Question

Solubility product of radium sulphate is \(4 \times 10^{-11}\) What will be the solubility of \(\mathrm{Ra}^{2+}\) in \(0.10 \mathrm{M}\) \(\mathrm{Na}_{2} \mathrm{SO}_{4} ?\) (a) \(4 \times 10^{-10} \mathrm{M}\) (b) \(2 \times 10^{-5} \mathrm{M}\) (c) \(4 \times 10^{-5} \mathrm{M}\) (d) \(2 \times 10^{-10} \mathrm{M}\)

Short answer

4 \times 10^{-10} \mathrm{M}

Step-by-step solution

- Write the Solubility Product Expression

For radium sulphate (\text{RaSO}_4), which dissociates into \text{Ra}^{2+} and \text{SO}_4^{2-}, its solubility product (K_{sp}) can be written as: \( K_{sp} = [\text{Ra}^{2+}][\text{SO}_4^{2-}] \). Given \( K_{sp} = 4 \times 10^{-11} \), we can represent the concentrations of \text{Ra}^{2+} and \text{SO}_4^{2-} in a saturated solution of \text{RaSO}_4.

- Consider Common Ion Effect

Since the solution already contains \(0.10 \mathrm{M}\) of \(\text{Na}_2\text{SO}_4\), \(\text{SO}_4^{2-}\) is provided by this source as well. The concentration of \(\text{SO}_4^{2-} \) is therefore constant and equal to \(0.10 \mathrm{M}\), due to the common ion effect. This reduces the solubility of \text{Ra}^{2+}.

- Calculate the Solubility of \text{Ra}^{2+}

Knowing the concentration of \(\text{SO}_4^{2-}\) is constant at \(0.10 \mathrm{M}\), we can solve for the concentration of \text{Ra}^{2+}:\[ K_{sp} = [\text{Ra}^{2+}][\text{SO}_4^{2-}] \Rightarrow [\text{Ra}^{2+}] = \frac{K_{sp}}{[\text{SO}_4^{2-}]} = \frac{4 \times 10^{-11}}{0.10} = 4 \times 10^{-10} \mathrm{M} \]. Thus, the solubility of \text{Ra}^{2+} in the presence of a common ion is \(4 \times 10^{-10} \mathrm{M} \).

Key concepts

Common Ion Effect

The common ion effect is a phenomenon in chemistry where the solubility of a salt is reduced when another salt that shares a common ion is already present in the solution. This effect is key in understanding how some salts dissolve more or less readily in the presence of other salts.

The common ion effect comes into play in our exercise, where the solubility of radium sulfate (\text{RaSO}\(_4\)) is influenced by the presence of sodium sulfate (\text{Na}\(_2\)\text{SO}\(_4\)), which dissociates to provide additional sulfate ions (\text{SO}\(_4^{2-}\)) in solution. The additional sulfate ions from \text{Na}\(_2\)\text{SO}\(_4\) shift the equilibrium of the dissolution of radium sulfate, suppressing the solubility of \text{Ra}\(^{2+}\) ions because the product of the ion concentrations has to remain constant at a value equal to the solubility product (\text{K}\(_{sp}\)).

In essence, the presence of a common ion in a solution can limit the solubility of a compound, which is particularly important for processes such as precipitation, ion exchange, and analytical chemistry.

Solubility Calculations

Solubility calculations involve determining the amount of a solute that can dissolve in a given amount of solvent at a specific temperature. The concept of the solubility product (\text{K}\(_{sp}\)) plays a crucial role in these calculations, particularly for sparingly soluble salts.

\text{K}\(_{sp}\) is the equilibrium constant for the dissolving process of a salt and indicates the extent to which a compound can dissolve in water. To calculate the solubility of an ion in a solution, you typically set up an expression based on the \text{K}\(_{sp}\) and the stoichiometry of the dissolving salt. In our exercise, the \text{K}\(_{sp}\) of radium sulfate dictates the maximum product of the concentrations of \text{Ra}\(^{2+}\) and \text{SO}\(_4^{2-}\) ions that can exist in a solution.

By knowing the \text{K}\(_{sp}\) and any changes introduced by the common ion effect, we can determine the new solubility of a given ion in the altered chemical environment. For example, we calculated the solubility of \text{Ra}\(^{2+}\) in the presence of a common \text{SO}\(_4^{2-}\) ion sourced from sodium sulfate.

Ion Dissociation

Ion dissociation refers to the process by which an ionic compound separates into its constituent ions when dissolved in a solvent, such as water. This separation is essential for the understanding of solubility and reactivity in solutions. Ionic compounds have different degrees of dissociation, and knowing how they dissociate is essential when predicting reactions in solution, understanding conductivity, or calculating pH.

In the case of radium sulfate, ion dissociation involves the compound \text{RaSO}\(_4\) breaking apart into \text{Ra}\(^{2+}\) and \text{SO}\(_4^{2-}\) ions. This process is reversible and reaches an equilibrium state described by its solubility product (\text{K}\(_{sp}\)). The extent of dissociation directly influences the ion concentrations and therefore the solubility of the compounds in solution. The exercise provided illustrates how to account for the presence of a common ion, which modifies the degree of ion dissociation due to the shift in equilibrium, a concept central to understanding solubility behavior in ionic solutions.