Question

Write balanced equations and solubility product expressions for the solubility equilibria of these compounds: (a) \(\mathrm{CuBr},\) (b) \(\mathrm{ZnC}_{2} \mathrm{O}_{4},\) (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

The solubility product expressions are \[K_{sp}=[\mathrm{Cu}^{+}][\mathrm{Br}^{-}]\], \[K_{sp}=[\mathrm{Zn}^{2+}][\mathrm{C}_{2} \mathrm{O}_{4}^{2-}]^{2}\], \[K_{sp}=[\mathrm{Ag}^{+}]^{2}[\mathrm{CrO}_{4}^{2-}]\], \[K_{sp}=[\mathrm{Hg}_{2}^{2+}][\mathrm{Cl}^{-}]^{2}\], \[K_{sp}=[\mathrm{Au}^{3+}][\mathrm{Cl}^{-}]^{3}\] and \[K_{sp}=[\mathrm{Mn}^{2+}]^{3}[\mathrm{PO}_{4}^{3-}]^{2}\].

Step-by-step solution

Establishing the General Equilibrium Equation for Solubility

The general equation for solubility equilibrium is: \(\mathrm{AB}_{(s)} \rightleftharpoons \mathrm{A}^{+}_{(aq)} + \mathrm{B}^{-}_{(aq)}\). Balancing is not needed here since every reactant is reacting in a ratio of 1:1.

Solubility Equilibrium for CuBr

Here's the balanced solubility equation for \(\mathrm{CuBr}\): \(\mathrm{CuBr}_{(s)} \rightleftharpoons \mathrm{Cu}^{+}_{(aq)} + \mathrm{Br}^{-}_{(aq)}\). The solubility product expression is: \[K_{sp}=[\mathrm{Cu}^{+}][\mathrm{Br}^{-}]\].

Solubility Equilibrium for ZnC2O4

The balanced solubility equation for \(\mathrm{ZnC}_{2} \mathrm{O}_{4}\) is: \(\mathrm{ZnC}_{2} \mathrm{O}_{4(s)} \rightleftharpoons \mathrm{Zn}^{2+}_{(aq)} + 2\mathrm{C}_{2} \mathrm{O}_{4}^{2-}_{(aq)}\). The solubility product expression is: \[K_{sp}=[\mathrm{Zn}^{2+}][\mathrm{C}_{2} \mathrm{O}_{4}^{2-}]^{2}\].

Solubility Equilibrium for Ag2CrO4

The balanced solubility equation for \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) is: \(\mathrm{Ag}_{2} \mathrm{CrO}_{4(s)} \rightleftharpoons 2\mathrm{Ag}^{+}_{(aq)} + \mathrm{CrO}_{4}^{2-}_{(aq)}\). The solubility product expression is: \[K_{sp}=[\mathrm{Ag}^{+}]^{2}[\mathrm{CrO}_{4}^{2-}]\].

Solubility Equilibrium for Hg2Cl2

The balanced solubility equation for \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}\) is: \(\mathrm{Hg}_{2} \mathrm{Cl}_{2(s)} \rightleftharpoons \mathrm{Hg}_{2}^{2+}_{(aq)} + 2\mathrm{Cl}^{-}_{(aq)}\). The solubility product expression is: \[K_{sp}=[\mathrm{Hg}_{2}^{2+}][\mathrm{Cl}^{-}]^{2}\].

Solubility Equilibrium for AuCl3

The balanced solubility equation for \(\mathrm{AuCl}_{3}\) is: \(\mathrm{AuCl}_{3(s)} \rightleftharpoons \mathrm{Au}^{3+}_{(aq)} + 3\mathrm{Cl}^{-}_{(aq)}\). The solubility product expression is: \[K_{sp}=[\mathrm{Au}^{3+}][\mathrm{Cl}^{-}]^{3}\].

Solubility Equilibrium for Mn3(PO4)2

The balanced solubility equation for \(\mathrm{Mn}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) is: \(\mathrm{Mn}_{3}\left(\mathrm{PO}_{4}\right)_{2(s)} \rightleftharpoons 3\mathrm{Mn}^{2+}_{(aq)} + 2\mathrm{PO}_{4}^{3-}_{(aq)}\). The solubility product expression is: \[K_{sp}=[\mathrm{Mn}^{2+}]^{3}[\mathrm{PO}_{4}^{3-}]^{2}\].

Key concepts

Balanced Equations

In chemistry, balanced equations are essential for understanding how different compounds interact in a reaction. A balanced equation ensures that the number of atoms for each element is the same on both sides of the equation.
This principle is crucial in solubility equilibria as it helps predict the products formed when a compound dissolves in water.

• For example, when copper(I) bromide (\(\mathrm{CuBr}\)) dissolves in water, it dissociates into copper ions (\(\mathrm{Cu}^{+}\)) and bromide ions (\(\mathrm{Br}^{-}\)). The balanced equation is: \(\)\mathrm{CuBr}_{(s)} \rightleftharpoons \mathrm{Cu}^{+}_{(aq)} + \mathrm{Br}^{-}_{(aq)}\(\).
• For an equation to be balanced, reactants and products should have equal numbers of each type of atom.

Understanding how to write balanced equations is the first step in solving solubility equilibrium problems.

Solubility Product Expressions

In terms of solubility equilibria, solubility product expressions are mathematical expressions that represent the equilibrium between a solid and its respective ions in solution.
The solubility product constant (\(K_{sp}\)) indicates the extent to which a compound can dissolve in water. It is specific to each compound and conditions such as temperature.

• For copper(I) bromide (\(\mathrm{CuBr}\)), the solubility product expression is given by: \[K_{sp}=[\mathrm{Cu}^{+}][\mathrm{Br}^{-}]\]
• This expression is derived from the balanced dissolution equation and shows the concentration of ions at equilibrium.
• Compounds with higher \(K_{sp}\) values are generally more soluble than those with lower values.

Using solubility product expressions allows chemists to calculate and predict solubility behavior under various conditions.

Chemical Equilibria

Chemical equilibria are a foundational concept in understanding reactions where reversible processes occur. In the case of solubility, equilibrium refers to the state where the rate of dissolution equals the rate of precipitation.
This balance means that a solution is saturated, and no more solid can dissolve at a given temperature.

• The equilibrium for \(\mathrm{Ag}_{2}\mathrm{CrO}_{4}\) can be represented as: \[\mathrm{Ag}_{2}\mathrm{CrO}_{4(s)} \rightleftharpoons 2\mathrm{Ag}^{+}_{(aq)} + \mathrm{CrO}_{4}^{2-}_{(aq)}\]
• The concentration of ions must satisfy the solubility product expression to remain at equilibrium.

Understanding chemical equilibria is vital for predicting how changes in conditions, such as temperature or concentration, can shift the equilibrium and affect solubility.

Ksp

The solubility product constant (\(K_{sp}\)) is a critical value in solubility equilibria. This constant helps predict whether a precipitate will form under specific conditions.
The \(K_{sp}\) is unique for each ionic compound and can vary based on factors like temperature.

• For example, the \(K_{sp}\) for \(\mathrm{Mn}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) is given by the expression: \[K_{sp}=[\mathrm{Mn}^{2+}]^{3}[\mathrm{PO}_{4}^{3-}]^{2}\]
• A low \(K_{sp}\) value indicates limited solubility, meaning the compound is less likely to dissociate completely in water.
• Conversely, high \(K_{sp}\) values suggest greater solubility.

By understanding \(K_{sp}\), students can better grasp how different factors affect the solubility of compounds in aqueous solutions.