Question

(a) Why is the concentration of undissolved solid not explicitly included in the expression for the solubility-product constant? (b) Write the expression for the solubility-product constant for cach of the following strong electrolytes: \(\mathrm{AgI}_{1} \mathrm{SrSO}_{4}, \mathrm{Fe}(\mathrm{OH})_{2}\), and \(\mathrm{Hg}_{2} \mathrm{Br}_{1}\). ?

Short answer

(a) The concentration of undissolved solid is not included in the solubility-product constant expression because it remains constant as long as some solid is present and does not affect the equilibrium of the dissolved ions. (b) The expressions for the solubility-product constants of the given strong electrolytes are: 1. For \(\mathrm{AgI}\): \(K_{sp} = [\mathrm{Ag}^+][\mathrm{I}^-]\) 2. For \(\mathrm{SrSO}_4\): \(K_{sp} = [\mathrm{Sr}^{2+}][\mathrm{SO}_4^{2-}]\) 3. For \(\mathrm{Fe(OH)}_{2}\): \(K_{sp} = [\mathrm{Fe}^{2+}][\mathrm{OH}^-]^2\) 4. For \(\mathrm{Hg}_2\mathrm{Br}_1\): \(K_{sp} = [\mathrm{Hg}_{2}^{2+}][\mathrm{Br}^-]\)

Step-by-step solution

Part (a): Understanding Solubility-Product Constant Expression

The solubility-product constant (Ksp) refers to the equilibrium constant of a slightly soluble ionic compound dissolved in water. When a sparingly soluble salt dissolves in water, it establishes an equilibrium between the dissolved ions and the solid. For example, consider the following equilibrium for a compound AB: \[AB_{(s)} \rightleftharpoons A^+_{(aq)} + B^-_{(aq)}\] The equilibrium constant (K) for this reaction would be: \[K = \frac{[A^+][B^-]}{[AB]}\] However, in the case of solubility-product constant (Ksp), the concentration of undissolved solid ([AB]) is not included in the expression. The reason for excluding the solid concentration is that it remains constant as long as some solid is present. Since the undissolved solid is not participating in the reaction, its concentration remains constant, and it won't affect the equilibrium of the dissolved ions. Therefore, the expression for Ksp becomes: \[K_{sp} = [A^+][B^-]\]

Part (b): Writing Solubility-Product Constant Expressions for Given Strong Electrolytes

Now, we will write the solubility-product constant expressions for the given strong electrolytes: \(\mathrm{AgI}, \mathrm{SrSO}_{4}, \mathrm{Fe}(\mathrm{OH})_{2}\), and \(\mathrm{Hg}_{2} \mathrm{Br}_{1}\). 1. For \(\mathrm{AgI}\): The dissolution equilibrium can be represented as: \[\mathrm{AgI}_{(s)} \rightleftharpoons \mathrm{Ag}^+_{(aq)} + \mathrm{I}^-_{(aq)}\] The expression for Ksp is: \[K_{sp} = [\mathrm{Ag}^+][\mathrm{I}^-]\] 2. For \(\mathrm{SrSO}_4\): The dissolution equilibrium can be represented as: \[\mathrm{SrSO}_{4(s)} \rightleftharpoons \mathrm{Sr}^{2+}_{(aq)} + \mathrm{SO}_4^{2-}_{(aq)}\] The expression for Ksp is: \[K_{sp} = [\mathrm{Sr}^{2+}][\mathrm{SO}_4^{2-}]\] 3. For \(\mathrm{Fe(OH)}_{2}\): The dissolution equilibrium can be represented as: \[\mathrm{Fe(OH)}_{2(s)} \rightleftharpoons \mathrm{Fe}^{2+}_{(aq)} + 2\mathrm{OH}^-_{(aq)}\] The expression for Ksp is: \[K_{sp} = [\mathrm{Fe}^{2+}][\mathrm{OH}^-]^2\] 4. For \(\mathrm{Hg}_2\mathrm{Br}_1\): The dissolution equilibrium can be represented as: \[\mathrm{Hg}_{2\mathrm{Br}_{(s)}} \rightleftharpoons \mathrm{Hg}_2^{2+}_{(aq)} + \mathrm{Br}^-_{(aq)}\] The expression for Ksp is: \[K_{sp} = [\mathrm{Hg}_{2}^{2+}][\mathrm{Br}^-]\]

Key concepts

Equilibrium Constant

The equilibrium constant, symbolized as K, is a fundamental concept in chemistry that quantifies the balance between products and reactants at equilibrium in a chemical reaction. It is a measure of the extent to which reactants are converted to products, describing the ratio of the concentration of products to the concentration of reactants, each raised to the power of their stoichiometric coefficients.

An equilibrium constant has a direct connection to the thermodynamics of a reaction: if K is high, the reaction favors the formation of products, whereas if K is low, the reaction favors the reactants. This concept holds true for all types of chemical equilibrium, including the dissolution of ionic solids in water, which leads us into the realm of the solubility-product constant (Ksp).

Dissolution Equilibrium

Dissolution equilibrium refers to the dynamic equilibrium that is established when a solute (usually a solid) dissolves in a solvent to form a solution. At this point, the rate at which the solute dissolves equals the rate at which the dissolved particles recombine to form the solid, resulting in no net change in concentration over time.

For ionic compounds, this process can be visualized as the solid compound disassociating into its respective ions. The equilibrium is represented by an equation showing the solid reactant on one side and its dissociated ions on the other. The equilibrium expression (like those shown in the solution for various ionic compounds) illustrates how the dissolved ions contribute to the solubility-product constant.

Ionic Compound Solubility

The solubility of an ionic compound refers to the extent to which it can dissolve in a solvent to form a solution. This solubility is affected by factors like temperature, the nature of the solvent and solute, and the presence of common ions.

When discussing solubility in terms of Ksp, we focus on sparingly soluble ionic compounds. These substances have low solubility and establish an equilibrium between the undissolved solid and the dissolved ions in solution. The Ksp provides a quantitative measure of this solubility, which can be used to calculate the maximum concentration of ions that can exist in a solution before precipitation occurs.

Concentration of Undissolved Solid

The concentration of undissolved solid in a saturated solution remains constant because the solid and its saturated solution form an equilibrium that does not change with the amount of solid present. As a result, this concentration is not factored into the Ksp expression.

The solid can be thought of as 'reservoir' for the ions; as long as there is some solid present and the system is at equilibrium, the concentration of the ions in solution will be consistent. Therefore, the Ksp expression involves only the concentrations of the dissolved ions, since it's these concentrations that change with varying conditions and are important for calculating solubility and predicting whether a precipitate will form.