Question

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

Short answer

The higher molar solubility of \(\mathrm{Ag}_{2}\mathrm{S}\) compared to \(\mathrm{CuS}\) can be attributed to the differing stoichiometry of their dissociation reactions. For \(\mathrm{Ag}_{2}\mathrm{S}\), \(K_{sp}^{Ag_2S} = 4s^3\) while for \(\mathrm{CuS}\), \(K_{sp}^{CuS} = s'^2\). The cube of molar solubility for \(\mathrm{Ag}_{2}\mathrm{S}\) has a more significant impact on its solubility product than the square of the molar solubility of \(\mathrm{CuS}\). Therefore, it is possible for \(\mathrm{Ag}_{2}\mathrm{S}\) to have a higher molar solubility (higher solubility in water) even with a smaller solubility product constant (\(K_\mathrm{sp}\)).

Step-by-step solution

Write the dissociation reactions of the given compounds

First, we write the balanced dissociation equations of both \(\mathrm{Ag}_{2}\mathrm{S}\) and \(\mathrm{CuS}\). For \(\mathrm{Ag}_{2}\mathrm{S}\): \[ \mathrm{Ag}_{2}\mathrm{S}(s) \rightleftharpoons 2 \mathrm{Ag}^{+}(aq) + \mathrm{S}^{2-}(aq) \] For \(\mathrm{CuS}\): \[ \mathrm{CuS}(s) \rightleftharpoons \mathrm{Cu}^{2+}(aq) + \mathrm{S}^{2-}(aq) \]

Write the solubility product expressions of the given compounds

Next, we find the solubility product expressions for both compounds. For \(\mathrm{Ag}_{2}\mathrm{S}\), the solubility product expression is: \[ K_\mathrm{sp}^{Ag_2S} = [\mathrm{Ag}^+]^2[\mathrm{S}^{2-}] \] For \(\mathrm{CuS}\), the solubility product expression is: \[ K_\mathrm{sp}^{CuS} = [\mathrm{Cu}^{2+}][\mathrm{S}^{2-}] \]

Determine the expression for molar solubility

In terms of molar solubility, let \(s\) represents the molar solubility of \(\mathrm{Ag}_{2}\mathrm{S}\) and \(s'\) represents the molar solubility of \(\mathrm{CuS}\). We can write the following expressions for ion concentrations: For \(\mathrm{Ag}_{2}\mathrm{S}\) ion concentrations: \([\mathrm{Ag}^+] = 2s\) and \([\mathrm{S}^{2-}] = s\) For \(\mathrm{CuS}\) ion concentrations: \([\mathrm{Cu}^{2+}] = s'\) and \([\mathrm{S}^{2-}] = s'\)

Write the solubility product expressions in terms of molar solubility

Now, let's write the solubility product expressions in terms of their molar solubilities: For \(\mathrm{Ag}_{2}\mathrm{S}\): \[ K_\mathrm{sp}^{Ag_2S} = (2s)^2(s) = 4s^3 \] For \(\mathrm{CuS}\): \[ K_\mathrm{sp}^{CuS} = (s')(s') = s'^2 \]

Explain the relationship between solubility and \(K_\mathrm{sp}\) values

The relationship between solubility (molar solubility) and the solubility product constant (\(K_\mathrm{sp}\)) can differ depending on the dissociation stoichiometry. As the formulas show, for \(\mathrm{Ag}_{2}\mathrm{S}\), the \(K_\mathrm{sp}\) is proportional to the cube of its molar solubility \(s\), whereas for \(\mathrm{CuS}\), the \(K_\mathrm{sp}\) is proportional to the square of its molar solubility \(s'\). Because of this difference in stoichiometry, it is possible for \(\mathrm{Ag}_{2}\mathrm{S}\) to have a larger molar solubility than \(\mathrm{CuS}\) despite having a smaller \(K_\mathrm{sp}\) value. Comparing the two expressions: \[ K_\mathrm{sp}^{Ag_2S} = 4s^3 \] \[ K_\mathrm{sp}^{CuS} = s'^2 \] In this case, the cube of the molar solubility of \(\mathrm{Ag}_{2}\mathrm{S}\) affects the \(K_\mathrm{sp}\) to a greater degree than the square of the molar solubility of \(\mathrm{CuS}\). Therefore, it is possible for \(\mathrm{Ag}_{2}\mathrm{S}\) to have a higher molar solubility (higher solubility in water) than \(\mathrm{CuS}\) despite having a smaller solubility product constant (\(K_\mathrm{sp}\)).

Key concepts

Chemical Equilibrium

Chemical equilibrium is a state where the rate of the forward reaction equals the rate of the reverse reaction, meaning that the concentrations of reactants and products remain constant over time. This concept is pivotal when exploring the solubility of ionic compounds in water. A saturated solution of a sparingly soluble salt has reached chemical equilibrium between the undissolved solid and the dissolved ions. Understanding this balance is key to interpreting why different compounds have varying molar solubilities, even if their solubility product constants are different.

Solubility Product Constant (Ksp)

The solubility product constant (Ksp) is a numerical representation of the solubility of a sparingly soluble ionic compound. It is defined as the product of the concentrations of the ions in a saturated solution, each raised to the power of its stoichiometric coefficient. The lower the Ksp value, the less soluble the compound is. However, because the stoichiometry of dissolution affects the expression of Ksp, two compounds with different Ksp values can exhibit counterintuitive solubility behavior, as seen in the comparison between \(\mathrm{Ag}_{2}\mathrm{S}\) and \(\mathrm{CuS}\).

It is important to note that Ksp is not the same as molar solubility, which measures the number of moles that dissolve to form a liter of solution. The Ksp provides insight into the extent to which a compound can dissolve but does not directly quantify the solubility.

Stoichiometry of Dissolution Reactions

Stoichiometry of dissolution reactions refers to the proportion of ions that an ionic compound produces when it dissolves in water. The stoichiometry is essential in determining the solubility product expression for the compound. For example, \(\mathrm{Ag}_{2}\mathrm{S}\) dissociates to produce two silver ions for every one sulfide ion, while \(\mathrm{CuS}\) produces one copper ion and one sulfide ion. As a result, the molar solubility of each compound contributes differently to their respective Ksp values.

The stoichiometric coefficients are crucial in the calculation of molar solubility from the Ksp value. One must carefully account for these coefficients to correctly express concentrations of the ions produced in the equilibrium state and further predict solubility behavior correctly.

Concentration of Ions in Saturated Solutions

The concentration of ions in a saturated solution plays a pivotal role in understanding the molar solubility of compounds. In the case of \(\mathrm{Ag}_{2}\mathrm{S}\) and \(\mathrm{CuS}\), molar solubility is described by the amount of the compound that can dissolve to form a saturated solution. The concentration of the ions in these solutions is directly dependent on the stoichiometry of the dissolution reaction.

As demonstrated in the steps to derive molar solubility, a compound like \(\mathrm{Ag}_{2}\mathrm{S}\), with a stoichiometry that produces more ions, can still be more soluble than \(\mathrm{CuS}\) in spite of a lower Ksp. This emphasizes that the solubility product is tied to the stoichiometry, which ultimately affects concentrations in the solution and determines the extent to which a solute can be dissolved.