Question

A solution contains \(2.0 \times 10^{-3} \mathrm{M} \mathrm{Ce}^{3+}\) and \(1.0 \times 10^{-2} \mathrm{M} \mathrm{IO}_{3}^{3-}\). Will \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}(s)\) precipitate? \(\left[K_{\mathrm{sp}}\right.\) for \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}\) is \(3.2 \times 10^{-10}\).]

Short answer

No, Ce(IO₃)₃ will not precipitate because the ion product Q (\(8.0 \times 10^{-15}\)) is less than the solubility product constant Kₛₚ (\(3.2 \times 10^{-10}\)).

Step-by-step solution

Write down the ion product formula for Ce(IO₃)₃

The ion product (Q) for Ce(IO₃)₃ is given by the formula: Q = [Ce³⁺]³[IO₃⁻]³

Substitute the given concentrations of ions into the formula to find Q

We are given the concentrations of Ce³⁺ and IO₃⁻: [Ce³⁺] = 2.0 x 10⁻³ M [IO₃⁻] = 1.0 x 10⁻² M Substitute these values into the ion product formula: Q = (2.0 x 10⁻³)³(1.0 x 10⁻²)³

Calculate Q

Simplify the expression to find the value of Q: Q = (8.0 x 10⁻⁹)(1.0 x 10⁻⁶) = 8.0 x 10⁻¹⁵

Compare Q to Kₛₚ

Now we will compare the value of Q (8.0 x 10⁻¹⁵) to the given Kₛₚ value (3.2 x 10⁻¹⁰) to determine if precipitation will occur. Since Q < Kₛₚ, the solution is unsaturated and Ce(IO₃)₃ will not precipitate.

Key concepts

Ion Product

The ion product (IP) is an essential concept in understanding solubility and precipitation reactions. It refers to the product of the molar concentrations of the ions that would form in a solution if a salt was to fully disassociate. These concentrations are raised to the power of their stoichiometric coefficients from the balanced chemical equation. When dealing with a solution that contains ions, such as cerium(III) and iodate ions, we can write the ion product expression for a salt like Ce(IO₃)₃(s).

Let's say you have a solution with certain concentrations of Ce³⁺ and IO₃⁻ ions. To calculate the ion product, we simply take the concentrations of these ions in the solution, raise them to the power corresponding to their respective coefficients in the balanced equation, and then multiply them together. For Ce(IO₃)₃, the ion product would be expressed as IP = [Ce³⁺]³[IO₃⁻]³.

Calculating the ion product helps us predict whether a precipitate will form. If the ion product exceeds the solubility product constant (Kₛₚ) of the salt, a precipitate will form. If the ion product is less than Kₛₚ, the solution is unsaturated and no precipitate will form, as it's the case with our workout example. Determining the ion product is crucial in laboratory settings and environmental science, where controlling precipitation is often necessary.

Solubility Equilibrium

Solubility equilibrium is a particular type of dynamic equilibrium that exists when a chemical compound in the solid state is in chemical equilibrium with a solution of that compound. The solubility equilibrium is determined by a constant, known as the solubility product constant (Kₛₚ). This constant is the product of the ions' concentrations from the dissolved solid, each raised to the power of its coefficient in the balanced equation.

In simple terms, this equilibrium describes how much of a substance can be dissolved in a solvent at a given temperature. When a salt like Ce(IO₃)₃ dissolves, it dissociates into Ce³⁺ and IO₃⁻ ions. When the maximum amount of the salt has dissolved, any additional solid will remain undissolved, and the system reaches a state of dynamic equilibrium, where the rate of dissolving and the rate of precipitation are equal.

The value of Kₛₚ is specific for each substance at a particular temperature. If the system is disturbed, such as by changing the concentration of one of the ions or the temperature, the equilibrium will shift to re-establish itself according to Le Chatelier's Principle. This could involve the forming of a precipitate or the dissolution of more solid into the solution.

Precipitation Reactions

Precipitation reactions occur when cations and anions that form a compound come together in a solution and the compound formed has a lower solubility than the individual ions in the solution. The result is that a solid forms, or precipitates, out of solution. This is commonly observed in the formation of sparingly soluble salts.

These reactions are vital in various fields such as chemistry, environmental science, and industrial processes. They can be used for removing unwanted ions from a solution, purifying substances, and in qualitative chemical analysis.

The key to predicting a precipitation reaction lies in knowing the solubility rules and the solubility product constants of the compounds involved. When two ionic solutions are mixed, if the calculated ion product is larger than the Kₛₚ of the resulting compound, a precipitate will form as the ions exceed their maximum solubility and begin to aggregate into solid form. However, if the ion product is smaller than the solubility product constant, no precipitate will form and all ions remain dissolved, maintaining a clear solution like the example given earlier. Understanding these reactions is fundamental in preventing or encouraging the formation of precipitates in various chemical applications.