Problem 94 - Chemistry and Chemical Reactivity

Question

There are many ionic compounds that dissolve in water to a very small extent. One example is lead(II) chloride. When it dissolves an equilibrium is established between the solid salt and its component ions. Suppose you stir some solid \(\mathrm{PbCl}_{2}\) into water. Explain how you would prove that the compound dissolves but to a small extent? Is the dissolving process product-favored or reactant-favored? $$\mathrm{PbCl}_{2}(\mathrm{s}) \rightleftharpoons \mathrm{Pb}^{2+}(\mathrm{aq})+2 \mathrm{Cl}^{-}(\mathrm{aq})$$

Step-by-Step Solution

Answer

\(\mathrm{PbCl}_{2}\) dissolves slightly, and the process is reactant-favored due to a low \(K_{sp}\).

Step 1: Understand the Dissolution Equation

The equilibrium expression for the dissolution of lead(II) chloride in water is given as \(\mathrm{PbCl}_{2}(\mathrm{s}) \rightleftharpoons \mathrm{Pb}^{2+}(\mathrm{aq})+2 \mathrm{Cl}^{-}(\mathrm{aq})\). This means that solid \(\mathrm{PbCl}_{2}\) dissociates into its ions \(\mathrm{Pb}^{2+}\) and \(\mathrm{Cl}^{-}\). Understanding this process is crucial to analyze whether the dissolution process is favored towards products or reactants.

Step 2: Establish the Equilibrium Constant Expression

To determine the favorability of the reaction, we look at the solubility product constant, \(K_{sp}\). The expression for \(K_{sp}\) is derived from the equilibrium: \[K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{Cl}^{-}]^2\]. A small \(K_{sp}\) value indicates limited solubility, meaning the solid doesn't dissolve significantly.

Step 3: Measure Ion Concentrations in Solution

To demonstrate that \(\mathrm{PbCl}_{2}\) indeed dissolves, measure the concentration of \(\mathrm{Pb}^{2+}\) or \(\mathrm{Cl}^{-}\) ions in the solution using techniques like spectrophotometry or ion-selective electrodes. Even small ion concentrations confirm some dissolution.

Step 4: Compare with Pure Water

Compare the measurements from the solution of \(\mathrm{PbCl}_{2}\) to pure water. In pure water, no \(\mathrm{Pb}^{2+}\) and \(\mathrm{Cl}^{-}\) ions should be present. Any detectable concentration of these ions in the \(\mathrm{PbCl}_{2}\) solution indicates dissolution.

Step 5: Determine Favorability

If the \(K_{sp}\) is much smaller than 1, equilibrium favors the reactants (solid form) over the products (ions in solution). This indicates that the dissolution process is reactant-favored.

Key Concepts

Equilibrium Constant

In chemistry, an equilibrium constant helps us understand how a reaction behaves when it reaches a state of balance. For ionic compounds like lead(II) chloride (\(\mathrm{PbCl}_{2}\)), this constant is crucial to analyzing whether the system favors the formation of ions or remains mostly as an undissolved solid.
The reaction for \(\mathrm{PbCl}_{2}\) dissolving in water is represented by: \(\mathrm{PbCl}_{2}(\mathrm{s}) \rightleftharpoons \mathrm{Pb}^{2+}(\mathrm{aq}) + 2 \mathrm{Cl}^{-}(\mathrm{aq})\).
The double arrow (\(\rightleftharpoons\)) shows that the reaction can go both ways—dissolving into ions and forming back into a solid.

• The equilibrium constant, denoted as \(K_{sp}\) for solubility product constant, helps quantify this balance.
• A small value of \(K_{sp}\) suggests that the solid doesn't dissolve much, meaning the equilibrium is reactant-favored.

Solubility Product Constant

The solubility product constant (\(K_{sp}\)) is a special type of equilibrium constant used for sparingly soluble salts. It provides insight into how much of the compound can dissolve in water before reaching equilibrium.

To find the \(K_{sp}\) for lead(II) chloride, we use:\[K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{Cl}^{-}]^2\]

• This equation is derived from the concentrations of the ions in solution.
• In our specific case: one \(\mathrm{Pb}^{2+}\) ion and two \(\mathrm{Cl}^{-}\) ions, hence the square of \([\mathrm{Cl}^{-}]\).

A low \(K_{sp}\) value indicates that few ions are present in water, confirming limited solubility. Thus, \(K_{sp}\) is essential for deciding whether the dissolution is product or reactant-favored.

Dissolution Process

The dissolution process for ionic compounds like \(\mathrm{PbCl}_{2}\) involves a dynamic balance between the solid and its ions in solution. While some of the compound's molecules dissociate into ions, many remain as a solid.
The chemical equation \(\mathrm{PbCl}_{2}(\mathrm{s}) \rightleftharpoons \mathrm{Pb}^{2+}(\mathrm{aq}) + 2 \mathrm{Cl}^{-}(\mathrm{aq})\) is a vital illustration.

• Solid \(\mathrm{PbCl}_{2}\) dissociates into \(\mathrm{Pb}^{2+}\) and \(\mathrm{Cl}^{-}\) ions.
• The balance reached between the solid and ions is called equilibrium.

This process clarifies that even if a compound is sparingly soluble, it does dissolve to some extent, contributing ions to the solution.

Ion Concentration Measurement

To confirm the dissolution of ionic compounds, measuring the ion concentration in the solution is essential. Techniques like spectrophotometry or ion-selective electrodes can detect even minute concentrations of ions.

When comparing the solution's ion concentration to pure water:

• In pure water, there are no \(\mathrm{Pb}^{2+}\) and \(\mathrm{Cl}^{-}\) ions.
• Even a small concentration of these ions in a \(\mathrm{PbCl}_{2}\) solution indicates that dissolution has occurred.

These measurements help affirm the limited solubility of the compound, supporting conclusions about the equilibrium state. Understanding these ion concentrations is a practical way to visualize how much a compound like \(\mathrm{PbCl}_{2}\) dissolves in water.