Problem 70 - Chemistry and Chemical Reactivity

Question

Neither \(\mathrm{PbCl}_{2}\) nor \(\mathrm{PbF}_{2}\) is appreciably soluble in water. If solid \(\mathrm{PbCl}_{2}\) and solid \(\mathrm{PbF}_{2}\) are placed in equal amounts of water in separate beakers, in which beaker is the concentration of \(\mathrm{Pb}^{2+}\) greater? Equilibrium constants for these solids dissolving in water are as follows: $$\begin{aligned} \mathrm{PbCl}_{2}(\mathrm{s}) & \rightleftharpoons \mathrm{Pb}^{2+}(\mathrm{aq})+2 \mathrm{Cl}^{-}(\mathrm{aq}) & & K_{c}=1.7 \times 10^{-5} \\ \mathrm{PbF}_{2}(\mathrm{s}) & \rightleftharpoons \mathrm{Pb}^{2+}(\mathrm{aq})+2 \mathrm{F}^{-}(\mathrm{aq}) & & K_{\mathrm{c}}=3.7 \times 10^{-8} \end{aligned}$$

Step-by-Step Solution

Answer

The concentration of \(\mathrm{Pb}^{2+}\) is greater in the \(\mathrm{PbCl}_2\) beaker.

Step 1: Understanding the Reaction Equilibria

Both \(\mathrm{PbCl}_2\) and \(\mathrm{PbF}_2\) dissolve in water reaching equilibrium, forming \(\mathrm{Pb}^{2+}\) ions. The reactions are: \(\mathrm{PbCl}_{2} \rightleftharpoons \mathrm{Pb}^{2+} + 2 \mathrm{Cl}^{-}\) and \(\mathrm{PbF}_{2} \rightleftharpoons \mathrm{Pb}^{2+} + 2 \mathrm{F}^{-}\). The equilibrium constants \(K_c\) indicate the extent of solubility. \(K_c\) for \(\mathrm{PbCl}_2\) is \(1.7 \times 10^{-5}\) and for \(\mathrm{PbF}_2\) is \(3.7 \times 10^{-8}\).

Step 2: Comparing Equilibrium Constants

The equilibrium constant \(K_c\) reflects how much of the solid dissolves. Larger \(K_c\) suggests greater solubility. Here, \(K_c\) for \(\mathrm{PbCl}_2\) is \(1.7 \times 10^{-5}\), which is much larger than \(3.7 \times 10^{-8}\) for \(\mathrm{PbF}_2\). Thus, \(\mathrm{PbCl}_2\) has higher solubility, leading to more \(\mathrm{Pb}^{2+}\) ions.

Step 3: Drawing Conclusions from \(K_c\) Values

Since \(\mathrm{PbCl}_2\) has a greater \(K_c\) value than \(\mathrm{PbF}_2\), the concentration of \(\mathrm{Pb}^{2+}\) ions will be higher in the beaker containing \(\mathrm{PbCl}_2\). Therefore, the concentration of \(\mathrm{Pb}^{2+}\) is greater in the \(\mathrm{PbCl}_2\) beaker compared to the \(\mathrm{PbF}_2\) beaker.

Key Concepts

Equilibrium Constant

In chemical reactions, the equilibrium constant, denoted as \( K_c \), helps determine how far a reaction will go towards products once it reaches equilibrium. Essentially, it provides a snapshot of a reaction's balance between reactants and products in a given system.

For a reaction at equilibrium, the equilibrium constant is calculated as a ratio of the concentrations of products to reactants, each raised to the power of their stoichiometric coefficients. For example, for the dissolution of \( \mathrm{PbCl}_2 \) in water, the equilibrium expression is:
\[ K_c = \frac{[\mathrm{Pb}^{2+}][\mathrm{Cl}^{-}]^2}{[\mathrm{PbCl}_2]} \]
Since \( \mathrm{PbCl}_2 \) and \( \mathrm{PbF}_2 \) are solids, their concentration does not change, simplifying the equilibrium constant expressions. Thus, \( K_c \) effectively measures the extent to which these solids can produce ions in solution.

Solubility Product

The solubility product constant, or \( K_{sp} \), is a specialized type of equilibrium constant used specifically for sparingly soluble salts. It describes the level at which a solute dissolves in solution to form ions. A larger \( K_{sp} \) corresponds to a more soluble compound.

For \( \mathrm{PbCl}_2 \), the solubility product expression is:
\[ K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{Cl}^{-}]^2 \]
And for \( \mathrm{PbF}_2 \), it's expressed as:
\[ K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{F}^{-}]^2 \]
With \( K_c \) values of \( 1.7 \times 10^{-5} \) for \( \mathrm{PbCl}_2 \) and \( 3.7 \times 10^{-8} \) for \( \mathrm{PbF}_2 \), it is clear that \( \mathrm{PbCl}_2 \) has a greater ability to dissolve in water, forming more \( \mathrm{Pb}^{2+} \) ions, compared to \( \mathrm{PbF}_2 \). This means the solubility, represented by the \( K_{sp} \), indicates a direct relation between dissolution capability and the amount of ions in solution.

Chemical Equilibria

Chemical equilibria refer to the state of a reaction in which the forward and reverse reactions occur at the same rate, resulting in constant concentrations of reactants and products over time. The concept is crucial in understanding reaction dynamics and solubility in solutions.

When \( \mathrm{PbCl}_2 \) and \( \mathrm{PbF}_2 \) are dissolved in water, they reach a point where the concentration of ions produced from these salts remains constant, achieving chemical equilibrium. In this state, the dissolution process (solid to ions) and the precipitation process (ions to solid) occur at the same rate.

This balance signifies no net change in the concentration of \( \mathrm{Pb}^{2+} \), \( \mathrm{Cl}^{-} \) from \( \mathrm{PbCl}_2 \), and \( \mathrm{F}^{-} \) from \( \mathrm{PbF}_2 \). The relevant \( K_c \) values guide us in determining which reaction produces more ions in solution, allowing us to infer where the equilibrium position lies and ultimately leading to a clearer understanding of how much solid will dissolve when equilibrium is achieved.