Problem 23 - Chemistry

Question

Write balanced equations for the dissolution reactions and the corresponding solubility product expressions for each of the following solids. a. \(A g C_{2} H_{3} O_{2}\) b. \(A l(O H)_{3}\) c. \(C a_{3}\left(\mathrm{PO}_{4}\right)_{2}\)

Step-by-Step Solution

Answer

a. Dissolution reaction: \(AgC_2H_3O_2 (s) \rightleftharpoons Ag^+ (aq) + C_2H_3O_2^- (aq)\); Solubility product expression: \(K_{sp}(AgC_2H_3O_2) = [Ag^+] [C_2H_3O_2^-]\) b. Dissolution reaction: \(Al(OH)_3 (s) \rightleftharpoons Al^{3+} (aq) + 3OH^- (aq)\); Solubility product expression: \(K_{sp}(Al(OH)_3) = [Al^{3+}] [OH^-]^3\) c. Dissolution reaction: \(Ca_3(PO_4)_2 (s) \rightleftharpoons 3Ca^{2+} (aq) + 2PO_4^{3-} (aq)\); Solubility product expression: \(K_{sp}(Ca_3(PO_4)_2) = [Ca^{2+}]^3 [PO_4^{3-}]^2\)

Step 1: Write the dissolution reaction for \(AgC_2H_3O_2\)

Dissolve the solid into its individual ions. \(AgC_2H_3O_2 (s) \rightleftharpoons Ag^+ (aq) + C_2H_3O_2^- (aq)\).

Step 2: Write the solubility product expression for \(AgC_2H_3O_2\)

Write the solubility product constant from the given equation as follows: \(K_{sp}(AgC_2H_3O_2) = [Ag^+] [C_2H_3O_2^-]\)

Step 3: Write the dissolution reaction for \(Al(OH)_3\)

Dissolve the aluminum hydroxide into its individual ions. \(Al(OH)_3 (s) \rightleftharpoons Al^{3+} (aq) + 3OH^- (aq)\)

Step 4: Write the solubility product expression for \(Al(OH)_3\)

Write the solubility product constant from the given equation as follows: \(K_{sp}(Al(OH)_3) = [Al^{3+}] [OH^-]^3\)

Step 5: Write the dissolution reaction for \(Ca_3(PO_4)_2\)

Dissolve the calcium phosphate into its individual ions. \(Ca_3(PO_4)_2 (s) \rightleftharpoons 3Ca^{2+} (aq) + 2PO_4^{3-} (aq)\)

Step 6: Write the solubility product expression for \(Ca_3(PO_4)_2\)

Write the solubility product constant from the given equation as follows: \(K_{sp}(Ca_3(PO_4)_2) = [Ca^{2+}]^3 [PO_4^{3-}]^2\)

Key Concepts

Dissolution Reactions

Dissolution reactions describe the process through which a solid dissolves in a solvent, usually water, to form a solution of ions. Imagine a sugar cube dissolving in a cup of tea; as the sugar dissolves, it spreads throughout the tea, leaving no visible solid particles. In chemistry, however, dissolution goes a step further as the solid breaks down into its individual ions.
This is crucial for understanding chemical reactions in aqueous solutions. Each component of the solid ionic lattice is separated and surrounded by solvent molecules, aiding chemical reactivity.
Let's delve into examples:

• For the compound \(AgC_2H_3O_2\), the dissolution reaction separates it into ions, \(Ag^+ (aq)\) and \(C_2H_3O_2^- (aq)\).
• In the case of \(Al(OH)_3\), the dissolution yields ions \(Al^{3+} (aq)\) and \(3OH^- (aq)\).
• For \(Ca_3(PO_4)_2\), this process results in ions \(3Ca^{2+} (aq)\) and \(2PO_4^{3-} (aq)\).

Understanding these reactions helps in predicting the behaviors of ionic compounds in solutions.
It also facilitates the calculation of solubility, which is integral in many chemical processes and industries.

Balanced Equations

Balanced equations are essential in chemistry because they portray the correct stoichiometry of a chemical reaction. Stoichiometry involves understanding the proportions of elements within a compound and balancing ensures the conservation of mass and charge. This simply means that the number of each type of atom on the reactant side equals the number of each type on the product side.
In dissolution reactions, a balanced equation accurately represents the breakdown of the solid compound into its constituent ions.
For example, the dissolution of \(Al(OH)_3\) can be represented as follows:
- Reactant: \(Al(OH)_3 (s)\)
- Products: \(Al^{3+} (aq)\), \(3OH^- (aq)\)
This shows that one formula unit of aluminum hydroxide produces one aluminum ion and three hydroxide ions.
Similarly, for \(Ca_3(PO_4)_2\), the equation is balanced to accommodate:

• Three ions of \(Ca^{2+}\)
• Two ions of \(PO_4^{3-}\)

Starting with a balanced equation aids in further calculations, such as those involving the solubility product, and it is fundamental in ensuring reaction predictions and quantitative analyses remain accurate.

Chemical Equilibrium

Chemical equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction. In the context of dissolution reactions, equilibrium is reached when the rate of the solid dissolving equals the rate of the dissolved ions recombining into the solid.
This can be described using the equilibrium constant known as the solubility product constant, \(K_{sp}\). It is represented by the concentrations of the dissolved ions raised to the power of their stoichiometric coefficients.
For example, the dissolution reaction of \(Al(OH)_3\) yields the expression:
\[ K_{sp} = [Al^{3+}] [OH^-]^3 \]
This formula indicates how likely the components are to stay dissolved in solution as opposed to forming the solid.
Similarly, for \(Ca_3(PO_4)_2\), the solubility product expression is:
\[ K_{sp} = [Ca^{2+}]^3 [PO_4^{3-}]^2 \]
These expressions are useful for predicting whether a precipitate will form in a solution under specific conditions.
Understanding chemical equilibrium and solubility products is vital for predicting the solubility and behavior of compounds in different environments, impacting fields ranging from pharmaceuticals to environmental science.