Question

The molar solubility of \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}\) in a \(0.10 \mathrm{M} \mathrm{NaIO}_{3}\) solution is \(2.4 \times 10^{-11} \mathrm{~mol} / \mathrm{L}\). What is \(K_{\mathrm{sp}}\) for \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2} ?\)

Short answer

The value of \(K_{sp}\) for \(Pb(IO_{3})_{2}\) is calculated to be \(2.4 \times 10^{-13}\).

Step-by-step solution

Establish the Dissociation Equations and Ksp Expression

First, let's consider the dissociation of \(Pb(IO_{3})_{2}\): \[Pb(IO_{3})_{2}(s) \leftrightarrow Pb^{2+}(aq) + 2IO_{3}^{-}(aq)\] The \(K_{sp}\) expression for this dissociation would be: \[K_{sp} = [Pb^{2+}][IO_{3}^{-}]^{2}\]

Consider the Effects of the Common Ion

Sodium iodate (\(NaIO_{3}\)) also dissociates to produce \(IO_{3}^{-}\) ions: \[NaIO_{3}(aq) \rightarrow Na^{+}(aq) + IO_{3}^{-}(aq)\] As it supplies \(0.10\, M\) of \(IO_{3}^{-}\) ions to the solution, the equilibrium concentration of \(IO_{3}^{-}\) equals the initial concentration of \(NaIO_{3}\) plus the molar solubility of \(Pb(IO_{3})_{2}\). But the molar solubility is so small that it can be ignored in comparison to 0.10. Therefore, [\(IO_{3}^{-}\)] is approximately 0.10 M.

Substitute into Ksp Expression

The molar solubility of \(Pb(IO_{3})_{2}\) is given as \(2.4 \times 10^{-11} \, mol/L\). This is also the concentration \( [Pb^{2+}]\). Hence, these values can be substituted into the \(K_{sp}\) expression to solve for \(K_{sp}\): \[K_{sp} = (2.4 \times 10^{-11}) * (0.10)^2\]

Key concepts

Ksp (Solubility Product Constant)

The solubility product constant, abbreviated as Ksp, is a useful value in chemistry that helps predict the solubility of ionic compounds in a solution. It is a type of equilibrium constant that applies to the dissolution of sparingly soluble salts. When an ionic compound dissociates in water, it separates into its constituent ions. The Ksp is the product of these ions' molar concentrations, each raised to the power of its stoichiometric coefficient.

For the compound Pb(IO₃)₂, the dissolution reaction can be represented as:
[Pb(IO₃)₂(s) ↔ Pb²⁺(aq) + 2IO₃⁻(aq)]
Thus, the Ksp expression is:
[K_sp = [Pb²⁺][IO₃⁻]²]

This expression indicates that the Ksp will increase if the concentration of ions in the solution increases. The value of Ksp is unique for each ionic compound at a given temperature and is typically determined experimentally. In practical applications, it can be used to estimate the maximum solubility of a compound under specific conditions, which is particularly important in industries where precipitation reactions are crucial.

Common Ion Effect

The common ion effect occurs when a compound dissolves in a solution that already contains one of the ions that the compound can produce. This has a significant impact on the solubility of the compound. The common ion effect is tied to Le Chatelier's Principle, which tells us that if a chemical system at equilibrium experiences a change in concentration, temperature, volume, or pressure, the system will shift its equilibrium position to counteract the effect of the applied change.

Let's consider sodium iodate (NaIO₃) being dissolved in a solution that already contains iodate ions (IO₃⁻). Since the iodate ion is a product of the dissociation of both NaIO₃ and Pb(IO₃)₂, adding NaIO₃ to the solution will increase the concentration of IO₃⁻ ions. According to Le Chatelier's Principle, this will shift the equilibrium of the Pb(IO₃)₂ dissolution to the left, decreasing its solubility. In simpler terms, the presence of a common ion suppresses the dissociation of an ionic compound, which is a critical factor to consider when predicting solubility in a solution and must be accounted for in solubility calculations.

Dissociation of Ionic Compounds

Dissociation is the process by which an ionic compound breaks apart into its constituent ions in a solution. The degree of dissociation is the extent to which the compound separates into ions and is influenced by factors such as the nature of the compound, temperature, and the presence of other ions in the solution.

In the context of Pb(IO₃)₂, the dissociation in water can be described by the following equation:
[Pb(IO₃)₂(s) ↔ Pb²⁺(aq) + 2IO₃⁻(aq)]
Here, Pb(IO₃)₂ is a sparingly soluble compound, meaning it will not dissociate completely in water. The process reaches an equilibrium point where the rate of the forward reaction (dissolving) equals the rate of the reverse reaction (precipitating). At this point, the concentrations of the ions in the solution remain constant. Understanding the dissociation behavior of ionic compounds is critical for predicting solubility, controlling precipitations, and planning syntheses in both laboratory and industrial settings.