Problem 24 - Chemistry

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. \(\mathrm{BaF}_{2}\)

Step-by-Step Solution

Answer

a. Dissolution: \(\mathrm{Ag}_{2} \mathrm{CO}_{3} \rightleftharpoons 2 \mathrm{Ag^+} + \mathrm{CO^{2-}_3}\), Solubility product: \(K_{sp} = [\mathrm{Ag^+}]^2 [\mathrm{CO^{2-}_3}]\) b. Dissolution: \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3} \rightleftharpoons \mathrm{Ce^{3+}} + 3 \mathrm{IO^{-}_3}\), Solubility product: \(K_{sp} = [\mathrm{Ce^{3+}}] [\mathrm{IO^{-}_3}]^3\) c. Dissolution: \(\mathrm{BaF}_{2} \rightleftharpoons \mathrm{Ba^{2+}} + 2 \mathrm{F^{-}}\), Solubility product: \(K_{sp} = [\mathrm{Ba^{2+}}] [\mathrm{F^{-}]^2\)

Step 1: Write balanced equations for the dissolution reactions

We will write balanced equations for the dissolution reactions of the given solids: a. \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\) \(\mathrm{Ag}_{2} \mathrm{CO}_{3} \rightleftharpoons 2 \mathrm{Ag^+} + \mathrm{CO^{2-}_3}\) b. \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}\) \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3} \rightleftharpoons \mathrm{Ce^{3+}} + 3 \mathrm{IO^{-}_3}\) c. \(\mathrm{BaF}_{2}\) \(\mathrm{BaF}_{2} \rightleftharpoons \mathrm{Ba^{2+}} + 2 \mathrm{F^{-}}\)

Step 2: Write the solubility product expressions

Now we will write the corresponding solubility product expressions for each equation: a. \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\) \(K_{sp} = [\mathrm{Ag^+}]^2 [\mathrm{CO^{2-}_3}]\) b. \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}\) \(K_{sp} = [\mathrm{Ce^{3+}}] [\mathrm{IO^{-}_3}]^3\) c. \(\mathrm{BaF}_{2}\) \(K_{sp} = [\mathrm{Ba^{2+}}] [\mathrm{F^{-}]^2\)

Key Concepts

Dissolution Reactions

Understanding dissolution reactions is a critical part of chemistry, especially when dealing with solid substances dissolving in liquids to form ions. A dissolution reaction shows how a solid compound breaks apart into its respective ions in an aqueous solution. This is important in analyzing how different substances dissolve in water, forming a homogenous mixture.
For instance:

• When silver carbonate (\( \mathrm{Ag}_2 \mathrm{CO}_3 \)) dissolves in water, it breaks apart into two silver ions (\(\mathrm{Ag^+}\)) and one carbonate ion (\(\mathrm{CO_3^{2-}}\)).
• Cerium iodate (\(\mathrm{Ce(IO_3)_3}\)) dissociates into one cerium ion (\(\mathrm{Ce^{3+}}\)) and three iodate ions (\(\mathrm{IO_3^-}\)).
• Barium fluoride (\(\mathrm{BaF_2}\)) breaks down into one barium ion (\(\mathrm{Ba^{2+}}\)) and two fluoride ions (\(\mathrm{F^-}\)).

These reversible reactions can be represented with a double-headed arrow (\( \rightleftharpoons \)), indicating that the reaction can go in both directions: from solid to ions and possibly back again if conditions change.

Balanced Equations

Balanced equations are essential in chemistry to ensure mass and charge are conserved during a chemical reaction. When writing a chemical equation for a dissolution reaction, each atom and charge must be accounted for on both the reactant and the product side.
In dissolution reactions:

• The compound formula gives a clear representation of what the solid is, while the breakdown shows how each element separates into its ionic components.
• The total numbers of each type of atom and total charge must be identical on both sides. For example, for \( \mathrm{Ag}_2 \mathrm{CO}_3 \), the balanced equation is: \( \mathrm{Ag}_2 \mathrm{CO}_3 \rightleftharpoons 2 \mathrm{Ag^+} + \mathrm{CO_3^{2-}} \).
• This ensures that both the mass of the silver and the charge are balanced with the carbonate ions produced.

It's like a chemical accounting system: what starts within the solid must be found among the ions represented in the aqueous solution.

Chemical Equilibrium

Chemical equilibrium refers to the state reached in a reversible reaction where the rate of the forward reaction equals the rate of the reverse reaction. This balanced state means that the concentrations of reactants and products remain constant over time, although they may not be equal.
In the context of dissolution:

• The dissolution and precipitation of the ions occurs simultaneously. When a solid like \(\mathrm{BaF_2}\) dissolves in water, a dynamic equilibrium is achieved between the dissolved ions and the undissolved solid.
• If you disrupt the equilibrium by adding more \(\mathrm{F^-}\) ions, for example, the system might shift to reduce this change, favoring the reverse reaction that forms more solid \(\mathrm{BaF_2}\).
• This behavior is described by Le Chatelier's principle. It indicates how equilibrium can shift in response to concentration changes, temperature, or pressure.

Equilibrium is a core concept in predicting how dissolving processes can be manipulated and controlled.

Solubility Expressions

To quantify how much of a compound can dissolve in a solution under equilibrium conditions, the solubility product constant (\(K_{sp}\)) is used. \(K_{sp}\) provides a powerful way to predict the extent of a dissolution reaction.
In this context:

• The \(K_{sp}\) value is determined by the concentrations of the resultant ions in a saturated solution.
• For \(\mathrm{Ag}_2 \mathrm{CO}_3\), \(K_{sp} = [\mathrm{Ag^+}]^2 [\mathrm{CO_3^{2-}}] \). This expression shows the dependence on the ionic concentrations that form from the dissolved solid.
• Higher \(K_{sp}\) values indicate higher solubility of the compound, while lower values suggest that the compound is less soluble.

By using \(K_{sp}\), chemists can predict whether a precipitate will form in a solution, aiding in the design and optimization of experiments and processes involving ionic compounds.