product too.