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Question: \({{K}_{sp}}\) of \(\text{CuS}\), \(\text{A}{{\text{g}}_{\text{2}}}\text{S}\) and \(\text{HgS}\) are...

Ksp{{K}_{sp}} of CuS\text{CuS}, Ag2S\text{A}{{\text{g}}_{\text{2}}}\text{S} and HgS\text{HgS} are10−31{{10}^{-31}}, 10−44{{10}^{-44}} and 10−54{{10}^{-54}} respectively. Select the correct order for their solubility in water:
(A) Ag2S\text{A}{{\text{g}}_{\text{2}}}\text{S}>HgS\text{HgS}>CuS\text{CuS}
(B) HgS\text{HgS}>CuS\text{CuS}>Ag2S\text{A}{{\text{g}}_{\text{2}}}\text{S}
(C) HgS\text{HgS}>Ag2S\text{A}{{\text{g}}_{\text{2}}}\text{S}>CuS\text{CuS}
(D) Ag2S\text{A}{{\text{g}}_{\text{2}}}\text{S}>CuS\text{CuS}>HgS\text{HgS}

Explanation

Solution

Ksp{{K}_{sp}} is the solubility product constant of a solid solute dissolving in a solvent. So, think about the relation between solubility product constants and solubility of a solute in a solvent. Then substitute the values of Ksp{{K}_{sp}} and find the solubility of each of the given solutes. Higher the value of solubility, higher will be the solubility of that solute in water.

Complete step by step solution:
-Ksp{{K}_{sp}} is the solubility product constant of a solid solute dissolving in a solvent.
-Let’s start by determining Ksp{{K}_{sp}} for each solute and then find their solubility.
-For CuS\text{CuS}, CuS→Cu2++S2−\text{CuS}\to C{{u}^{2+}}+{{S}^{2-}} so, Ksp=[Cu2+][S2−]=S×S=S2{{K}_{sp}}=\left[ C{{u}^{2+}} \right]\left[ {{S}^{2-}} \right]=S\times S={{S}^{2}} where S is the solubility. Therefore, Ksp=S2{{K}_{sp}}={{S}^{2}}.
-Now, let’s calculate its solubility for CuS\text{CuS}. So, S=(10−31)1/2=3.162×10−16S={{\left( {{10}^{-31}} \right)}^{{}^{1}/{}_{2}}}=3.162\times {{10}^{-16}}
-Similarly for HgS\text{HgS}, HgS→Hg2++S2−\text{HgS}\to H{{g}^{2+}}+{{S}^{2-}} so, Ksp=[Hg2+][S2−]=S×S=S2{{K}_{sp}}=\left[ H{{g}^{2+}} \right]\left[ {{S}^{2-}} \right]=S\times S={{S}^{2}} where S is the solubility. Therefore, Ksp=S2{{K}_{sp}}={{S}^{2}}.
-Hence, solubility of HgS\text{HgS}, S=(10−54)1/2=10−27S={{\left( {{10}^{-54}} \right)}^{{}^{1}/{}_{2}}}={{10}^{-27}}
-No, let’s calculate for Ag2S\text{A}{{\text{g}}_{\text{2}}}\text{S}. Ag2S→2Ag++S2−\text{A}{{\text{g}}_{\text{2}}}\text{S}\to 2\text{A}{{\text{g}}^{+}}+{{S}^{2-}} so, Ksp=[Ag+]2[S2−]=(2S)2×S=4S3{{K}_{sp}}={{\left[ \text{A}{{\text{g}}^{+}} \right]}^{2}}\left[ {{S}^{2-}} \right]={{\left( 2S \right)}^{2}}\times S=4{{S}^{3}} where S is the solubility. Here concentration of Ag is 2S because two silver ions are formed after dissociation of silver sulphide.
-Thus, solubility of Ag2S\text{A}{{\text{g}}_{\text{2}}}\text{S}, S=((10−44)4)1/3=1.357×10−15S={{\left( \dfrac{\left( {{10}^{-44}} \right)}{4} \right)}^{{}^{1}/{}_{3}}}=1.357\times {{10}^{-15}}.
-Therefore, silver sulphide has the highest solubility value followed by copper sulphide and then mercury sulphide.
-We know that, higher the solubility value, higher is the solubility.
-Therefore, solubility order is Ag2S\text{A}{{\text{g}}_{\text{2}}}\text{S}>CuS\text{CuS}>HgS\text{HgS}.

Therefore, the correct option is (D).

Note: Remember solubility product constant and solubility are different terminologies. Higher the solubility value, higher is the solubility. Solubility values are calculated using solubility product constant with the help of the dissociation reaction of solute.