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 solubility product constant, Ksp, for Ce(IO3)3 is approximately $6.364\times {10}^{-21}$.

Step-by-step solution

Write the balanced chemical equation for Ce(IO3)3 dissolution

Cе(IO3)3(s) ⇌ Ce^3+(aq) + 3 IO3^-(aq)

Write down the given concentrations and solubilities

Initial concentration of KIO3: 0.20 M Solubility of Cе(IO3)3 in KIO3 solution: 4.4 x 10^-8 mol/L

Set up an ICE table

ICE table for the reaction: | | Ce^3+(aq) | IO3^-(aq) | | ------ | --------- | --------- | | Initial| 0 | 0.20 | | Change | +s | +3s | | Equilibrium| s | 0.20 + 3s | The solubility of Ce(IO3)3 is given as 4.4 x 10^-8 mol/L. So, s = 4.4 x 10^-8 M.

Find the equilibrium concentrations

At equilibrium, the concentrations will be: Ce^3+(aq): 4.4 × 10^-8 M IO3^-(aq): 0.20 + 3 × (4.4 × 10^-8) = 0.20 + 1.32 × 10^-7 = 0.200000132 M

Write the Ksp expression for the balanced equation

For the balanced equation, Ksp = [Ce^3+][IO3^-]^3

Calculate the Ksp value for Ce(IO3)3

Plug the equilibrium concentrations into the Ksp expression: Ksp = (4.4 × 10^-8)(0.200000132)^3 Ksp = 6.364 × 10^-21 The Ksp for Ce(IO3)3 is approximately 6.364 × 10^-21.

Key concepts

Solubility Product Constant

The solubility product constant, often represented as $K_{sp}$, is a critical concept in understanding the dissolution of sparingly soluble compounds. Essentially, $K_{sp}$ is the product of the molar concentrations of the ions in a saturated solution, each raised to the power of their stoichiometric coefficients in the balanced chemical equation.

In the example of cerium iodate, $\text{Ce}({\text{IO}}_3{)}_3$, this would translate to a dissolution process that produces ${\text{Ce}}^{3+}$ and ${\text{IO}}_3^{-}$ ions. Here, $$K_{sp}=[{\text{Ce}}^{3+}][{\text{IO}}_3^{-}{]}^3$$.

Understanding $K_{sp}$ is crucial because it allows us to calculate the extent of solubility of a compound under given conditions, such as specific ionic strengths or common ion effects. For instance, you might find that adding a common ion to the solution will lower the solubility of the compound, reflected in the $K_{sp}$ calculation.

Dissolution Equilibrium

Dissolution equilibrium describes the state at which a solid solute dissolves in a solvent at a fixed rate equal to the rate at which it precipitates out of the solution. This means that there is no net change in the amounts of solute and solvent as the dissolution occurs.

Let's consider the dissolution of $\text{Ce}({\text{IO}}_3{)}_3$ in a solution. The equation $\text{Ce}({\text{IO}}_3{)}_3(s)\rightleftharpoons {\text{Ce}}^{3+}(aq)+3{\text{IO}}_3^{-}(aq)$ shows the system reaching equilibrium. At this point, the concentrations of the ions in the solution remain constant.

This equilibrium is vital in calculating the $K_{sp}$. Understanding the concept of equilibrium helps in comprehending how changes in concentration, temperature, and pressure can affect the dissolution process and the perceived solubility of a compound.

ICE Table

An ICE table (Initial, Change, Equilibrium) is a simple yet powerful tool used to track the changes in concentrations of reactants and products as a chemical reaction approaches equilibrium.

Consider the dissolution of $\text{Ce}({\text{IO}}_3{)}_3$. An ICE table might look like this:

• Initial: Concentration of ${\text{Ce}}^{3+}$ = 0, ${\text{IO}}_3^{-}$ = 0.20 M
• Change: The solubility $s$ adds $+s$ and $+3s$ to the concentrations of ${\text{Ce}}^{3+}$ and ${\text{IO}}_3^{-}$ respectively.
• Equilibrium: At equilibrium, $[{\text{Ce}}^{3+}]=s$ and $[{\text{IO}}_3^{-}]=0.20+3s$.

Using this setup, it's easy to visualize how changes in initial conditions can affect the final equilibrium state, and it simplifies the calculation of $K_{sp}$.

Concentration Calculation

Calculating concentrations at equilibrium is another pivotal skill in chemistry, especially when dealing with solubility products. Once the ICE table is organized, you can determine the equilibrium concentrations of ions produced by the dissolution process.

For $\text{Ce}({\text{IO}}_3{)}_3$, the solubility is given as $4.4\times {10}^{-8}$ mol/L. Setting $s=4.4\times {10}^{-8}$, this implies a concentration of $[{\text{Ce}}^{3+}]=s=4.4\times {10}^{-8}$ M. For ${\text{IO}}_3^{-}$, the equilibrium concentration becomes $0.20+3\times 4.4\times {10}^{-8}=0.200000132$ M.

Once you have these values, substitute them into the $K_{sp}$ expression to find $$K_{sp}=[{\text{Ce}}^{3+}][{\text{IO}}_3^{-}{]}^3$$, which in this case amounts to $6.364\times {10}^{-21}$. This showcases how precise and small changes at microscopic levels influence the macroscopic behavior of solutions, enriching your understanding of chemical interactions.