Problem 52 - General Chemistry: Principles and Modern Applications

Question

Use thermodynamic data from Appendix D to calculate values of \(K_{\mathrm{sp}}\) for the following sparingly soluble solutes: (a) \(\operatorname{AgBr} ;\) (b) \(\operatorname{CaSO}_{4} ;\) (c) \(\operatorname{Fe}(\text { OH })_{3}\). [Hint: Begin by writing solubility equilibrium expressions.

Step-by-Step Solution

Answer

To calculate the \(K_{sp}\) for \(AgBr\), \(CaSO_4\) and \(Fe(OH)_3\), write the solubility equilibrium expressions for them, and calculate \(\Delta G^\circ\) using Gibbs Free energy of formation values from Appendix D. Substitute \(\Delta G^\circ\) into the relationship equation with \(K_{sp}\) to find their values. The values would depend on the specific \(\Delta G_f^\circ\) values given in the Appendix D.

Step 1: Write Solubility Equilibrium Expressions

For a general solute \(AB\), the dissociation can be represented as \( AB \leftrightarrow A^+ + B^-\).\nHence, for each solute:\n(a) \(AgBr \leftrightarrow Ag^+ + Br^-\)\n(b) \(CaSO_4 \leftrightarrow Ca^{2+}+ SO_4^{2-}\)\n(c) \(Fe(OH)_3 \leftrightarrow Fe^{3+}+ 3OH^-\)

Step 2: Express \(K_{sp}\) in terms of Activities

The solubility product expression is the product of the activities of the ions raised to their respective stoichiometric coefficients. Hence,\n(a) \(K_{sp, AgBr} = a_{Ag^+} \cdot a_{Br^-} \)\n(b) \(K_{sp, CaSO_4} = a_{Ca^{2+}} \cdot a_{SO_4^{2-}}\)\n(c) \(K_{sp, Fe(OH)_3} = a_{Fe^{3+}} \cdot (a_{OH^-})^3 \). At infinite dilution, the activity of a species can be replaced with its concentration.

Step 3: Use Thermodynamics to Calculate \(K_{sp}\)

The relationship between \(\Delta G^\circ\) and \(K_{sp}\) is given by the equation \( \Delta G^\circ = -RT \ln K_{sp} \). Re-arranging this equation gives \( K_{sp} = \exp(-\Delta G^\circ/RT) \). Here, \(R\) is the gas constant (8.314 J/mol.K) and \(T\) is the absolute temperature (normally 298.15 K). Appendix D provides the standard Gibbs Free energy of formation, \(\Delta G_f^\circ\) for different species. The \(\Delta G^\circ\) for the dissociation reaction is the sum of \(\Delta G_f^\circ\) of products minus \(\Delta G_f^\circ\) of reactants. This value can be substituted into the equation above to calculate \(K_{sp}\) for each solute.

Key Concepts

Solubility Product Constant

The solubility product constant, often denoted as \(K_{sp}\), is an equilibrium constant specifically designed to describe the solubility of sparingly soluble ionic compounds. In essence, it provides a quantitative measure of how much of an ionic solid can dissolve in water, resulting in a saturated solution.
This constant helps predict whether a precipitate will form when two solutions are mixed.
The larger the \(K_{sp}\) value, the more soluble the compound is in solution.To determine the \(K_{sp}\), you begin by writing the solubility equilibrium expression for the compound of interest. For instance:

• For silver bromide, the dissolution is represented as: \(AgBr \leftrightarrow Ag^+ + Br^-\).
• This yields the expression: \(K_{sp, AgBr} = [Ag^+][Br^-]\).

Here, the concentration of ions in a saturated solution is used to calculate \(K_{sp}\). At equilibrium, the product of these ion concentrations, each raised to the power of their stoichiometric coefficient, equals \(K_{sp}\).
Understanding \(K_{sp}\) is vital in predicting dissolving behavior and potential reactions in a solution.

Gibbs Free Energy

Gibbs free energy, symbolized as \(\Delta G \), is a thermodynamic potential used to compute the maximum amount of work a thermodynamic system can perform at constant temperature and pressure. It acts as an indicator to predict which chemical reactions can occur spontaneously.

When \(\Delta G < 0\), the process is spontaneous because the system can perform work. Conversely, when \(\Delta G > 0\), the system requires energy input. When \(\Delta G = 0\), the system is at equilibrium. In the context of solubility, the relationship between \(\Delta G^\circ\) and \(K_{sp}\) is expressed through \[\Delta G^\circ = -RT \ln K_{sp}.\]

• \(\Delta G^\circ\) for a reaction can be derived using the free energies of formation (\(\Delta G_f^\circ\)) for the products and reactants.
• The Gibbs free energy change \(\Delta G^\circ\) equals the sum of the product's \(\Delta G_f^\circ\) values subtracted by the sum of the reactant's \(\Delta G_f^\circ\) values.

By rearranging the equation above, we can compute \(K_{sp}\): \[K_{sp} = \exp\left(-\frac{\Delta G^\circ}{RT}\right).\]Each component of this equation fits together to illustrate how energy changes in a chemical system can determine reaction feasibility.

Chemical Equilibrium

Chemical equilibrium refers to the state in a chemical reaction where the concentrations of reactants and products remain unchanged over time, achieving a balance in the reaction. This state is characterized by the rates of the forward and reverse reactions becoming equal.In a dynamic equilibrium, while reactions continue to occur in both directions, the overall composition of the system does not change. The stability of this equilibrium can be disrupted by changes in concentration, temperature, or pressure, as described by Le Chatelier's Principle.
When it comes to the solubility of sparingly soluble solutes:

• The dissolution of a solid compound in a solvent until the ionic species in solution achieve equilibrium is representative of a chemical equilibrium.
• For the solute silver bromide (\(AgBr\)), when it dissolves, an equilibrium is reached: \(AgBr \leftrightarrow Ag^+ + Br^-\).
• This equilibrium state can be expressed using the \(K_{sp}\), which provides crucial insights into the relationship between the dissolved ions and the undissolved solid.

Equilibrium concepts are pivotal for understanding reactions in closed systems, where equilibrium constants like \(K_{sp}\) further quantify the extent of reactions that involve ionic solutes.