Question

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

Short answer

(a) The concentration of undissolved solid is not included in the solubility product constant expression because it remains constant in a saturated solution, and its concentration does not change or affect the equilibriums of the dissolved ions. (b) Solubility-product constant expressions for given strong electrolytes: 1. AgI: \(K_{sp} = [\mathrm{Ag}^{+}][\mathrm{I}^{-}]\) 2. SrSO4: \(K_{sp} = [\mathrm{Sr}^{2+}][\mathrm{SO}_{4}^{2-}]\) 3. Fe(OH)2: \(K_{sp} = [\mathrm{Fe}^{2+}][\mathrm{OH}^{-}]^{2}\) 4. Hg2Br2: \(K_{sp} = [\mathrm{Hg}^{+}]^{2}[\mathrm{Br}^{-}]^{2}\)

Step-by-step solution

(a) Explanation of the absence of undissolved solid in solubility product constant expression

In a saturated solution, the solid and the dissolved ions are in equilibrium. The concentration of undissolved solid does not change and remains constant as long as the solution is saturated. Therefore, it is not included in the solubility product expression. When a chemical reaction reaches equilibrium, we only consider the concentrations of the products and reactants, which change during the course of the reaction. The solid does not actively participate in the chemical reaction equilibrium, and its concentration remains unchanged. Thus, it does not have an effect on the solubility product constant.

(b) Solubility-product constant expressions for given strong electrolytes

1. Silver iodide (AgI): Dissociation equation: \(\mathrm{AgI} \rightleftharpoons \mathrm{Ag}^{+} + \mathrm{I}^{-}\) Solubility-product constant expression: \(K_{sp} = [\mathrm{Ag}^{+}][\mathrm{I}^{-}]\) 2. Strontium sulfate (SrSO4): Dissociation equation: \(\mathrm{SrSO}_{4} \rightleftharpoons \mathrm{Sr}^{2+} + \mathrm{SO}_{4}^{2-}\) Solubility-product constant expression: \(K_{sp} = [\mathrm{Sr}^{2+}][\mathrm{SO}_{4}^{2-}]\) 3. Iron(II) hydroxide (Fe(OH)2): Dissociation equation: \(\mathrm{Fe}(\mathrm{OH})_{2} \rightleftharpoons \mathrm{Fe}^{2+} + 2 \mathrm{OH}^{-}\) Solubility-product constant expression: \(K_{sp} = [\mathrm{Fe}^{2+}][\mathrm{OH}^{-}]^{2}\) 4. Mercury(I) bromide (Hg2Br2): Dissociation equation: \(\mathrm{Hg}_{2} \mathrm{Br}_{2} \rightleftharpoons 2 \mathrm{Hg}^{+} + 2 \mathrm{Br}^{-}\) Solubility-product constant expression: \(K_{sp} = [\mathrm{Hg}^{+}]^{2}[\mathrm{Br}^{-}]^{2}\)

Key concepts

Chemical Equilibrium

Chemical equilibrium is a critical concept in understanding solubility product constants. It occurs when a chemical reaction has reached a point where the rate of the forward reaction equals the rate of the reverse reaction. This means the concentrations of reactants and products remain constant over time.

In the context of solubility, equilibrium involves the dissolved ions and undissolved solid in a saturated solution. At equilibrium, there is a dynamic balance between the ions dissolving into the solution and the solid precipitating out. Because of this ongoing exchange, any measurable change is seen in the dissolved ions rather than the solid. The equilibrium state allows us to consider only the concentrations of the ions in solution often without needing to account for the solid itself. It is why the solubility product constant, or \(K_{sp}\), focuses solely on ion concentrations.

Electrolyte Dissociation

Electrolyte dissociation is the process by which an ionic compound splits into its component ions in solution. Strong electrolytes dissociate completely, separating into individual cations and anions as they dissolve.

In solubility calculations, the equation representing this dissociation is crucial. It gives insight into how many ions result from the dissociation process, which directly influences the solubility product expression. For instance, when silver iodide (AgI) dissociates, it forms Ag⁺ and I⁻ ions. Knowing this, we can write the expression for the \(K_{sp}\) of AgI to include the concentrations of Ag⁺ and I⁻.

This dissociation is vital because it shows how the solid interacts with the solution environment, highlighting the presence of free-moving ions critical for chemical reactions and electrical conductivity.

Saturated Solutions

A saturated solution is one in which the maximum amount of solute is dissolved under the given conditions of temperature and pressure. In such solutions, any additional solute will remain undissolved, maintaining equilibrium between the dissolved ions and undissolved solid.

The concept of saturation is fundamental in solubility product constant expressions. When a solution is saturated, it informs us that the solution has reached its capacity for dissolving a solute, and is thereby at equilibrium.

• At this point, the rate at which ions leave the solid equals the rate at which they redeposit back into the solid.
• This balance ensures no net change in the amount of dissolved solute.
• Saturated solutions provide the basis for calculating \(K_{sp}\), reflecting solely the concentration of the constituent ions.

Ksp Expressions

The solubility product constant, \(K_{sp}\), is a mathematical representation of the equilibrium between an insoluble compound and its ions in a saturated solution. It quantifies the solubility by showing the product of the molar concentrations of the dissociated ions, each raised to the power of their stoichiometric coefficients from the balanced equation.

For example, consider silver iodide (AgI), which has a simple dissociation formula: \(\text{AgI} \rightleftharpoons \text{Ag}^{+} + \text{I}^{-}\). The \(K_{sp}\) expression for AgI is entirely dependent on the concentrations of Ag⁺ and I⁻, given by \(K_{sp} = [\text{Ag}^{+}][\text{I}^{-}]\).

Understanding how to express \(K_{sp}\) allows chemists to predict the solubility of compounds.

• It is especially important for determining the conditions necessary for precipitation or dissolution.
• The smaller the \(K_{sp}\), the less soluble the compound.
• The \(K_{sp}\) values vary with temperature, which can shift equilibrium conditions.

Therefore, \(K_{sp}\) provides crucial insight into the chemical behavior of sparingly soluble salts.