Question

\(\mathrm{CaSO}_{4}\left(K_{\mathrm{sp}}=2.4 \times 10^{-5}\right)\) has a larger \(K_{\mathrm{sp}}\) value than that of \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\left(K_{\mathrm{sp}}=1.4 \times 10^{-5}\right)\). Does it necessarily follow that \(\mathrm{CaSO}_{4}\) also has greater solubility \((\mathrm{g} / \mathrm{L}) ?\) Explain.

Short answer

No, \( \text{CaSO}_4 \) does not necessarily have greater solubility in g/L; \( \text{Ag}_2\text{SO}_4 \) has a greater solubility.

Step-by-step solution

Understand Ksp and Solubility Relationship

The solubility product constant, \(K_{sp}\), is a measure of how much of a compound can dissolve in water to reach a saturation state. However, the solubility \( (g/L) \) also depends on the formula unit and the dissociation products of the compound.

Dissociation Equations

Write the dissociation equations for both compounds:- For \( \text{CaSO}_4 \), the dissociation is: \[ \text{CaSO}_4 (s) \rightarrow \text{Ca}^{2+} (aq) + \text{SO}_4^{2-} (aq) \]- For \( \text{Ag}_2\text{SO}_4 \), the dissociation is: \[ \text{Ag}_2\text{SO}_4 (s) \rightarrow 2 \text{Ag}^{+} (aq) + \text{SO}_4^{2-} (aq) \]

Establish the Relation with Ksp

The \(K_{sp}\) expressions for the compounds are:- For \( \text{CaSO}_4 \): \[ K_{sp} = [\text{Ca}^{2+}][\text{SO}_4^{2-}] = s \times s = s^2 \] where \( s \) is the solubility in mol/L.- For \( \text{Ag}_2\text{SO}_4 \): \[ K_{sp} = [\text{Ag}^+]^2[\text{SO}_4^{2-}] = (2s)^2 \times s = 4s^3 \] where \( s \) is again the solubility in mol/L of \( \text{Ag}_2\text{SO}_4 \).

Solve for Solubility

For \( \text{CaSO}_4 \): \[ s^2 = 2.4 \times 10^{-5} \] \[ s = \sqrt{2.4 \times 10^{-5}} \approx 4.9 \times 10^{-3} \text{ mol/L} \]For \( \text{Ag}_2\text{SO}_4 \): \[ 4s^3 = 1.4 \times 10^{-5} \] \[ s^3 = \frac{1.4 \times 10^{-5}}{4} = 3.5 \times 10^{-6} \] \[ s = \sqrt[3]{3.5 \times 10^{-6}} \approx 1.5 \times 10^{-2} \text{ mol/L} \]

Convert to g/L and Compare

Convert the solubilities from mol/L to g/L:- Molar Mass of \( \text{CaSO}_4 \) is about 136.14 g/mol. \[ 4.9 \times 10^{-3} \text{ mol/L} \times 136.14 \text{ g/mol} = 0.667 \text{ g/L} \]- Molar Mass of \( \text{Ag}_2\text{SO}_4 \) is about 311.79 g/mol. \[ 1.5 \times 10^{-2} \text{ mol/L} \times 311.79 \text{ g/mol} = 4.68 \text{ g/L} \]Thus, \( \text{Ag}_2\text{SO}_4 \) has a greater solubility in g/L despite having a smaller \(K_{sp}\).

Key concepts

Saturation State

When a solution reaches its saturation state, it means that the maximum amount of a solute has dissolved in the solvent, forming a stable equilibrium. Any additional solute would remain undissolved, as the solution cannot hold more. In terms of solubility product constant (\(K_{sp}\)), this concept helps us understand the extent to which a compound can dissolve under equilibrium.
This saturation point is directly influenced by various factors, including temperature and the nature of the solvent. For instance, \(CaSO_4\) and \(Ag_2 SO_4\) both have specific \(K_{sp}\) values that dictate their saturation levels in a solution. \(CaSO_4\) has a \(K_{sp}\) of \(2.4 \times 10^{-5}\), while \(Ag_2 SO_4\) has \(1.4 \times 10^{-5}\), suggesting different saturation states despite one having a greater \(K_{sp}\) value than the other.

Dissociation Equation

A dissociation equation represents how a solid ionic compound separates into its constituent ions in a solution. This is crucial for calculating solubility, as the breakdown of these compounds contributes to the concentration of ions in the solution.
For \(CaSO_4\), the dissociation equation is:

• \(CaSO_4 (s) \rightarrow Ca^{2+} (aq) + SO_4^{2-} (aq)\)

For \(Ag_2SO_4\), it is:

• \(Ag_2SO_4 (s) \rightarrow 2 Ag^{+} (aq) + SO_4^{2-} (aq)\)

These equations highlight how many ions each compound produces when dissolved. This influences not just the solubility, but also how \(K_{sp}\) values are used to find solubility, as seen in the different formations of products from each compound.

Molar Solubility

Molar solubility refers to the number of moles of a solute that can dissolve per liter of solution, reaching saturation. It's a crucial measure for predicting how much of the compound will dissolve.
For instance, \(CaSO_4\) has a solubility of approximately \(4.9 \times 10^{-3}\) mol/L. Meanwhile, \(Ag_2SO_4\), despite having a lower \(K_{sp}\), computes to a solubility of approximately \(1.5 \times 10^{-2}\) mol/L.
This illustrates that more \(Ag_2SO_4\) dissolves to saturate the solution, due to how its dissociation creates a different equilibrium state. Comparing these values, it's clear that \(K_{sp}\) alone doesn't determine solubility; the overall dissolution and products formed heavily influence the outcome.

Ksp Expression

The \(K_{sp}\) expression is a mathematical representation of the solubility product constant. It's vital for deducing how ions separate in solution and their equilibrium states.
For \(CaSO_4\), the \(K_{sp}\) expression is:\[K_{sp} = [Ca^{2+}][SO_4^{2-}] = s \times s = s^2\]Where \(s\) is the solubility in mol/L.
In contrast, the expression for \(Ag_2SO_4\) is:\[K_{sp} = [Ag^+]^2[SO_4^{2-}] = (2s)^2 \times s = 4s^3\]These expressions determine how \(K_{sp}\) translates into soluble quantities of the compounds. By analyzing these equations, we derive insights into how solubilities relate to the distinct stoichiometries and equilibria of the dissociating ions, paving the way towards understanding varied solubilities despite contrasting \(K_{sp}\) values.