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Chapter 16: Problem 56

Predict what effect each of the following has on the position of the equilibrium $$ \mathrm{PbCl}_{2}(s) \rightleftharpoons \mathrm{Pb}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q) \quad \Delta H=23.4 \mathrm{~kJ} $$ (a) addition of \(1 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) solution (b) increase in temperature (c) addition of \(\mathrm{Ag}^{+}\), forming \(\mathrm{AgCl}\) (d) addition of \(1 M\) hydrochloric acid

Short Answer

Expert verified
Question: Predict the effect of the following changes on the equilibrium reaction: $$\mathrm{PbCl}_{2}(s) \rightleftharpoons \mathrm{Pb}^{2+}(a q)+2\mathrm{Cl}^{-}(a q) \quad \Delta H=23.4 \mathrm{~kJ}$$ (a) Addition of 1 M \(\mathrm{Pb}(\mathrm{NO}_{3})_{2}\) solution; (b) Increase in temperature; (c) Addition of \(\mathrm{Ag}^{+}\), forming \(\mathrm{AgCl}\) ; (d) Addition of 1 M hydrochloric acid. Answer: (a) The equilibrium will shift to the left, increasing the amount of \(\mathrm{PbCl}_{2}(s)\) and decreasing the concentrations of \(\mathrm{Pb}^{2+}(a q)\) and \(\mathrm{Cl}^{-}(a q)\); (b) The equilibrium will shift to the right, decreasing the amount of \(\mathrm{PbCl}_{2}(s)\) and increasing the concentrations of \(\mathrm{Pb}^{2+}(a q)\) and \(\mathrm{Cl}^{-}(a q)\); (c) The equilibrium will shift to the right, consuming solid \(\mathrm{PbCl}_{2}\) and producing more \(\mathrm{Pb}^{2+}(a q)\) and \(\mathrm{Cl}^{-}(a q)\) ions; (d) The equilibrium will shift to the left, increasing the amount of \(\mathrm{PbCl}_{2}(s)\) and reducing the concentrations of the \(\mathrm{Pb}^{2+}(a q)\) and \(\mathrm{Cl}^{-}(a q)\) ions.

Step by step solution

01

(a) Addition of 1 M \(\mathrm{Pb}(\mathrm{NO}_{3})_{2}\) solution

According to Le Châtelier's principle, when a system at equilibrium is subjected to a change in concentration, it will shift its equilibrium position to counteract the change. In this case, adding 1 M \(\mathrm{Pb}(\mathrm{NO}_{3})_{2}\) solution will increase the concentration of \(\mathrm{Pb}^{2+}(a q)\). To counteract this change, the equilibrium will shift to the left, resulting in the formation of more \(\mathrm{PbCl}_{2}(s)\) and a decrease in the concentrations of \(\mathrm{Pb}^{2+}(a q)\) and \(\mathrm{Cl}^{-}(a q)\).
02

(b) Increase in temperature

The given reaction has a positive \(\Delta H\), which means it is an endothermic reaction. According to Le Châtelier's principle, when an endothermic reaction at equilibrium experiences an increase in temperature, the equilibrium will shift towards the products to absorb the excess heat. Therefore, the position of the equilibrium will shift to the right, increasing the concentrations of \(\mathrm{Pb}^{2+}(a q)\) and \(\mathrm{Cl}^{-}(a q)\), while decreasing the amount of \(\mathrm{PbCl}_{2}(s)\).
03

(c) Addition of \(\mathrm{Ag}^{+}\), forming \(\mathrm{AgCl}\)

When \(\mathrm{Ag}^{+}\) ions are added to the system, they will react with the \(\mathrm{Cl}^{-}\) ions to form \(\mathrm{AgCl}\), a solid precipitate. This will cause a decrease in the concentration of \(\mathrm{Cl}^{-}(a q)\). According to Le Châtelier's principle, the equilibrium will shift to the right to compensate for this decrease, producing more \(\mathrm{Pb}^{2+}(a q)\) and \(\mathrm{Cl}^{-}(a q)\) ions, and consuming solid \(\mathrm{PbCl}_{2}\) in the process.
04

(d) Addition of 1 M hydrochloric acid

Adding 1 M hydrochloric acid (\(\mathrm{HCl}\)) will increase the concentration of \(\mathrm{Cl}^{-}(a q)\) ions in the solution. According to Le Châtelier's principle, the equilibrium will shift to the left, reducing the concentration of the \(\mathrm{Pb}^{2+}(a q)\) and \(\mathrm{Cl}^{-}(a q)\) ions to counteract this change. As a result, the amount of \(\mathrm{PbCl}_{2}(s)\) will increase.

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Key Concepts

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

Le Châtelier’s Principle
Le Châtelier's Principle is a fundamental concept in chemistry that explains how a chemical system at equilibrium reacts to external changes. This principle states that if a system at equilibrium is disturbed by altering conditions like concentration, temperature, or pressure, the system will readjust to a new equilibrium state as a way to counteract that disturbance.

For instance, when the concentration of a reactant or product in a reversible reaction is increased, the equilibrium will shift to consume more of the substance that was added. Conversely, if a substance is removed from the system, the equilibrium will shift to produce more of it. This is why, in our exercise, adding a solution containing \(\mathrm{Pb}^{2+}\) ions causes the equilibrium to shift left, reducing the concentration of \(\mathrm{Pb}^{2+}\) ions in the solution by forming solid \(\mathrm{PbCl}_{2}\).

