Question

Write balanced equations and solubility product expressions for the solubility equilibria of the following compounds: (a) \(\mathrm{CuBr}\), (b) \(\mathrm{ZnC}_{2} \mathrm{O}_{4},(\mathrm{c}) \mathrm{Ag}_{2} \mathrm{CrO}_{4},\) (d) \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}\), (e) \(\mathrm{AuCl}_{3}\) (f) \(\mathrm{Mn}_{3}\left(\mathrm{PO}_{4}\right)_{2}\).

Short answer

Write dissociation equations for each compound and use their ions to construct Ksp expressions.

Step-by-step solution

Understand Solubility Product

The solubility product, represented as K_sp, is used to quantify the solubility of sparingly soluble ionic compounds. It describes the equilibrium between a solid and its constituent ions in a saturated solution. The equilibrium reaction is written for the dissociation of the ionic compound into ions, and K_sp is the product of the ion concentrations each raised to the power of their stoichiometric coefficients.

Write the Equilibrium Equation for CuBr

The compound CuBr dissociates as follows: \[ \mathrm{CuBr (s)} \rightleftharpoons \mathrm{Cu}^{+} (aq) + \mathrm{Br}^{-} (aq) \]The solubility product expression for CuBr is: \[ K_{sp} = [\mathrm{Cu}^{+}][\mathrm{Br}^{-}] \]

Write the Equilibrium Equation for ZnC2O4

The compound ZnC2O4 dissociates as follows:\[ \mathrm{ZnC}_{2}\mathrm{O}_{4} (s) \rightleftharpoons \mathrm{Zn}^{2+} (aq) + \mathrm{C}_{2} \mathrm{O}_{4}^{2-} (aq) \]The solubility product expression for ZnC2O4 is:\[ K_{sp} = [\mathrm{Zn}^{2+}][\mathrm{C}_{2} \mathrm{O}_{4}^{2-}] \]

Write the Equilibrium Equation for Ag2CrO4

The compound Ag2CrO4 dissociates as follows:\[ \mathrm{Ag}_{2}\mathrm{CrO}_{4} (s) \rightleftharpoons 2\mathrm{Ag}^{+} (aq) + \mathrm{CrO}_{4}^{2-} (aq) \]The solubility product expression for Ag2CrO4 is:\[ K_{sp} = [\mathrm{Ag}^{+}]^2[\mathrm{CrO}_{4}^{2-}] \]

Write the Equilibrium Equation for Hg2Cl2

The compound Hg2Cl2 can be thought of as dissociating partially due to its dimerization:\[ \mathrm{Hg}_{2}\mathrm{Cl}_{2} (s) \rightleftharpoons \mathrm{Hg}_{2}^{2+} (aq) + 2\mathrm{Cl}^{-} (aq) \]The solubility product expression for Hg2Cl2 is:\[ K_{sp} = [\mathrm{Hg}_{2}^{2+}][\mathrm{Cl}^{-}]^2 \]

Write the Equilibrium Equation for AuCl3

The compound AuCl3 dissociates as follows:\[ \mathrm{AuCl}_{3} (s) \rightleftharpoons \mathrm{Au}^{3+} (aq) + 3\mathrm{Cl}^{-} (aq) \]The solubility product expression for AuCl3 is:\[ K_{sp} = [\mathrm{Au}^{3+}][\mathrm{Cl}^{-}]^3 \]

Write the Equilibrium Equation for Mn3(PO4)2

The compound Mn3(PO4)2 dissociates as follows:\[ \mathrm{Mn}_{3}(\mathrm{PO}_{4})_{2} (s) \rightleftharpoons 3\mathrm{Mn}^{2+} (aq) + 2\mathrm{PO}_{4}^{3-} (aq) \]The solubility product expression for Mn3(PO4)2 is:\[ K_{sp} = [\mathrm{Mn}^{2+}]^3[\mathrm{PO}_{4}^{3-}]^2 \]

Key concepts

Solubility Product (K_sp)

The solubility product constant, denoted as \( K_{sp} \), is an important concept in understanding solubility equilibria. It helps predict the extent to which a sparingly soluble compound will dissolve in water. Unlike common soluble salts, sparingly soluble compounds only slightly dissociate into their ions when they dissolve.
The solubility product is calculated as the product of the concentrations of the ions, each raised to the power of their stoichiometric coefficients from the balanced equilibrium equation. For example, if you have a compound \( AB \) that dissolves into \( A^+ \) and \( B^- \), the solubility product expression is \( K_{sp} = [A^+][B^-] \).

• \( K_{sp} \) helps determine whether a precipitate will form in a solution.
• A small \( K_{sp} \) value indicates low solubility and therefore, a higher likelihood of precipitation.

Understanding \( K_{sp} \) allows chemists to predict the behavior of ions in solution.

Ionic Dissociation

Ionic dissociation is a process where an ionic compound separates into its constituent ions when dissolved in water. For sparingly soluble compounds, the extent of dissociation into ions is minimal. When you write an equilibrium equation for a compound, you represent how it dissociates into its ions in an aqueous solution.
Consider \( \text{CuBr} \): when it dissolves, it separates into \( \text{Cu}^+ \) and \( \text{Br}^- \), demonstrating ionic dissociation. The equilibrium equation is:
\[ \text{CuBr (s)} \rightleftharpoons \text{Cu}^{+} \text{(aq)} + \text{Br}^{-} \text{(aq)} \] This is crucial because it shows the formation of ions that affect the rates and extents of chemical reactions.

• Dissociation strength affects the solubility of the compound.
• Proper representation of dissociation is essential for calculating \( K_{sp} \).

Equilibrium Equations

Equilibrium equations represent the state where the dissolution of a sparingly soluble compound and the precipitation form a balanced dynamic system. At equilibrium, the rate at which the solid dissolves is equal to the rate at which ions precipitate back into the solid form.
For example, consider the dissolution of \( \text{Ag}_2\text{CrO}_4 \): it dissociates into \( 2\text{Ag}^+ \) and \( \text{CrO}_4^{2-} \): \[ \text{Ag}_2\text{CrO}_4 (s) \rightleftharpoons 2\text{Ag}^{+} \text{(aq)} + \text{CrO}_4^{2-} \text{(aq)} \] This illustrates the balance of dissolution and precipitation at equilibrium.

• The coefficients from the balanced chemical equation become exponents in the \( K_{sp} \) expression.
• Understanding equilibrium helps in predicting the concentration of ions in a saturated solution.

Sparingly Soluble Compounds

Sparingly soluble compounds are compounds that dissolve in water to only a small extent. Unlike highly soluble salts that completely dissociate into ions, these compounds reach a point where no more solute can dissolve in the solution, known as the saturation point.
Some examples include \( \text{ZnC}_2\text{O}_4 \) and \( \text{Mn}_3(\text{PO}_4)_2 \). Upon dissolution, they establish an equilibrium between the solid phase and its ionic components in the solution.

• They usually have very low \( K_{sp} \) values.
• In practice, these compounds slightly affect the ionic concentration in solutions, making them relevant in precipitation reactions.

Understanding these compounds is crucial in various fields, such as predicting mineral solubility in geological formations.