Question

For each of the following, write the equilibrium-constant expression for \(K_{\mathrm{sp}}\) : (a) \(\mathrm{Ca}(\mathrm{OH})_{2}\) (b) \(\mathrm{Ag}_{3} \mathrm{PO}_{4}\) (c) \(\mathrm{BaCO}_{3}\) (d) \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH}\)

Short answer

(a) \( K_{sp} = [\mathrm{Ca}^{2+}][\mathrm{OH}^-]^2 \); (b) \( K_{sp} = [\mathrm{Ag}^+]^3[\mathrm{PO}_4^{3-}] \); (c) \( K_{sp} = [\mathrm{Ba}^{2+}][\mathrm{CO}_3^{2-}] \); (d) \( K_{sp} = [\mathrm{Ca}^{2+}]^5[\mathrm{PO}_4^{3-}]^3[\mathrm{OH}^-] \).

Step-by-step solution

- Understanding the Compounds and Their Dissociation

Begin by understanding each compound and how it dissociates in water. Dissociation means breaking down into its constituent ions. For (a) \( \mathrm{Ca(OH)}_2 \), it dissociates into \( \mathrm{Ca}^{2+} \) and \( 2\mathrm{OH}^- \). For (b) \( \mathrm{Ag}_3\mathrm{PO}_4 \), it dissociates into \( 3\mathrm{Ag}^+ \) and \( \mathrm{PO}_4^{3-} \). For (c) \( \mathrm{BaCO}_3 \), it dissociates into \( \mathrm{Ba}^{2+} \) and \( \mathrm{CO}_3^{2-} \). Lastly, for (d) \( \mathrm{Ca}_5(\mathrm{PO}_4)_3\mathrm{OH} \), it dissociates into \( 5\mathrm{Ca}^{2+} \), \( 3\mathrm{PO}_4^{3-} \), and \( \mathrm{OH}^- \).

- Write the Dissociation Equations

Based on the dissociation:- (a) \( \mathrm{Ca(OH)}_2 \rightarrow \mathrm{Ca}^{2+} + 2\mathrm{OH}^- \)- (b) \( \mathrm{Ag}_3\mathrm{PO}_4 \rightarrow 3\mathrm{Ag}^+ + \mathrm{PO}_4^{3-} \)- (c) \( \mathrm{BaCO}_3 \rightarrow \mathrm{Ba}^{2+} + \mathrm{CO}_3^{2-} \)- (d) \( \mathrm{Ca}_5(\mathrm{PO}_4)_3\mathrm{OH} \rightarrow 5\mathrm{Ca}^{2+} + 3\mathrm{PO}_4^{3-} + \mathrm{OH}^- \)

- Write the General Form of \( K_{sp} \) Expression

The general form of the solubility product constant \( K_{sp} \) is a product of the molar concentrations of the ions, each raised to the power of its coefficient in the balanced equation.

- Formulate the \( K_{sp} \) Expression for Each Compound

Using the dissociation equations:- (a) \( K_{sp} = [\mathrm{Ca}^{2+}][\mathrm{OH}^-]^2 \)- (b) \( K_{sp} = [\mathrm{Ag}^+]^3[\mathrm{PO}_4^{3-}] \)- (c) \( K_{sp} = [\mathrm{Ba}^{2+}][\mathrm{CO}_3^{2-}] \)- (d) \( K_{sp} = [\mathrm{Ca}^{2+}]^5[\mathrm{PO}_4^{3-}]^3[\mathrm{OH}^-] \)

- Verify Each Expression

Make sure the powers of each ion concentration in each \( K_{sp} \) expression correspond to their stoichiometric coefficients in their original dissociation reactions. This ensures that each equilibrium expression accurately reflects the stoichiometry of the dissolution.

Key concepts

Solubility Product Constant

The solubility product constant, commonly represented as \( K_{sp} \), is a crucial concept in chemistry, particularly when dealing with sparingly soluble ionic compounds. It provides a quantitative measure of the solubility of a compound under specific conditions.

When a sparingly soluble salt dissolves in water, it reaches a dynamic equilibrium where the rate of dissolution equals the rate of precipitation. The \( K_{sp} \) is the equilibrium constant for this reaction, involving the solid and its dissolved ions in solution.

- It is calculated by multiplying the molar concentrations of the dissolved ions, each raised to the power of their stoichiometric coefficients in the balanced equation.
- This constant is useful for predicting whether a precipitate will form in a solution. If the ion product exceeds \( K_{sp} \), the solution is supersaturated, leading to precipitation.

Understanding the solubility product constant is fundamental for mastering concepts in chemical equilibria and predicting outcomes in solution chemistry.

Dissociation Reactions

The process of dissociation is a key aspect in understanding how ionic compounds interact with water. When these compounds dissolve, they separate into individual ions, a process that is represented by dissociation equations.

For example, calcium hydroxide \( \mathrm{Ca(OH)}_2 \) dissociates into one calcium ion \( \mathrm{Ca}^{2+} \) and two hydroxide ions \( \mathrm{OH}^- \). This equilibrium in a solution can be represented as:
\[ \mathrm{Ca(OH)}_2 (s) \rightarrow \mathrm{Ca}^{2+} (aq) + 2\mathrm{OH}^- (aq) \]

These equations are not just simple balances of reactants and products; they express how a compound's dissolution affects the concentration of ions in a solution. Knowing the dissociation reactions is essential for writing the \( K_{sp} \) expression accurately.

Each ion in the product side of the equation contributes to the ionic strength of a solution and is reflected in the final solubility product, demonstrating the interconnectivity between dissociation and equilibrium constants.

Ionic Compounds

Ionic compounds are composed of positively and negatively charged ions held together by electrostatic forces. They typically form crystalline solids, and their solubility in water varies greatly.

These compounds have distinct characteristics, one being the ability to dissociate into their constituent ions when dissolved in water. For example, barium carbonate \( \mathrm{BaCO}_3 \) separates into barium ions \( \mathrm{Ba}^{2+} \) and carbonate ions \( \mathrm{CO}_3^{2-} \):
\[ \mathrm{BaCO}_3 (s) \rightarrow \mathrm{Ba}^{2+} (aq) + \mathrm{CO}_3^{2-} (aq) \]

The behavior of ionic compounds in aqueous solutions is vital to predicting chemical behavior, understanding solubility, and manipulating conditions to favor crystallization or dissolution.

Recognizing the nature and dissolution of ionic compounds helps in writing solubility product expressions, which are used in calculating the extent of solubility of these compounds.

Ksp Expression

The \( K_{sp} \) expression is derived from the dissociation reaction of an ionic compound in a saturated solution. It reflects the maximum concentration of ions that can exist in equilibrium with the undissolved solid.

To construct a \( K_{sp} \) expression, follow these steps:

• Write the balanced dissociation equation for the compound.
• Identify the ions and their stoichiometric coefficients from the equation.
• Construct the \( K_{sp} \) by multiplying the concentrations of the ions, raised to the power of their coefficients.

For example, consider silver phosphate \( \mathrm{Ag}_3\mathrm{PO}_4 \):
\[ \mathrm{Ag}_3\mathrm{PO}_4 (s) \rightarrow 3\mathrm{Ag}^+ (aq) + \mathrm{PO}_4^{3-} (aq) \]
The \( K_{sp} \) is written as:
\[ K_{sp} = [\mathrm{Ag}^+]^3 [\mathrm{PO}_4^{3-}] \]

The \( K_{sp} \) expression is instrumental for predicting whether a solution will result in precipitation given certain ion concentrations.