Which one of the following sulphides has the lowest solubility product ?
A. $FeS$
B. $MnS$
C. $PbS$
D. $ZnS$
Answer
415.8k+ views
Hint: We can find the answer by considering their value of solubility product. $PbS$ has the lowest solubility product of all the sulphides. $P{{b}^{2+}}$ cation belongs to group-2 of salt analysis sulphides of group-2 and has the lower solubility products.
Complete Step by Step Answer:
The solubility product constant is the equilibrium constant for the dissociation of a solid substance into an aqueous solution. We can find the solubility product of all the given compounds:
Equilibrium reaction for dissolution reaction of $FeS$ is:
$FeS\rightleftharpoons [F{{e}^{2+}}]+[{{S}^{2-}}]$
So ${{K}_{sp}}=[F{{e}^{2+}}][{{S}^{2-}}]={{10}^{-18}}$
Similarly the Equilibrium reaction for dissolution reaction of $MnS$ is:
$MnS\rightleftharpoons [M{{n}^{2+}}]+[{{S}^{2-}}]$
So ${{K}_{sp}}=[M{{n}^{2+}}][{{S}^{2-}}]={{10}^{-11}}$
Similarly the Equilibrium reaction for dissolution reaction of $PbS$ is:
$PbS\rightleftharpoons [P{{b}^{2+}}]+[{{S}^{2-}}]$
So ${{K}_{sp}}=[P{{b}^{2+}}][{{S}^{2-}}]={{10}^{-28}}$
Similarly the Equilibrium reaction for dissolution reaction of $ZnS$ is:
$ZnS\rightleftharpoons [Z{{n}^{2+}}]+[{{S}^{2-}}]$
So ${{K}_{sp}}=[Z{{n}^{2+}}][{{S}^{2-}}]={{10}^{-23}}$
Thus we see that $PbS$ has the lowest solubility among all the sulphides, then $ZnS$ has the lowest solubility, then $FeS$ and then the $MnS$.
So $PbS$ has the lowest solubility product among all sulphides.
Thus, Option (C) is correct.
Note: Solubility is a chemical property which refers to the ability of the substance, the solute, to dissolve in a solvent. It is measured as the maximum amount of solute dissolved in a solvent at equilibrium and the resulting solution is a saturated solution. And the solubility product ${{K}_{sp}}$ is the mathematical product of its dissolved ion concentration raised to the power of their coefficients. Solubility products are relevant when a sparingly soluble ionic compound breaks ions into solution.
Complete Step by Step Answer:
The solubility product constant is the equilibrium constant for the dissociation of a solid substance into an aqueous solution. We can find the solubility product of all the given compounds:
Equilibrium reaction for dissolution reaction of $FeS$ is:
$FeS\rightleftharpoons [F{{e}^{2+}}]+[{{S}^{2-}}]$
So ${{K}_{sp}}=[F{{e}^{2+}}][{{S}^{2-}}]={{10}^{-18}}$
Similarly the Equilibrium reaction for dissolution reaction of $MnS$ is:
$MnS\rightleftharpoons [M{{n}^{2+}}]+[{{S}^{2-}}]$
So ${{K}_{sp}}=[M{{n}^{2+}}][{{S}^{2-}}]={{10}^{-11}}$
Similarly the Equilibrium reaction for dissolution reaction of $PbS$ is:
$PbS\rightleftharpoons [P{{b}^{2+}}]+[{{S}^{2-}}]$
So ${{K}_{sp}}=[P{{b}^{2+}}][{{S}^{2-}}]={{10}^{-28}}$
Similarly the Equilibrium reaction for dissolution reaction of $ZnS$ is:
$ZnS\rightleftharpoons [Z{{n}^{2+}}]+[{{S}^{2-}}]$
So ${{K}_{sp}}=[Z{{n}^{2+}}][{{S}^{2-}}]={{10}^{-23}}$
Thus we see that $PbS$ has the lowest solubility among all the sulphides, then $ZnS$ has the lowest solubility, then $FeS$ and then the $MnS$.
So $PbS$ has the lowest solubility product among all sulphides.
Thus, Option (C) is correct.
Note: Solubility is a chemical property which refers to the ability of the substance, the solute, to dissolve in a solvent. It is measured as the maximum amount of solute dissolved in a solvent at equilibrium and the resulting solution is a saturated solution. And the solubility product ${{K}_{sp}}$ is the mathematical product of its dissolved ion concentration raised to the power of their coefficients. Solubility products are relevant when a sparingly soluble ionic compound breaks ions into solution.
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