Question

Write balanced equations for the dissolution reactions and the corresponding solubility product expressions for each of the following solids. a. \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\) b. \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}\) c. \(\mathrm{BaF}_{2}\)

Short answer

The balanced equations for the dissolution reactions and solubility product expressions for the given solids are as follows: a. \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\): \[ \mathrm{Ag}_{2} \mathrm{CO}_{3} \rightleftharpoons 2\mathrm{Ag^{+}} + \mathrm{CO}_{3}^{2-}\] \[ K_{sp} = [\mathrm{Ag^{+}}]^{2}[\mathrm{CO}_{3}^{2-}] \] b. \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}\): \[ \mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3} \rightleftharpoons \mathrm{Ce}^{3+} + 3\mathrm{IO}_{3}^{-}\] \[ K_{sp} = [\mathrm{Ce}^{3+}][\mathrm{IO}_{3}^{-}]^{3} \] c. \(\mathrm{BaF}_{2}\): \[ \mathrm{BaF}_{2} \rightleftharpoons \mathrm{Ba}^{2+} + 2\mathrm{F}^{-}\] \[ K_{sp} = [\mathrm{Ba}^{2+}][\mathrm{F}^{-}]^{2} \]

Step-by-step solution

Write the balanced equations

We will break down each solid into its constituent ions during dissolution: a. For \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\), we have: \[ \mathrm{Ag}_{2} \mathrm{CO}_{3} \rightleftharpoons 2\mathrm{Ag^{+}} + \mathrm{CO}_{3}^{2-}\] b. For \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}\), we have: \[ \mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3} \rightleftharpoons \mathrm{Ce}^{3+} + 3\mathrm{IO}_{3}^{-}\] c. For \(\mathrm{BaF}_{2}\), we have: \[ \mathrm{BaF}_{2} \rightleftharpoons \mathrm{Ba}^{2+} + 2\mathrm{F}^{-}\]

Find solubility product expressions

Now, we can find the solubility product expressions by multiplying the concentrations of the resulting ions, raised to the power of their stoichiometric coefficients in the balanced equation. a. For \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\): \[ K_{sp} = [\mathrm{Ag^{+}}]^{2}[\mathrm{CO}_{3}^{2-}] \] b. For \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}\): \[ K_{sp} = [\mathrm{Ce}^{3+}][\mathrm{IO}_{3}^{-}]^{3} \] c. For \(\mathrm{BaF}_{2}\): \[ K_{sp} = [\mathrm{Ba}^{2+}][\mathrm{F}^{-}]^{2} \] These are the solubility product expressions for the given solids.

Key concepts

Balanced Chemical Equations

Balanced chemical equations are crucial for understanding the dissolution process. When a solid is dissolved in water, it breaks down into its constituent ions. This process is represented by a balanced chemical equation.
For instance, when you dissolve silver carbonate \( \mathrm{Ag}_{2} \mathrm{CO}_{3} \) in water, the solid dissociates into two silver ions \(2\mathrm{Ag}^{+}\) and one carbonate ion \(\mathrm{CO}_{3}^{2-}\).
A balanced chemical equation shows that the number of atoms for each element is the same on both sides of the equation. This alignment ensures conservation of mass, meaning the mass remains constant before and after the reaction. It's a tool to visualize what happens at a molecular level during dissolution.

Dissolution Reactions

Dissolution reactions refer to the process where a solid substance (often a salt) dissolves in a solvent to form ions. This is essential to comprehend solubility and reactions in aqueous solutions.
When cerium iodate \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}\) dissolves, it separates into one cerium ion \(\mathrm{Ce}^{3+}\) and three iodate ions \(\mathrm{IO}_{3}^{-}\).
The reaction can be summarized by observing how the lattice structure of the solid breaks down to release individual ions into the solution, allowing the ions to interact freely with the solvent molecules.

Stoichiometry

Stoichiometry involves the quantitative analysis of products and reactants in chemical reactions, allowing predictions of product amounts from given reactants. It's a method that uses the ratios of molecules participating in the chemical reaction, as seen in a balanced equation.
For dissolution, stoichiometry provides the mole ratio of ions formed. For example, the dissolution of barium fluoride \(\mathrm{BaF}_{2}\) yields one molecule of barium ion \(\mathrm{Ba}^{2+}\) for every two molecules of fluoride ion \(2\mathrm{F}^{-}\).
These ratios are critical when calculating solubility products and understanding the extent to which a substance can dissolve in a solvent.

Ion Concentration

Ion concentration refers to the amount of a given ion in a solution, usually expressed in moles per liter (M). Understanding this concept is key to calculating properties like the solubility product constant \(K_{sp}\).
The \(K_{sp}\) expression relies on the ion concentration of each dissolved species. For \(\mathrm{BaF}_{2}\), the concentration of barium ions \(\mathrm{Ba}^{2+}\) multiplied by the concentration of fluoride ions \(\mathrm{F}^{-}\) squared gives its \(K_{sp}\). Understanding ion concentrations helps predict the solubility and stability of compounds in various conditions.