Question

\(\mathrm{Ag}_{2} \mathrm{~S}(s)\) has a larger molar solubility than CuS even though \(\mathrm{Ag}_{2} \mathrm{~S}\) has the smaller \(K_{\mathrm{sp}}\) value. Explain how this is possible.

Short answer

The molar solubility of \(\mathrm{Ag}_{2} \mathrm{~S}(s)\) and CuS are related to their respective \(K_{\mathrm{sp}}\) values differently, with \(\mathrm{Ag}_{2} \mathrm{~S}(s)\) having a cubic relationship (\(K_{\mathrm{sp}}^{Ag_2S} \propto s_{\mathrm{Ag_2S}}^3\)) and CuS having a square relationship (\(K_{\mathrm{sp}}^{CuS} \propto s_{\mathrm{CuS}}^2\)). Therefore, despite having a smaller \(K_{\mathrm{sp}}\) value, the disproportionality factor between \(K_{\mathrm{sp}}\) and molar solubility of \(\mathrm{Ag}_{2} \mathrm{~S}(s)\) is sufficiently large to result in a higher molar solubility compared to CuS.

Step-by-step solution

Write the solubility product constant expression for both compounds

For \(\mathrm{Ag}_{2} \mathrm{~S}(s)\), the dissociation reaction can be given by: \[\mathrm{Ag}_{2} \mathrm{~S}(s) \rightleftharpoons 2\mathrm{Ag}^{+}(aq) + \mathrm{S}^{2-}(aq)\] Thus, the solubility product constant expression for this reaction would be: \[K_{\mathrm{sp}}^{Ag_2S} = [\mathrm{Ag}^{+}]^2[\mathrm{S}^{2-}]\] For CuS(s), the dissociation reaction can be given by: \[\mathrm{CuS}(s) \rightleftharpoons \mathrm{Cu}^{2+}(aq) + \mathrm{S}^{2-}(aq)\] Thus, the solubility product constant expression for this reaction would be: \[K_{\mathrm{sp}}^{CuS} = [\mathrm{Cu}^{2+}][\mathrm{S}^{2-}]\]

Define molar solubility and relate it to solubility product constant

Molar solubility is defined as the number of moles of a solute that can dissolve per liter of solution to form a saturated solution. Let's denote the molar solubility of \(\mathrm{Ag}_{2} \mathrm{~S}(s)\) as \(s_{\mathrm{Ag_2S}}\) and that of CuS(s) as \(s_{\mathrm{CuS}}\). For \(\mathrm{Ag}_{2} \mathrm{~S}(s)\), from the dissociation reaction, its molar solubility can be expressed as: \[s_{\mathrm{Ag_2S}} = [\mathrm{S}^{2-}]\] and since two moles of \(\mathrm{Ag}^{+}\) ions are formed per mole of \(\mathrm{Ag}_{2} \mathrm{~S}(s)\) dissolved, \[[\mathrm{Ag}^{+}] = 2s_{\mathrm{Ag_2S}}\] Now substitute the expressions of molar solubility in the \(K_{\mathrm{sp}}^{Ag_2S}\) expression: \[K_{\mathrm{sp}}^{Ag_2S} = (2s_{\mathrm{Ag_2S}})^2s_{\mathrm{Ag_2S}} = 4s_{\mathrm{Ag_2S}}^3\] For CuS(s), from the dissociation reaction, its molar solubility can be expressed as: \[s_{\mathrm{CuS}} = [\mathrm{Cu}^{2+}] = [\mathrm{S}^{2-}]\] Now substitute the expressions of molar solubility in the \(K_{\mathrm{sp}}^{CuS}\) expression: \[K_{\mathrm{sp}}^{CuS} = s_{\mathrm{CuS}}^2\]

Explain why \(\mathrm{Ag}_{2} \mathrm{~S}(s)\) has a larger molar solubility than CuS

