What is the solubility of Hg2S in mol/L in a solution that contains an S2- concentration of 0.0500 M? (Ksp of Hg2S = 1.0×10^-47)
The Correct Answer and Explanation is:
To calculate the solubility of mercury(I) sulfide (Hg2S) in a solution where the concentration of sulfide ions (S2-) is 0.0500 M, we use the solubility product constant (Ksp) expression for Hg2S.
The dissociation reaction of Hg2S in water is as follows:Hg2S(s)⇌Hg22+(aq)+S2−(aq)Hg_2S(s) \rightleftharpoons Hg_2^{2+}(aq) + S^{2-}(aq)Hg2S(s)⇌Hg22+(aq)+S2−(aq)
For this reaction, the Ksp expression is:Ksp=[Hg22+][S2−]K_{sp} = [Hg_2^{2+}][S^{2-}]Ksp=[Hg22+][S2−]
Given:
- KspK_{sp}Ksp of Hg2S = 1.0×10−471.0 \times 10^{-47}1.0×10−47
- The concentration of S2−S^{2-}S2− is 0.0500 M (given in the problem)
Let the solubility of Hg2S in mol/L be represented by “s”. When Hg2S dissociates, it produces “s” moles of Hg2^2+ per liter (since the stoichiometry of Hg2S dissociation is 1:1). Therefore, the concentration of Hg22+Hg_2^{2+}Hg22+ will be “s”, and the concentration of S2−S^{2-}S2− will be 0.0500 M (as given).
Now, substituting the values into the Ksp expression:Ksp=(s)(0.0500)K_{sp} = (s)(0.0500)Ksp=(s)(0.0500)1.0×10−47=s×0.05001.0 \times 10^{-47} = s \times 0.05001.0×10−47=s×0.0500
Solving for “s”:s=1.0×10−470.0500s = \frac{1.0 \times 10^{-47}}{0.0500}s=0.05001.0×10−47s=2.0×10−46 mol/Ls = 2.0 \times 10^{-46} \, \text{mol/L}s=2.0×10−46mol/L
Thus, the solubility of Hg2S in the given solution is 2.0×10−462.0 \times 10^{-46}2.0×10−46 mol/L.
Explanation:
- The very small Ksp value indicates that Hg2S is extremely insoluble in water, meaning that only a tiny amount of it will dissolve.
- The presence of a significant concentration of sulfide ions (0.0500 M) further reduces the solubility of Hg2S, as it shifts the equilibrium towards the solid Hg2S, making it even less soluble. The sulfide ions from the solution essentially “saturate” the solution, reducing the amount of Hg2S that can dissolve.
