Question

Would the solubility of \(\mathrm{PbI}_{2}(s)\) be greater in water or in an aqueous solution of NaI? Explain your answer. (Hint: If \(\left[\mathrm{Pb}^{2+}\right] \times\left[\mathrm{I}^{-}\right]^{2}>K_{\mathrm{sp}}\) ' precipitation will occur.)

Short answer

The solubility of PbI₂ would be greater in an aqueous solution of NaI due to the common ion effect. The presence of additional iodide ions from the dissolution of NaI shifts the solubility equilibrium, allowing for a greater amount of PbI₂ to dissolve in the NaI solution compared to water.

Step-by-step solution

Write the solubility equilibrium expression for PbI₂

We know that the solubility equilibrium expression for PbI₂ can be written as: \(PbI_{2}(s) \rightleftharpoons Pb^{2+}(aq) + 2I^{-}(aq)\) The solubility product constant (Kₛₚ) expression for this reaction can be written as: \(K_{sp} = [Pb^{2+}][I^{-}]^{2}\)

Solubility of PbI₂ in water

In water, the solubility of PbI₂ will be limited by its Kₛₚ value. Let's call the solubility of PbI₂ in water as \(s_{w}\). Since the stoichiometry of the reaction is 1:1:2, we can say that: [Pb²⁺] = \(s_{w}\) [I⁻] = 2\(s_{w}\) Substituting these values into the Kₛₚ expression, we have: \(K_{sp} = s_{w} (2s_{w})^{2}\)

Solubility of PbI₂ in an aqueous solution of NaI

In an aqueous solution of NaI, the concentration of iodide ions (I⁻) is already increased due to the dissociation of NaI. Let's call the solubility of PbI₂ in NaI solution as \(s_{N}\). Since the stoichiometry of the reaction is 1:1:2, we can say that: [Pb²⁺] = \(s_{N}\) [I⁻] = 2\(s_{N}\) + [I⁻] from NaI Substituting these values into the Kₛₚ expression, we get: \(K_{sp} = s_{N} (2s_{N} + [I^{-}]_{NaI})^{2}\)

Comparing the solubility of PbI₂ in both solutions

Based on the expressions derived in steps 2 and 3, it becomes clear that the solubility of PbI₂ in an aqueous solution of NaI is higher than in water. The presence of additional iodide ions from NaI causes a shift in the solubility equilibrium, resulting in a greater ability to dissolve PbI₂ in the NaI solution. Therefore, the solubility of PbI₂ would be greater in an aqueous solution of NaI. This process is known as the common ion effect, where the presence of a common ion in a solution can affect the solubility of a sparingly soluble salt.

Key concepts

Common Ion Effect

The concept of the common ion effect plays a critical role in understanding the behavior of sparingly soluble salts in the presence of other compounds. This effect is a particular example of Le Chatelier's principle applied to solubility equilibria. It states that if a salt is dissolved in a solution which already contains one of the ions in the salt, the solubility of the salt is reduced as compared to its solubility in pure water.

To illustrate, consider the solubility of lead iodide ((PbI_{2})) in an aqueous solution of sodium iodide (NaI). The solution already contains iodide ions (I⁻), which are common to both PbI₂ and NaI. According to the common ion effect, the solubility of PbI₂ would decrease in the NaI solution because the existing iodide ions shift the equilibrium towards the solid PbI₂, reducing its tendency to dissolve.

Exercise Improvement Advice:

It's beneficial for students to clarify this concept by contrasting scenarios with and without the additional common ion. Visual aids such as graphs of solubility versus concentration of the common ion can also enhance understanding. In calculations, it's essential to carefully account for the additional common ion concentration when determining the solubility of the salt.

Chemical Equilibrium

Chemical equilibrium is a state in which the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the concentrations of reactants and products over time. This concept is crucial in the context of solubility equilibria because a saturated solution of a substance is in dynamic equilibrium with the undissolved substance.

In this state, individual ions continue to dissolve and precipitate, but their overall concentrations remain constant. The equilibrium constant (K_{sp}) for the dissolution of a sparingly soluble salt, like PbI₂, is a way to quantitatively describe this equilibrium point.

Understanding Kₛₚ:

K_{sp} (solubility product constant) is specific to each compound and is influenced by temperature. Knowing this value helps predict how much salt can dissolve in a solution before reaching saturation. It's also important for students to recognize that the value of K_{sp} does not change with the presence of a common ion, even though the salt's solubility does change.

Solubility Equilibrium

Solubility equilibrium refers to the dynamic balance between the dissolution and precipitation of a substance. The solubility product, K_{sp}, is the mathematical representation of this equilibrium in a saturated solution. For lead iodide, the expression is K_{sp} = [Pb^{2+}][I^-]^{2}, where the concentrations of the ions are at equilibrium.

For compounds with differing stoichiometry, understanding the relationship between the solubility (s) and the stoichiometric coefficients becomes crucial. For every mole of PbI₂ that dissolves, one mole of Pb²⁺ ions and two moles of I⁻ ions are produced. These relationships must be accounted for when calculating the solubility of the salt in water versus in a solution containing a common ion.

Factors Affecting Solubility:

Temperature, pressure (for gases), the presence of other substances (as seen with the common ion effect), and even the pH of the solution can influence the solubility of a substance. Students should explore these factors and consider how they can manipulate conditions to alter the solubility of a substance, both in experimental design and in practical applications like wastewater treatment.