Le Châtelier's Principle allows chemists to predict how changes in conditions can alter the outcome of chemical reactions, making it a vital tool for industrial applications where control over reaction products is essential.
Equilibrium Shifts
Equilibrium shifts refer to the directional change of a reversible reaction when it passes from one state of equilibrium to another due to a change in its system. An equilibrium shift can occur as a result of alterations in reactant or product concentrations, changes in pressure or volume for gaseous reactions, or variations in temperature.

In the context of our exercise, different actions such as the addition of a \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) solution, increasing temperature, or introduction of \(\mathrm{Ag}^{+}\) ions, each caused a shift in equilibrium. For instance, the addition of \(\mathrm{Ag}^{+}\) ions led to the formation of \(\mathrm{AgCl}\) precipitate, thereby decreasing the \(\mathrm{Cl}^{-}\) ion concentration and causing a shift to the right, favoring the production of more ions from solid \(\mathrm{PbCl}_{2}\).

Understanding shifts helps in controlling and directing chemical reactions, which is particularly important when trying to maximize yields in chemical manufacturing.
Endothermic Reaction
Endothermic reactions are chemical processes that absorb heat from their surroundings. This contrasts with exothermic reactions which release heat. In endothermic reactions, the energy required to break bonds in the reactants is greater than the energy released when new bonds form in the products, resulting in a net absorption of energy.

In our exercise, the dissolving of \(\mathrm{PbCl}_{2}\) is an endotheric process, as indicated by a positive \(\Delta H\). When temperature is increased, it provides extra heat that the system can absorb. According to Le Châtelier’s Principle, the equilibrium for an endothermic reaction will shift in the direction that absorbs heat—in this case, toward the products or right side of the equation. Consequently, more \(\mathrm{Pb}^{2+}\) and \(\mathrm{Cl}^{-}\) ions are expected in the solution when the temperature rises.

Recognizing whether a reaction is endothermic or exothermic is key to manipulating the reaction conditions to achieve a desired outcome, such as in thermochemical processes used in various industries.

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Most popular questions from this chapter

\(\mathrm{K}_{\mathrm{sp}}\) for silver acetate \(\left(\mathrm{AgC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)\) at \(80^{\circ} \mathrm{C}\) is estimated to be \(2 \times 10^{-2}\). Ten grams of silver acetate are added to \(1.0 \mathrm{~L}\) of water at \(25^{\circ} \mathrm{C}\). (a) Will all the silver acetate dissolve at \(25^{\circ} \mathrm{C} ?\) (b) If the solution (assume the volume to be \(1.0 \mathrm{~L}\) ) is heated to \(80^{\circ} \mathrm{C}\), will all the silver acetate dissolve?

Solid lead nitrate is added to a solution that is \(0.020 \mathrm{M}\) in \(\mathrm{OH}^{-}\) and \(\mathrm{SO}_{4}^{2-}\). Addition of the lead nitrate does not change the volume of the solution. (a) Which compound, \(\mathrm{PbSO}_{4}\) or \(\mathrm{Pb}(\mathrm{OH})_{2}\left(K_{s p}=2.8 \times 10^{-16}\right)\), will precipitate first? (b) What is the \(\mathrm{pH}\) of the solution when \(\mathrm{PbSO}_{4}\) first starts to precipitate?

Calcium ions in blood trigger clotting. To prevent that in donated blood, sodium oxalate, \(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\), is added to remove calcium ions according to the following equation. $$ \mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q)+\mathrm{Ca}^{2+}(a q) \longrightarrow \mathrm{CaC}_{2} \mathrm{O}_{4}(s) $$ Blood contains about \(0.10 \mathrm{mg} \mathrm{Ca}^{2+} / \mathrm{mL}\). If a 250.0-mL sample of donated blood is treated with an equal volume of \(0.160 \mathrm{M} \mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\), estimate \(\left[\mathrm{Ca}^{2+}\right]\) after precipitation. \(\left(K_{s p} \mathrm{CaC}_{2} \mathrm{O}_{4}=4 \times 10^{-9}\right)\)

Consider the following solubility data for calcium oxalate \(\left(\mathrm{Ca} \mathrm{C}_{2} \mathrm{O}_{4}\right)\) : $$ \begin{aligned} &K_{s p} \text { at } 25^{\circ} \mathrm{C}=4 \times 10^{-9} \\ &K_{s p} \text { at } 95^{\circ} \mathrm{C}=1 \times 10^{-8} \end{aligned} $$ Five hundred mL of a saturated solution is prepared at \(95^{\circ} \mathrm{C}\). How many milligrams of \(\mathrm{CaC}_{2} \mathrm{O}_{4}\) will precipitate when the solution is cooled to \(25^{\circ} \mathrm{C}\) ? (Assume that supersaturation does not take place.)

What is the solubility of \(\mathrm{CaF}_{2}\) in a buffer solution containing \(0.30 \mathrm{M}\) \(\mathrm{HCHO}_{2}\) and \(0.20 \mathrm{M} \mathrm{NaCHO}_{2}\) ? Hint: Consider the equation $$ \mathrm{CaF}_{2}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Ca}^{2+}(a q)+2 \mathrm{HF}(a q) $$ and solve the equilibrium problem.
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