Question

The solubility of \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}\) in a \(0.20-\mathrm{M} \mathrm{KIO}_{3}\) solution is \(4.4 \times 10^{-8} \mathrm{~mol} / \mathrm{L}\). Calculate \(K_{\mathrm{sp}}\) for \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3} .\)

Short answer

The Kₛₚ for Ce(IO₃)₃ is approximately 3.52 × 10⁻¹⁰.

Step-by-step solution

Write the balanced chemical equation

Write the balanced chemical equation for the dissolution of cerium(III) iodate in water: Ce(IO₃)₃(s) ↔ Ce³⁺(aq) + 3IO₃⁻(aq)

Set up an ICE table

Set up an ICE table to represent the initial concentration, change in concentration, and equilibrium concentration of the ions in this reaction: \( \begin{array}{c|c|c|c} & [Ce^{3+}] & [IO_{3}^-] \\ \hline Initial & 0 & 0.20 \\ Change & +x & +3x \\ Equilibrium& x & 0.20+3x \end{array} \) Where x represents the solubility, we know that the equilibrium concentration of Ce³⁺ is \(4.4 \times 10^{-8} \mathrm{~mol} / \mathrm{L}\), so for IO₃⁻, the final concentration would be \(0.20 + 3 \times (4.4 \times 10^{-8})\).

Calculate equilibrium concentrations

Calculate the equilibrium concentration of IO₃⁻: [IO₃⁻] = 0.20 + 3 * (4.4 × 10⁻⁸) [IO₃⁻] = 0.20 + 1.32 × 10⁻⁷ [IO₃⁻] ≈ 0.200 Since the calculated value is very close to 0.2 M, we can safely assume that the equilibrium concentration of IO₃⁻ will remain approximately at 0.2 M.

Calculate Ksp

Now we can use the equilibrium concentrations of Ce³⁺ and IO₃⁻ to calculate the Kₛₚ for Ce(IO₃)₃. The expression for Kₛₚ is: Kₛₚ = [Ce³⁺] * [IO₃⁻]³ Plug in the equilibrium concentrations: Kₛₚ = (4.4 × 10⁻⁸) * (0.200)³ Kₛₚ = (4.4 × 10⁻⁸) * 0.008 Kₛₚ ≈ 3.52 × 10⁻¹⁰ The Kₛₚ for Ce(IO₃)₃ is approximately 3.52 × 10⁻¹⁰.

Key concepts

Ksp Calculation

Calculating the solubility product constant, commonly denoted as Ksp, is a fundamental exercise in chemistry when dealing with the solubility of ionic compounds in solution. It provides a quantitative measure of how much of a sparingly soluble compound will dissolve in water to reach a state of dynamic equilibrium.

The Ksp expression is derived from the equilibrium constant for the dissociation reaction of a solid ionic compound into its ions. For example, for a general sparingly soluble compound AB that dissociates as follows: AB(s) ↔ A⁺(aq) + B⁻(aq), the Ksp expression would be: Ksp = [A⁺] * [B⁻], where [A⁺] and [B⁻] are the molar concentrations of the ions at equilibrium. When calculating Ksp, it's important to remember that pure solids are not included in the equilibrium expression, as their activity is considered to be constant.

The existing solution provided the Ksp by considering the dissociation of Ce(IO₃)₃ into Ce³⁺ and IO₃⁻ ions and determining their equilibrium concentrations. By multiplying these concentrations and raising to the power respective to each ion's stoichiometric coefficient in the balanced equation, the Ksp value is obtained, which in this case is approximately 3.52 × 10⁻¹⁰.

ICE Table

An ICE table, standing for Initial, Change, Equilibrium, is a tool commonly used to organize data and solve equilibrium problems in chemistry. It helps visualize the changes in concentration of reactants and products as a reaction progresses from its initial state to equilibrium.

Here's a brief overview of how to set up an ICE table:

• Initial: Write down the initial concentrations of each species in the reaction.
• Change: Indicate the changes in concentrations for each species as the reaction progresses towards equilibrium. These are usually represented by a variable, often 'x', and correspond to the stoichiometry of the balanced equation.
• Equilibrium: Combine the initial concentrations with the changes to determine the equilibrium concentrations.

In the provided exercise, the concentration of Ce³⁺ started at zero because, as a solid, it doesn't contribute to the concentration in the solution. The change in concentration is represented by 'x' for Ce³⁺ and '3x' for IO₃⁻ based on the stoichiometry of the dissolution reaction. At equilibrium, 'x' corresponds to the known solubility of Ce(IO₃)₃, and '3x' is added to the initial concentration of IO₃⁻ from KIO₃ present in the solution.

Equilibrium Concentration

Equilibrium concentrations refer to the concentrations of reactants and products in a chemical reaction when the rate of the forward reaction equals the rate of the reverse reaction, and no further change in concentration of reactants and products occurs.

To determine the equilibrium concentrations in a solubility context, one can use the stoichiometry of the dissociation equation and the Ksp value. Sometimes, when dealing with common ion effects or in other complicated systems, approximations are made to simplify calculations. For instance, if the contribution of a common ion to the overall concentration is very small compared with an ion from a strong electrolyte, the contribution can be neglected, as seen in the exercise where the extra concentration added by the soluble Ce(IO₃)₃ was so minor that the equilibrium concentration of IO₃⁻ was approximated to remain at 0.2 M.

Solubility in Common Ion Solutions

The solubility of a compound in a solution that already contains one of the ions it would produce upon dissolving is affected by the common ion effect. According to Le Chatelier's principle, if a reaction at equilibrium is subjected to a change in concentration, temperature, volume, or pressure, the equilibrium shifts to counteract the imposed change and a new equilibrium is established.

When a common ion is present, the system will shift towards the left, meaning that less of the solid will dissolve than if the common ion weren't present. This is because the ions' product will likely exceed the Ksp sooner given the common ion's initial concentration. In the exercise, the solubility of Ce(IO₃)₃ is decreased due to the presence of IO₃⁻ ions from KIO₃. Thus, the solubility is represented by a smaller 'x' value than would be expected if the solution did not have a common ion.

This concept is particularly important in understanding how to control the precipitation of ions in various chemical processes and has significant implications in both industrial applications and biological systems.