Question

Consider a beaker containing a saturated solution of \(\mathrm{PbI}_{2}\) in equilibrium with undissolved \(\mathrm{PbI}_{2}(s) .\) Now solid KI is added to this solution. (a) Will the amount of solid \(\mathrm{PbI}_{2}\) at the bottom of the beaker increase, decrease, or remain the same? (b) Will the concentration of \(\mathrm{Pb}^{2+}\) ions in solution increase or decrease? (c) Will the concentration of I \(^{-}\) ions in solution increase or decrease?

Short answer

(a) The amount of solid \(PbI_{2}\) at the bottom of the beaker will increase, because the equilibrium shifts to the left due to the increased I⁻ ion concentration from the added KI. (b) The concentration of \(Pb^{2+}\) ions in the solution will decrease as the equilibrium shifts to the left. (c) The concentration of I⁻ ions in the solution will initially increase due to the addition of KI but may decrease as more solid \(PbI_{2}\) forms; however, the final concentration might still be higher than the initial one.

Step-by-step solution

Write the equation for the system

The system is given by the equilibrium between solid PbI₂ and its ions in water: \[PbI_{2}(s) \rightleftharpoons Pb^{2+}(aq) + 2 I^{-}(aq)\]

Write the equation for the addition of KI

Adding solid KI to the solution implies the dissolution of KI in water: \[KI(s) \rightarrow K^{+}(aq) + I^{-}(aq)\]

Apply the principle of Le Châtelier

The principle of Le Châtelier states that when a system in equilibrium is subjected to a change, the system will adjust the positions of the equilibrium to counteract the change. In this case, the addition of KI increases the concentration of I⁻ ions.

Analyze the effect on the amount of solid PbI2

As the concentration of I⁻ ions increases, the equilibrium for the PbI₂ system will shift to the left to counteract the change, according to Le Châtelier's principle. This means more solid PbI₂ will be formed, so (a) the amount of solid PbI₂ at the bottom of the beaker will increase.

Analyze the effect on the concentration of Pb²⁺ ions

Since the system is shifting to the left, due to the increased I⁻ ion concentration from the added KI, the concentration of Pb²⁺ ions will decrease to maintain the equilibrium. So, (b) the concentration of Pb²⁺ ions in the solution will decrease.

Analyze the effect on the concentration of I⁻ ions

Initially, the concentration of I⁻ ions will increase due to the addition of KI. However, as the system tries to counteract this change, the equilibrium shifts to the left, promoting the formation of solid PbI₂ by consuming I⁻ ions. Therefore, (c) the concentration of I⁻ ions in the solution will increase initially, and then likely decrease as more solid PbI₂ forms but may still be higher than the initial concentration.

Key concepts

Chemical Equilibrium

Chemical equilibrium occurs when a reversible reaction in a closed system reaches a point where the rates of the forward and reverse reactions are equal. At this stage, the concentrations of reactants and products remain constant over time. This concept is crucial for understanding how chemical reactions behave under different conditions.
When we talk about the equilibrium between solid PbI₂ and its ions in a solution, we refer to the dynamic balance described by the equation: \[PbI_{2}(s) \rightleftharpoons Pb^{2+}(aq) + 2 I^{-}(aq)\] This shows us how solid lead(II) iodide (PbI₂) can dissolve into its constituent ions and how these ions can recombine to form the solid. The equilibrium can shift to the left or right if conditions change.
According to Le Châtelier's Principle, if a change is applied to a system at equilibrium, the system adjusts to minimize that change, restoring equilibrium at a new position.

Solubility Product

The solubility product, often represented as \(K_{sp}\), is a constant for a given compound that reflects its solubility at a particular temperature. For PbI₂, its solubility product expression is: \[K_{sp} = [Pb^{2+}][I^{-}]^2\] This equation indicates that in a saturated solution, the product of the concentrations of the ions (\(Pb^{2+}\) and \(I^{-}\)) at equilibrium will remain consistent as long as temperature does not change.
When KI is added to the equilibrium system, it affects the solubility product by increasing the concentration of \(I^{-}\).
According to Le Châtelier's Principle, the system reacts to the increased ionic concentration by shifting the equilibrium to the left, increasing the formation of PbI₂ solid to restore the equilibrium balance. As a result, the apparent solubility of PbI₂ is reduced in the solution.

Ion Concentration

Ion concentration can drastically impact the position of equilibrium in a chemical system. In a solution containing PbI₂, the concentration of its ions \(Pb^{2+}\) and \(I^{-}\) in equilibrium are crucial to maintaining balance. When excess \(I^{-}\) ions are introduced by adding KI, this changes the initial concentration balance.
The excess \(I^{-}\) initially leads to a higher ion concentration in the solution, but the system will try to restore equilibrium by shifting to the left, thus forming more solid PbI₂. Consequently, the concentration of\(Pb^{2+}\) ions decreases because fewer lead ions are available in the solution.
This behavior emphasizes the relationship between solubility and ion concentration. Understanding how ion concentration affects equilibrium enables us to predict and manipulate the solubility and reaction outcomes in chemical processes.