Question

Use Le Chatelier's principle to explain why \(\mathrm{PbCl}_{2}\) is less soluble in \(0.010-\mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) than in pure water.

Short answer

\(\mathrm{Pb}^{2+}\) ions in the solution shift the equilibrium left, reducing \(\mathrm{PbCl}_2\) solubility.

Step-by-step solution

Identify the Dissolution Reaction

The dissolution of the lead(II) chloride in water can be represented by the equilibrium reaction: \[ \mathrm{PbCl}_2 (s) \rightleftharpoons \mathrm{Pb}^{2+} (aq) + 2\mathrm{Cl}^- (aq) \]

Consider Le Chatelier's Principle

Le Chatelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change and restore equilibrium.

Apply the Principle to the Solution

In a \(0.010\)-M \(\mathrm{Pb(NO_3)}_2\) solution, the concentration of \(\mathrm{Pb}^{2+}\) ions is increased compared to pure water. According to Le Chatelier's principle, the equilibrium will shift left to reduce the added \(\mathrm{Pb}^{2+}\), thereby decreasing the solubility of \(\mathrm{PbCl}_2\).

Conclude The Effect on Solubility

Since the equilibrium shifts toward the reactants (solid \(\mathrm{PbCl}_2\)) due to the added \(\mathrm{Pb}^{2+}\) ions, less \(\mathrm{PbCl}_2\) dissolves. Thus, \(\mathrm{PbCl}_2\) is less soluble in \(0.010\)-M \(\mathrm{Pb(NO_3)}_2\) than in pure water due to this shift.

Key concepts

Solubility

Solubility refers to the capacity of a substance to dissolve in a solvent, forming a homogeneous solution. It's an essential concept in chemistry because it tells us how much of a material can dissolve in a particular amount of solvent at a certain temperature. When a salt dissolves in water, it separates into ions. The solubility product constant, or Ksp, is a useful value that helps us understand the solubility of ionic compounds like lead(II) chloride ( When we talk about lower solubility, it means fewer ions from the substance, such as

• Temperature
• Common ion effect
• Nature of the solute and solvent

In the case of lead(II) chloride, increasing the concentration of lead ions ( This decrease in solubility occurs because the added ions promote the formation of more solid lead(II) chloride, shifting the balance back towards the solid state.

Equilibrium

Chemical equilibrium occurs when the rates of the forward and reverse reactions are equal, leading to no net change in the concentration of reactants and products over time. For a dissolution reaction such as the dissolving of lead(II) chloride in water, the equilibrium can be represented as follows: \[ \mathrm{PbCl}_2 (s) \rightleftharpoons \mathrm{Pb}^{2+} (aq) + 2\mathrm{Cl}^- (aq) \] This indicates that the dissolution of lead(II) chloride is a dynamic state where solid crystals dissolve to form ions in solution, while ions in the solution can also recombine to form the solid.The position of equilibrium can be affected by several factors, which include:

• Changes in concentration of reactants or products
• Temperature variations
• Addition of catalysts (though catalysts do not shift the equilibrium, they affect how quickly it is reached)

Le Chatelier's principle gives us a straightforward method to predict changes in the system. If we add more This is the principle at play in our scenario with lead(II) chloride solubility in different solutions.

Dissolution Reaction

A dissolution reaction involves a solid solute mixing with a solvent to form a solution. For ionic compounds like lead(II) chloride, this process involves the solute separating into its ions, which then become surrounded by solvent molecules.The dissolution of lead(II) chloride can be represented by its equilibrium equation: \[ \mathrm{PbCl}_2 (s) \rightleftharpoons \mathrm{Pb}^{2+} (aq) + 2\mathrm{Cl}^- (aq) \] Here, the solid lead(II) chloride breaks apart into lead ions and chloride ions upon dissolving. This system is dynamic, meaning ions are continually dissolving and precipitating at the same rate in equilibrium.In saturated solutions, the system has reached its maximum capacity for dissolution under the given conditions. Any disturbance, like adding more lead ions as in the introduction of Understanding dissolution reactions is crucial, as they explain how substances dissolve, precipitate, and interact with their environment, contributing to the chemistry of everyday processes.