Question

Consider a beaker containing a saturated solution of \(\mathrm{PbI}_{2}\) in equilibrium with undissolved \(\mathrm{PbI}_{2}(s)\). Now solid \(\mathrm{K} \mathrm{I}\) 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 \(\mathrm{PbI}_{2}\) at the bottom of the beaker will increase due to the common ion effect and Le Chatelier's principle, as the \(I^-\) concentration increases from the dissolution of \(\mathrm{KI}\). (b) The concentration of \(\mathrm{Pb}^{2+}\) ions in the solution will decrease as more \(\mathrm{PbI}_{2}\) precipitates. (c) The concentration of \(I^-\) ions in the solution will increase despite the precipitation of \(\mathrm{PbI}_{2}\), due to the dissociation of \(\mathrm{KI}\).

Step-by-step solution

Write the solubility equilibrium for \(\mathrm{PbI}_{2}\)

The solubility equilibrium for \(\mathrm{PbI}_{2}\) can be expressed as: \[\mathrm{PbI}_{2} \rightleftharpoons \mathrm{Pb}^{2+} + 2\mathrm{I}^-\] The \(K_{sp}\) expression for this equilibrium is: \[K_{sp} = [\mathrm{Pb}^{2+}] [I^-]^2\] Remember that for this particular situation, the solution is saturated.

Adding solid \(\mathrm{KI}\) to the beaker

When solid \(\mathrm{KI}\) is added to the saturated solution, it will dissolve and dissociate into \(\mathrm{K}^+\) and \(I^-\) ions. The equilibrium will be affected by the increase in concentration of \(I^-\) ions from the dissociation of \(\mathrm{KI}\). The reaction for the dissolution of \(\mathrm{KI}\) is: \[\mathrm{KI} \rightarrow \mathrm{K}^+ + I^-\] Since \(\mathrm{PbI}_{2}\) and \(\mathrm{KI}\) have a common ion (\(I^-\)), the common ion effect comes into play.

Analyzing the effect on the solubility of \(\mathrm{PbI}_{2}\)

The increase in the concentration of \(I^-\) ions in the solution causes the equilibrium to shift according to Le Chatelier's principle. In this case, the equilibrium shifts to the left (reactants), which means the solubility of \(\mathrm{PbI}_{2}\) decreases. As a result, more solid \(\mathrm{PbI}_{2}\) will precipitate, increasing the amount of solid \(\mathrm{PbI}_{2}\) at the bottom of the beaker. (a) The amount of solid \(\mathrm{PbI}_{2}\) at the bottom of the beaker will increase.

Analyzing the change in the \(\mathrm{Pb}^{2+}\) concentration

Since the equilibrium shifts to the left, the concentration of \(\mathrm{Pb}^{2+}\) will decrease as more \(\mathrm{PbI}_{2}\) precipitates. (b) The concentration of \(\mathrm{Pb}^{2+}\) ions in the solution will decrease.

Analyzing the change in the \(I^-\) concentration

As the concentration of \(I^-\) ions increases due to the dissolution of \(\mathrm{KI}\), the solubility of \(\mathrm{PbI}_{2}\) decreases. However, the overall concentration of the \(I^-\) ions in the solution will still increase because the dissociation of \(\mathrm{KI}\) contributes more \(I^-\) ions than the decrease in \(I^-\) ions due to the precipitation of \(\mathrm{PbI}_{2}\). (c) The concentration of \(I^-\) ions in the solution will increase.

Key concepts

Ksp

In the realm of chemistry, understanding the concept of solubility equilibrium is essential for solving many problems. One important measure in this context is the solubility product constant, known as **Ksp**. This constant provides valuable information about the solubility of a compound in a given solution. For instance, let's consider lead(II) iodide, or \(\mathrm{PbI}_{2}\), which exists in equilibrium as expressed by the equation: \[\mathrm{PbI}_{2} \rightleftharpoons \mathrm{Pb}^{2+} + 2\mathrm{I}^-\]. The solubility product expression is thus given as:\[K_{sp} = [\mathrm{Pb}^{2+}] [I^-]^2\]. In a saturated solution, the product of the concentrations of the ions, each raised to the power of their respective coefficients in the balanced equation, is equal to the \(K_{sp}\). A higher \(K_{sp}\) value implies greater solubility of the compound in water, whereas a lower value indicates limited solubility. Ksp describes how much of the ionic solid can dissolve in water before the solution becomes saturated. It is a crucial factor for predicting whether a precipitate will form or dissolve under certain conditions. When more of an ion is added to the solution, the equilibrium shifts to maintain the \(K_{sp}\) value, affecting solubility as seen in various scenarios.

Le Chatelier's Principle

Le Chatelier's Principle is a fundamental concept in chemistry that helps us understand how a system at equilibrium responds to external changes. When a change in concentration, pressure, or temperature occurs, this principle dictates how the equilibrium will shift to counteract the change and re-establish balance.
Regarding the situation with \(\mathrm{PbI}_{2}\), consider the addition of \(\mathrm{KI}\) to the beaker. Le Chatelier's Principle tells us that adding \(I^-\) ions from the dissociation of \(\mathrm{KI}\) will disturb the equilibrium of \(\mathrm{PbI}_{2}\). Consequently, the equilibrium will shift to the left, toward the solid form \(\mathrm{PbI}_{2}(s)\), attempting to "reduce" the added \(I^-\) ions by forming more solid.
This shift reduces the concentration of \(\mathrm{Pb}^{2+}\) ions because more \(\mathrm{PbI}_{2}\) precipitates out of the solution. Hence, in line with Le Chatelier's Principle, the system reacts in a way that counteracts the initial change, leading to more solid \(\mathrm{PbI}_{2}\) at the bottom of the beaker and less \(\mathrm{Pb}^{2+}\) in solution.

Common Ion Effect

The common ion effect is another important principle to understand when dealing with solubility equilibria. It explains how the addition of an ion common to two solutes can decrease the solubility of a solute in solution. This happens because the presence of a common ion will shift the equilibrium position of a solubility equilibrium reaction.
Let's revisit our beaker containing \(\mathrm{PbI}_{2}\). By introducing \(\mathrm{KI}\), we add \(I^-\) ions to the solution, which are common to the equilibrium of \(\mathrm{PbI}_{2}\). The addition of these extra \(I^-\) ions causes the equilibrium to shift leftward, as per Le Chatelier's Principle. This results in more \(\mathrm{PbI}_{2}\) precipitating out of the solution, thereby decreasing its solubility.
But why does this happen? The common ion effect can be understood simply: adding more of an ion already involved in the equilibrium causes the system to try and "remove" some of the ions to return to equilibrium. This results in an increase in the formation of the solid phase from the dissolved ions, increasing the amount of precipitate. Understanding the common ion effect is crucial for anticipating how various reactions will behave in mixed ion solutions.