Problem 1 - General Chemistry: Principles and Modern Applications

Question

Write \(K_{\text {sp }}\) expressions for the following equilibria. For example, for the reaction \(\mathrm{AgCl}(\mathrm{s}) \rightleftharpoons \mathrm{Ag}^{+}(\mathrm{aq})+\) \(\mathrm{Cl}^{-}(\mathrm{aq}), K_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]\). (a) \(\mathrm{Ag}_{2} \mathrm{SO}_{4}(\mathrm{s}) \rightleftharpoons 2 \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{SO}_{4}^{2-}(\mathrm{aq})\) (b) \(\operatorname{Ra}\left(\mathrm{IO}_{3}\right)_{2}(\mathrm{s}) \rightleftharpoons \mathrm{Ra}^{2+}(\mathrm{aq})+2 \mathrm{IO}_{3}^{-}(\mathrm{aq})\) (c) \(\mathrm{Ni}_{3}\left(\mathrm{PO}_{4}\right)_{2}(\mathrm{s}) \rightleftharpoons 3 \mathrm{Ni}^{2+}(\mathrm{aq})+2 \mathrm{PO}_{4}^{3-}(\mathrm{aq})\) (d) \(\mathrm{PuO}_{2} \mathrm{CO}_{3}(\mathrm{s}) \rightleftharpoons \mathrm{PuO}_{2}^{2+}(\mathrm{aq})+\mathrm{CO}_{3}^{2-}(\mathrm{aq})\)

Step-by-Step Solution

Answer

The \(K_{sp}\) expressions for the given equilibria are: (a) \(K_{sp}=[Ag^+]^2[SO_{4}^{2-}]\), (b) \(K_{sp}=[Ra^{2+}][IO_{3}^{-}]^2\), (c) \(K_{sp}=[Ni^{2+}]^3[PO_{4}^{3-}]^2\), (d) \(K_{sp}=[PuO_{2}^{2+}][CO_{3}^{2-}]\)

Step 1: Write expression for Ag2SO4

The equilibrium reaction for the dissociation of \(Ag_2SO_4\) is: \(Ag_{2}SO_{4}(s) \rightleftharpoons 2Ag^{+}(aq)+SO_{4}^{2-}(aq)\). From this, the \(K_{sp}\) expression can be written as: \(K_{sp}=[Ag^+]^2[SO_{4}^{2-}]\)

Step 2: Write expression for Ra(IO3)2

The equilibrium reaction for the dissociation of \(Ra(IO_3)_2\) is: \(Ra(IO_3)_2(s) \rightleftharpoons Ra^{2+}(aq)+2IO_{3}^{-}(aq)\). The \(K_{sp}\) expression can be written as: \(K_{sp}=[Ra^{2+}][IO_{3}^{-}]^2\)

Step 3: Write expression for Ni3(PO4)2

The equilibrium reaction for the dissociation of \(Ni_3(PO_4)_2\) is: \(Ni_{3}(PO_{4})_{2}(s) \rightleftharpoons 3Ni^{2+}(aq)+2PO_{4}^{3-}(aq)\). Using this information, we can write the \(K_{sp}\) expression as: \(K_{sp}=[Ni^{2+}]^3[PO_{4}^{3-}]^2\)

Step 4: Write expression for PuO2CO3

The equilibrium reaction for the dissociation of \(PuO_2CO_3\) is: \(PuO_{2} CO_{3}(s) \rightleftharpoons PuO_{2}^{2+}(aq)+CO_{3}^{2-}(aq)\). From this, we can write the \(K_{sp}\) expression as: \(K_{sp}=[PuO_{2}^{2+}][CO_{3}^{2-}]\)

Key Concepts

Chemical Equilibrium

In chemistry, equilibrium represents a state where the forward and reverse reactions occur at the same rate. This balance is achieved when the concentrations of reactants and products remain constant over time, although both reactions continue to occur. Understanding equilibrium is crucial because it dictates how reactions proceed and how we can influence them.
In the context of solubility and dissolution, the equilibrium concept is expressed through the solubility product constant, or \(K_{sp}\). This constant quantifies the extent to which a compound will dissolve in water. The principle is that, in a chemical reaction, equilibrium is when the rate of dissolution equals the rate of precipitation.
The expressions for solubility product constants, such as those for \(Ag_2SO_4\), are derived from these equilibrium principles. Each species' concentration is raised to the power of its coefficient in the balanced equation, illustrating the law of mass action's application to solubility products.

Dissolution Reactions

Dissolution reactions describe the process where a solid substance dissolves into a solvent, becoming a solute. In these reactions, the solid breaks into its constituent ions. For example, when \(Ag_2SO_4\) dissolves, it separates into two \(Ag^+\) ions and one \(SO_4^{2-}\) ion.
These dissolution reactions are represented by an equation showing the solid transforming into its ions, such as: - \(Ag_2SO_4(s) \rightleftharpoons 2Ag^+(aq) + SO_4^{2-}(aq)\) Here, the double-headed arrow indicates that the process is reversible, a key aspect of equilibrium.
Dissolution is significant because it helps us understand the solubility of different compounds. By studying dissolution reactions, we can predict how much of a substance will dissolve in a given amount of solvent, crucial for many practical applications like water purification and medication formulation.

Precipitation Reactions

Precipitation reactions are the opposite of dissolution reactions. They involve the formation of a solid from ions in a solution. When the conditions of a solution change, such as through concentration or temperature, a solid may form and settle out of the liquid, becoming what is known as a precipitate.
In the context of chemical equilibria, precipitation occurs when the product of the concentrations of the ions in solution exceeds the \(K_{sp}\) value. Once this threshold is surpassed, the ions combine to form a new solid substance.
Understanding precipitation reactions helps us manage chemical concentrations, like in water treatment, where unwanted ions must be removed as solids. This concept also assists in identifying unknown ions in a solution, as different substances precipitate under different conditions. Recognizing these reactions is essential for controlling and predicting outcomes in industrial and laboratory processes.