Question

For \(\mathrm{PbCl}_{3}^{-}, K_{\text {form }}=2.5 \times 10^{1}\). If a solution containing this complex ion is diluted with water, \(\mathrm{PbCl}_{2}\) precipitates. Write the equations for the equilibria involved and use them together with Le Châtelier's principle to explain how this happens.

Short answer

Upon dilution, the dissolution equilibrium of \(\mathrm{PbCl}_{3}^{-}\) shifts to the right, increasing the concentration of \(\mathrm{PbCl}_{2}\), which precipitates once it exceeds its solubility product.

Step-by-step solution

Write the Dissociation Equation for the Complex Ion

To analyze the equilibria, first write the dissociation equation for the complex ion \(\mathrm{PbCl}_{3}^{-}\) in water: \[\mathrm{PbCl}_{3}^{-}(aq) \rightleftharpoons \mathrm{PbCl}_{2}(s) + \mathrm{Cl}^{-}(aq)\] This equation shows that the complex ion can dissociate into solid lead(II) chloride and a chloride ion.

Apply Le Châtelier's Principle on Dilution

When the solution is diluted with water, according to Le Châtelier's principle, the equilibrium will shift to counteract the change. In this case, the dilution increases the volume, effectively reducing the ion concentrations. To restore equilibrium, the reaction will shift to the right, favoring the production of \(\mathrm{PbCl}_{2}\) and \(\mathrm{Cl}^{-}\).

Precipitation of PbCl2

As the equilibrium shifts to the right due to dilution, the concentration of \(\mathrm{PbCl}_{2}\) increases. When the concentration of \(\mathrm{PbCl}_{2}\) in the solution exceeds its solubility product \(K_{sp}\), it will no longer be able to stay dissolved in the solution and \(\mathrm{PbCl}_{2}\) will precipitate out of the solution.

Key concepts

Dissociation Equation

Understanding the dissociation equation is crucial for studying reaction mechanisms in aqueous solutions. The dissociation equation represents the reversible breakdown of a compound into its constituents under certain conditions. For the complex ion \(\mathrm{PbCl}_3^{-}\), its dissociation in water can be represented as:\[\mathrm{PbCl}_3^{-}(aq) \rightleftharpoons \mathrm{PbCl}_2(s) + \mathrm{Cl}^{-}(aq)\]
This illustrates that the complex ion \(\mathrm{PbCl}_3^{-}\) in an aqueous solution can separate into lead(II) chloride, which is a solid, and a chloride ion dissolved in the water. Through this process, the relative concentrations of these species are dictated by the equilibrium constant associated with the reaction, which can be altered by changes in the conditions—such as dilution, as seen in our exercise.

Complex Ion Equilibrium

Complex ion equilibrium concerns itself with the balance between a complex ion in solution and its dissociated form. The formation or dissociation of complex ions like \(\mathrm{PbCl}_3^{-}\) can be influenced by changes in the system, such as concentration variations, temperature, and presence of other ions. \(K_{\text{form}}\) represents the equilibrium constant for the formation of the complex ion, and it signifies the stability of the complex ion in solution. In the given exercise, the value of \(K_{\text{form}}\) is quite high, indicating a stable complex ion under normal conditions, but when the equilibrium is disturbed by dilution, the reaction shifts to regain balance.

Precipitation Reactions

Precipitation reactions occur when certain ions in solution form an insoluble compound, resulting in a precipitate. The occurrence of such reactions is greatly dependent on the product concentrations and their solubility limits. When a solution containing a soluble complex ion like \(\mathrm{PbCl}_3^{-}\) is diluted, it can lead to the formation of an insoluble substance, in this case, \(\mathrm{PbCl}_2\). This process exemplifies a precipitation reaction wherein the solubility threshold is exceeded and the excess solid separates from the solution.

Solubility Product

The solubility product (\(K_{sp}\)) is a special type of equilibrium constant that applies to the solubility of ionic compounds. It represents the maximum amount of the compound that can dissolve in the solution at a given temperature. For compounds like \(\mathrm{PbCl}_2\), this constant is essential in predicting whether a precipitate will form under certain conditions. In the exercise, the dilution causes an increase in the relative concentration of \(\mathrm{PbCl}_2\), which exceeds its \(K_{sp}\), leading to the precipitation of \(\mathrm{PbCl}_2\) from the solution.

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. Le Châtelier's principle asserts that if an external change is applied to a system at equilibrium, the system adjusts in a way that counteracts the change. In the context of our exercise, when water is added to dilute the solution, the equilibrium shifts toward the right to produce more \(\mathrm{PbCl}_2\) and \(\mathrm{Cl}^{-}\), as explained previously, demonstrating the dynamic nature of chemical reactions at equilibrium.