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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: $$\begin{aligned} \mathrm{PbCl}_{2}(\mathrm{s}) & \rightleftharpoons \mathrm{Pb}^{2+}(\mathrm{aq})+2 \mathrm{Cl}^{-}(\mathrm{aq}) & & K_{c}=1.7 \times 10^{-5} \\ \mathrm{PbF}_{2}(\mathrm{s}) & \rightleftharpoons \mathrm{Pb}^{2+}(\mathrm{aq})+2 \mathrm{F}^{-}(\mathrm{aq}) & & K_{\mathrm{c}}=3.7 \times 10^{-8} \end{aligned}$$

Short Answer

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

Step by step solution

01

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 \(3.7 \times 10^{-8}\).
02

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.
03

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Consider the following equilibrium: \(\operatorname{COBr}_{2}(g) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{Br}_{2}(\mathrm{g}) \quad K_{\mathrm{c}}=0.190\) at \(73^{\circ} \mathrm{C}\) (a) \(A\) 0.50 mol sample of \(\operatorname{COBr}_{2}\) is transferred to a 9.50-L. flask and heated until equilibrium is attained. Calculate the equilibrium concentrations of each species. (b) The volume of the container is decreased to 4.5 L and the system allowed to return to equilibrium. Calculate the new equilibrium concentrations. (Hint: The calculation will be easier if you view this as a new problem with 0.5 mol of \(\mathrm{COBr}_{2}\) transferred to a 4.5 -L flask. (c) What is the effect of decreasing the container volume from 9.50 L. to 4.50 L?

\(K_{c}=5.6 \times 10^{-12}\) at \(500 \mathrm{K}\) for the dissociation of iodine molecules to iodine atoms. $$ \mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{I}(\mathrm{g})$$ A mixture has \(\left|\mathrm{I}_{2}\right|=0.020 \mathrm{mol} / \mathrm{L}\) and \(|\mathrm{I}|=\) \(2.0 \times 10^{-8} \mathrm{mol} / \mathrm{L} .\) Is the reaction at equilibrium (at \(500 \mathrm{K}\) )? If not, which way must the reaction proceed to reach equilibrium?

The dissociation of calcium carbonate has an equilibrium constant of \(K_{\mathrm{p}}=1.16\) at \(800^{\circ} \mathrm{C}\) $$ \mathrm{CaCO}_{3}(\mathrm{s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) $$ (a) What is \(K_{c}\) for the reaction? (b) If you place \(22.5 \mathrm{g}\) of \(\mathrm{CaCO}_{3}\) in a 9.56 -L container at \(800^{\circ} \mathrm{C},\) what is the pressure of \(\mathrm{CO}_{2}\) in the container? (c) What percentage of the original 22.5 -g sample of \(\mathrm{CaCO}_{3}\) remains undecomposed at equilibrium?

Assume 3.60 mol of ammonia is placed in a 2.00 - \(L\) vessel and allowed to decompose to the elements at \(723 \mathrm{K}\) $$ 2 \mathrm{NH}_{3}(\mathrm{g}) \rightleftharpoons \mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) $$ If the experimental value of \(K_{r}\) is 6.3 for this reaction at the temperature in the reactor, calculate the equilibrium concentration of each reagent. What is the total pressure in the flask?

The reaction $$ 2 \mathrm{NO}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) $$ has an equilibrium constant, \(K_{c}\) of 170 at \(25^{\circ} \mathrm{C}\). If \(2.0 \times 10^{-3}\) mol of \(\mathrm{NO}_{2}\) is present in a \(10 .\) -L. flask along with \(1.5 \times 10^{-3}\) mol of \(\mathrm{N}_{2} \mathrm{O}_{4}\), is the system at equilibrium? If it is not at equilibrium, does the concentration of \(\mathrm{NO}_{2}\) increase or decrease as the system proceeds to equilibrium?

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