Consider a beaker containing a saturated solution of \(\mathrm{Pbl}_{2}\) in equilibrium with undissolved \(\mathrm{Pbl}_{2}(s) .\) (a) If solid KI is added to this solution, 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 \(\mathrm{I}^{-}\) ions in solution increase or decrease?

Le Chatelier's principle is a key aspect of chemical equilibrium that helps predict how a system will respond to disturbances. It states that if an external change is imposed on a system at equilibrium, the system adjusts to partially counteract that change. Think of it as the system's way of 'pushing back' against the change.

For example, let's imagine a room that always wants to maintain a certain number of people inside. If more people enter, the room 'responds' by finding ways for some people to leave, maintaining its desired number. In chemistry, equilibria can be disrupted by changes in concentration of reactants or products, temperature, or pressure. The reaction will shift to either the left or right to re-establish equilibrium, reducing the impact of the change.

When KI is added to the saturated solution of PbI₂, as in the exercise, Le Chatelier's principle instructs us that the system will respond. It tells us that the reaction will shift left, causing more solid PbI₂ to form, which is exactly what we observed in the step-by-step solution provided.

The solubility product constant (Ksp) is a special type of equilibrium constant that applies to the dissolution of sparingly soluble compounds. It's a numerical value reflecting how much of a compound can dissolve in water to form a saturated solution. The Ksp expression for a compound is written by multiplying the concentrations of the ions in its dissolved state, raised to the power of their coefficients in the balanced dissolution equation.

For our PbI₂ example, Ksp is represented by \[K_{sp} = [Pb^{2+}][I^{-}]^{2}\]This equation doesn't just tell us the amount that dissolves; it's also the key to understanding what happens when equilibrium is disturbed. If any product in this equation changes (like when we add KI and increase the concentration of I⁻ ions), the Ksp value remains constant, but the concentrations of other ions must adjust accordingly, leading to precipitation or dissolution until equilibrium is re-established.

The common ion effect occurs when a compound containing an ion in common with a dissolved species is added to the solution, affecting the solubility of the original compound. This is directly tied to Le Chatelier’s principle. When more of a common ion is added, the equilibrium shifts in a direction that reduces the concentration of that ion.

In the provided exercise, when KI is added to the solution containing PbI₂, it supplies additional I⁻ ions. Because these ions are common to the dissolved PbI₂, this causes the equilibrium to shift towards the formation of more solid PbI₂, effectively decreasing the solubility of PbI₂. This shift occurs because the system is seeking to lower the concentration of I⁻ ions to maintain the Ksp constant. As a result, less Pb²⁺ remains in solution, as more is incorporated into solid PbI₂ to counteract the change.

Understanding the common ion effect is crucial not only for predicting the solubility changes but also in various practical applications, including controlling the pH in buffer solutions and purifying chemicals through recrystallization.