Question

The solubility of \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}\) in a \(0.20-M\) 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 \(3.52 \times 10^{-14}\).

Step-by-step solution

Write the balanced chemical equation

First, we need to write down the balanced equation for the dissociation of cerium iodate in aqueous solution: \[Ce(IO_3)_3(s) \rightleftharpoons Ce^{3+}(aq) + 3IO^−_3(aq)\]

Determine the equilibrium concentrations of Ce3+ and IO3- in the solution

Given the solubility of Ce(IO3)3 is \(4.4 \times 10^{-8} mol/L\), we can determine the equilibrium concentration changes for Ce3+ and IO3- ions in the solution: For every mole of Ce(IO3)3 that dissolves, one mole of Ce3+ ions and three moles of IO3- ions are formed in the solution. The initial concentration of KIO3 in the solution is 0.20 M, which also represents the initial IO3- ion concentration. At equilibrium: \[ [Ce^{3+}] = 4.4 \times 10^{-8} M\] \[ [IO^−_3] = [IO^−_3]_{initial} + 3 \times [Ce^{3+}]\] \[ [IO^−_3] = 0.20 M + 3 \times 4.4 \times 10^{-8} M\] Now, we can calculate the equilibrium concentration of IO3- ions: \[ [IO^−_3] = 0.20 M + 3 \times 4.4 \times 10^{-8} M = 0.20 M + 1.32 \times 10^{-7} M \approx 0.20 M\] Since the solubility of Ce(IO3)3 is very low, the change in IO3- ion concentration due to its dissociation can be considered negligible, and hence the equilibrium concentration of IO3- remains approximately constant at 0.20 M.

Calculate the solubility product constant (Ksp)

Using the equilibrium concentrations of Ce3+ and IO3- ions, we can now calculate the solubility product constant (Ksp) for Ce(IO3)3: \[K_{sp} = [Ce^{3+}] [IO^−_3]^3\] Plugging in the equilibrium concentrations: \[K_{sp} = (4.4 \times 10^{-8} M) (0.20 M)^3\] Now, calculate the Ksp: \[K_{sp} = (4.4 \times 10^{-8} M) (0.20 M)^3 = 3.52 \times 10^{-14}\] Therefore, the solubility product constant (Ksp) for Ce(IO3)3 is approximately \(3.52 \times 10^{-14}\).

Key concepts

Equilibrium Concentrations

Equilibrium concentrations refer to the amounts of substances present in a solution at equilibrium. When a solute like cerium iodate (\(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}\)) dissolves in water, it dissociates into ions until the rate of dissolution equals the rate of precipitation. At this point, the solution reaches equilibrium.
This means that the concentration of ions in the solution becomes constant. In the exercise we've discussed, the solubility of cerium iodate is given as (4.4 \times 10^{-8} \, \text{mol/L}), which helps us to calculate the equilibrium concentrations of the resulting ions.
For cerium iodate:

• The equilibrium concentration of \(\mathrm{Ce^{3+}}\) ions is directly equal to the solubility: \([\mathrm{Ce^{3+}}] = 4.4 \times 10^{-8} \, \text{M}\).
• The concentration of iodate ions \([\mathrm{IO_3^-}]\) needs to account for the initial iodate concentration and the change due to dissociation.

The resulting equilibrium concentration can often be simplified if the solubility is much smaller compared to the initial concentration, allowing us to approximate more easily.

Dissociation Equation

A dissociation equation represents how a compound splits into its ions in a solution. For ionic compounds like cerium iodate \(\mathrm{Ce(IO_3)_3}\), dissociation is key to understanding how it behaves in water.
The dissociation of (\mathrm{Ce(IO_3)_3}) can be represented as:\[Ce(IO_3)_3(s) \rightleftharpoons Ce^{3+}(aq) + 3IO^-_3(aq)\]
This equation shows that when a single molecule of cerium iodate dissolves, it breaks down into one cerium ion (\mathrm{Ce^{3+}}) and three iodate ions (\mathrm{IO_3^-}).
Understanding the dissociation process is crucial because it lays the groundwork for determining solubility and calculating equilibrium concentrations and the solubility product constant (Ksp).
During calculations, we take into account how the lettered coefficients (e.g., the '3' in front of (\mathrm{IO_3^-})) indicate the ratio of ions produced, which helps in setting up expressions for Ksp calculations.

Chemical Equation

Chemical equations are fundamental in chemistry, representing the transformation of reactants to products. In the context of solubility and equilibrium, balanced chemical equations like the dissociation we used earlier are incredibly essential.
For our case, the dissociation of cerium iodate is:\[Ce(IO_3)_3(s) \rightleftharpoons Ce^{3+}(aq) + 3IO^-_3(aq)\]
This equation tells us what species are involved in the reaction and the mole ratios between them.
Key components of a chemical equation include:

• Reactants and products: Substances started with and derived from the equation.
• Coefficients: Numbers placed in front of formulas to balance atoms, charges, and help derive stoichiometric relationships.
• States of matter: Indicated by symbols (s, aq) to show solid, aqueous, etc.

These equations serve as a blueprint, allowing us to understand reactions, predict products, and calculate constants like Ksp by establishing relationships between different ionic species in the solution.