Given that both \(K_{\mathrm{sp}}^{Ag_2S}\) and \(K_{\mathrm{sp}}^{CuS}\) are constants for the respective salt, we can see that the relationship between molar solubility and \(K_{\mathrm{sp}}\) differs for both compounds. Notice that, \(K_{\mathrm{sp}}^{Ag_2S}\) is proportional to the cube of the molar solubility of \(\mathrm{Ag}_{2} \mathrm{~S}(s)\) whereas \(K_{\mathrm{sp}}^{CuS}\) is proportional to the square of the molar solubility of CuS(s). \[\begin{cases} K_{\mathrm{sp}}^{Ag_2S} \propto s_{\mathrm{Ag_2S}}^3\\ K_{\mathrm{sp}}^{CuS} \propto s_{\mathrm{CuS}}^2 \end{cases}\] This means, even if the value of \(K_{\mathrm{sp}}^{Ag_2S}\) is smaller compared to \(K_{\mathrm{sp}}^{CuS}\), it does not necessarily imply that \(s_{\mathrm{Ag_2S}}\) must be smaller compared to \(s_{\mathrm{CuS}}\). In this case, the disproportionality factor that links the \(K_{\mathrm{sp}}\) and molar solubility is sufficiently large, leading to larger molar solubility of \(\mathrm{Ag}_{2} \mathrm{~S}(s)\) compared to CuS, despite having a smaller \(K_{\mathrm{sp}}\) value.

Key concepts

Molar Solubility

Molar solubility is an essential concept in the study of solubility equilibrium, particularly when dealing with sparingly soluble salts. It is the number of moles of a solute that will dissolve in a liter of solution to reach a saturated state. This measurement isn't simply a matter of the solute's ability to dissolve; it's intricately tied to the solubility product constant, or Ksp. This value, specific to each compound, provides insight into just how much of a substance can be dissolved before the solution becomes saturated.

In the given exercise, by defining the molar solubility, students can understand the direct relationship it has with the corresponding Ksp values. For instance, the silver sulfide (Ag2S) example highlights that the molar solubility of a compound is not always directly proportional to the solubility product; in fact, it varies with the stoichiometry of the dissociation reaction. Thus, a substance with a smaller Ksp can still have a higher molar solubility if it produces more moles of ions per mole of the compound dissolved in solution. Understanding molar solubility aids in predicting and explaining the solubility behaviors of various compounds in different contexts.

Chemical Equilibrium

The concept of chemical equilibrium is at the heart of understanding reactions in a closed system. Equilibrium occurs when the rates of the forward and reverse reactions are equal, resulting in no net change in the concentration of reactants and products over time. It is a dynamic state, meaning that while the concentrations remain constant, the reactions continue to take place.

Chemical equilibrium plays a pivotal role when discussing solubility product constants (Ksp) because a saturated solution represents a system at equilibrium. It implies that the rate of dissolution of the solid solute is equal to the rate at which the dissolved ions combine to form the undissolved solute. Ksp is a special instance of the equilibrium constant which applies to these solid-liquid equilibria. When learning about solubility, comprehending equilibrium principles will enable students to understand why a system might favor the formation of a precipitate or the dissolution of a solid at any given point.

Dissociation Reactions

Dissociation reactions describe the breaking apart of compounds into their constituent ions when they dissolve in water. These reactions are central to the discussion of solubility because they provide the fundamental basis for defining the solubility product constant (Ksp). The precise nature of these reactions varies with each substance, often leading to different numbers of ions being produced.

In the context of the provided exercise, understanding the dissociation reactions of Ag2S and CuS is crucial. For Ag2S, two moles of positive silver ions (Ag+) and one mole of negative sulfide ions (S2-) are produced per mole of Ag2S that dissolves. In contrast, the dissociation of CuS results in equal moles of Cu2+ and S2- ions. Therefore, the difference in the stoichiometry of these two dissociation reactions can lead to different molar solubilities, which is counter-intuitive given the smaller Ksp value for Ag2S compared to that of CuS. By exploring these reactions, students can gain a richer understanding of how ionic compounds interact with solvents and how their solubility is both qualitatively and quantitatively assessed.