Problem 4 - Chemistry: Principles and Reactions

Question

Write the equilibrium equations on which the following \(K_{s p}\) expressions are based. (a) \(\left[\mathrm{Ca}^{2+}\right]\left[\mathrm{CO}_{3}{\underline{\phantom{xx}}}^{2-}\right]\) (b) \(\left[\mathrm{Co}^{3+}\right]\left[\mathrm{OH}^{-}\right]^{3}\) (c) \(\left[\mathrm{Ag}^{+}\right]^{2}\left[\mathrm{~S}^{2-}\right]\) (d) \(\left[\mathrm{Pb}^{2+}\right]\left[\mathrm{Cl}^{-}\right]^{2}\)

Step-by-Step Solution

Answer

Question: Write the equilibrium equations for the following solubility product constant expressions: a) Ksp = [Ca²⁺][CO₃²⁻] b) Ksp = [Co³⁺][OH⁻]³ c) Ksp = [Ag⁺]²[S²⁻] d) Ksp = [Pb²⁺][Cl⁻]² Answer: a) CaCO₃ ⇌ Ca²⁺ + CO₃²⁻ b) Co(OH)₃ ⇌ Co³⁺ + 3OH⁻ c) Ag₂S ⇌ 2Ag⁺ + S²⁻ d) PbCl₂ ⇌ Pb²⁺ + 2Cl⁻

Step 1: (a) Identify the chemical species

We have two chemical species here: \(\mathrm{Ca^{2+}}\) and \(\mathrm{CO_3^{2-}}\). These ions are formed through the dissociation of calcium carbonate (CaCO3).

Step 2: (a) Write the dissociation reaction

The dissociation reaction of calcium carbonate (CaCO3) is given by: \(\mathrm{CaCO_3} \leftrightarrows \mathrm{Ca^{2+}} + \mathrm{CO}_{3}^{2-}\)

Step 3: (a) Write the equilibrium equation

The equilibrium equation for this reaction is: Ksp = \(\left[\mathrm{Ca}^{2+}\right]\left[\mathrm{CO}_{3}{\underline{\phantom{xx}}}^{2-}\right]\)

Step 4: (b) Identify the chemical species

We have two chemical species here: \(\mathrm{Co^{3+}}\) and \(\mathrm{OH^-}\). These ions are formed through the dissociation of cobalt(III) hydroxide (Co(OH)3).

Step 5: (b) Write the dissociation reaction

The dissociation reaction of cobalt(III) hydroxide (Co(OH)3) is given by: \(\mathrm{Co(OH)}_3 \leftrightarrows \mathrm{Co^{3+}} + 3\mathrm{OH^-}\)

Step 6: (b) Write the equilibrium equation

The equilibrium equation for this reaction is: Ksp = \(\left[\mathrm{Co}^{3+}\right]\left[\mathrm{OH}^{-}\right]^{3}\)

Step 7: (c) Identify the chemical species

We have two chemical species here: \(\mathrm{Ag^+}\) and \(\mathrm{S^{2-}}\). These ions are formed through the dissociation of silver(I) sulfide (Ag2S).

Step 8: (c) Write the dissociation reaction

The dissociation reaction of silver(I) sulfide (Ag2S) is given by: \(\mathrm{Ag}_2\mathrm{S} \leftrightarrows 2\mathrm{Ag^+} + \mathrm{S}^{2-}\)

Step 9: (c) Write the equilibrium equation

The equilibrium equation for this reaction is: Ksp = \(\left[\mathrm{Ag}^{+}\right]^{2}\left[\mathrm{~S}^{2-}\right]\)

Step 10: (d) Identify the chemical species

We have two chemical species here: \(\mathrm{Pb^{2+}}\) and \(\mathrm{Cl^-}\). These ions are formed through the dissociation of lead(II) chloride (PbCl2).

Step 11: (d) Write the dissociation reaction

The dissociation reaction of lead(II) chloride (PbCl2) is given by: \(\mathrm{PbCl}_2 \leftrightarrows \mathrm{Pb^{2+}} + 2\mathrm{Cl^-}\)

Step 12: (d) Write the equilibrium equation

The equilibrium equation for this reaction is: Ksp = \(\left[\mathrm{Pb}^{2+}\right]\left[\mathrm{Cl}^{-}\right]^{2}\)

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a fascinating concept in chemistry that describes a state in which the concentrations of reactants and products remain constant over time. This occurs when the forward and reverse reactions in a closed system happen at the same rate. At this point, the system has reached an equilibrium state.

Understanding chemical equilibrium is crucial when studying solubility products. The solubility product constant, often denoted as \(K_{sp}\), is derived from the equilibrium constant for the dissociation of a sparingly soluble ionic compound in water. This constant indicates how much of the compound can dissolve in a solution under equilibrium conditions.

In essence, chemical equilibrium helps us predict the maximum amount of compound that can exist in a dissolved state, providing valuable insights into various chemical processes.

Dissociation Reactions

Dissociation reactions play a pivotal role in understanding solubility and chemical equilibrium. These reactions involve the breaking apart of a compound into smaller ions or molecules, usually in a solvent such as water.

In the context of ionic compounds, dissociation reactions are fundamental for calculating the solubility product constant \(K_{sp}\). For example, when calcium carbonate \(\text{CaCO}_3\) dissociates, it forms calcium ions \(\text{Ca}^{2+}\) and carbonate ions \(\text{CO}_3^{2-}\). This can be represented by the equation: \[\text{CaCO}_3 \leftrightarrows \text{Ca}^{2+} + \text{CO}_3^{2-}\]The equilibrium reached in such reactions allows us to define \(K_{sp}\), providing a quantitative measure of solubility.

Ionic Compounds

Ionic compounds are made up of positively charged ions (cations) and negatively charged ions (anions). These compounds are generally formed when metals react with non-metals, transferring electrons to achieve stable electronic configurations.

In solutions, ionic compounds can dissociate into their constituent ions, as seen with examples like silver sulfide \(\text{Ag}_2\text{S}\). Upon dissociation in water, silver sulfide breaks into silver ions \(\text{Ag}^+\) and sulfide ions \(\text{S}^{2-}\):\[\text{Ag}_2\text{S} \leftrightarrows 2\text{Ag}^+ + \text{S}^{2-}\]The ability of an ionic compound to dissolve and reach equilibrium with its ions in solution is crucial for determining its solubility product.

Equilibrium Constant Expressions

Equilibrium constant expressions provide a mathematical way to express the concentrations of reactants and products at equilibrium in a reversible reaction. For dissolved ionic compounds, the solubility product constant \(K_{sp}\) is a specific type of equilibrium constant.

The expression is formed by taking the product of the concentrations of the dissociated ions, each raised to the power of their stoichiometric coefficients in the balanced equation. For instance, lead(II) chloride \(\text{PbCl}_2\) dissociates in water as follows:\[\text{PbCl}_2 \leftrightarrows \text{Pb}^{2+} + 2\text{Cl}^-\]The \(K_{sp}\) expression for this dissociation is:\[K_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2\]These expressions help predict how much solute can dissolve in the solution before reaching saturation, maintaining equilibrium.