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\(\mathrm{PbCl}_{2}(\mathrm{s})\) is considerably more soluble in \(\mathrm{HCl}(\mathrm{aq})\) than in pure water, but its solubility in \(\mathrm{HNO}_{3}(\mathrm{aq})\) is not much different from what it is in water. Explain this difference in behavior.

Short Answer

Expert verified
The increased solubility of PbCl2 in HCl is due to the increase in Cl- ions from HCl dissociation leading to the common ion effect, which shifts the equilibrium to favor the formation of more PbCl2. However, because HNO3 does not contain a common ion (Cl-) with PbCl2, it doesn't significantly affect its solubility, hence its solubility remains similar to that in water.

Step by step solution

01

Analyze the solubility of PbCl2

Lead chloride, PbCl2 is sparingly soluble in water. When dissolved, it dissociates into Pb2+ ions and Cl- ions. The equilibrium can be represented as: \(PbCl2 \leftrightarrow Pb^{2+} + 2Cl^{-}\)
02

Explain the solubility in HCl solution

When PbCl2 is in solution with HCl, it will increase the concentration of Cl- ions since HCl dissociates completely in water to give H+ and Cl-. This high concentration of Cl- shifts the PbCl2 dissolution equilibrium to left (Common ion effect), and favors the formation of more PbCl2, thus increases its solubility.
03

Explain the solubility in HNO3 solution

HNO3 does not contain the Cl- ion. Therefore, adding HNO3 to the solution doesn't add more common ions, from Le Chatelier's principle, it will not influence the dissolution equilibrium of PbCl2 as much. Hence, its solubility in HNO3 is not much different from what it is in water.

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Key Concepts

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

Common Ion Effect
The common ion effect is a concept that explains how the presence of an ion that is shared between two solutes can influence the solubility of a sparingly soluble salt. When a common ion is introduced into a solution where an equilibrium exists, it can shift the equilibrium position. This typically decreases the solubility of the compound in that particular solvent.

Let's take the example of lead chloride (\( \mathrm{PbCl}_{2} \)). It has limited solubility in water, dissociating into lead ions (\( \mathrm{Pb^{2+}} \)) and chloride ions (\( \mathrm{Cl^{-}} \)). In pure water, there is an equilibrium between the dissolved ions and the undissolved solid:

\[ \mathrm{PbCl_2} \leftrightarrow \mathrm{Pb^{2+}} + 2\mathrm{Cl^{-}} \]

When \( \mathrm{HCl} \) is added to the solution, it dissociates completely to produce \( \mathrm{H^+} \) and \( \mathrm{Cl^-} \) ions.

Adding more \( \mathrm{Cl^-} \) ions shifts the equilibrium toward the left, favoring the formation of more solid \( \mathrm{PbCl_2} \), effectively reducing the overall concentration of dissolved \( \mathrm{Pb^{2+}} \).

This phenomenon explains why \( \mathrm{PbCl}_{2} \) is observed to have increased solubility in \( \mathrm{HCl} \) compared to pure water.
Le Chatelier's Principle
Le Chatelier's Principle provides insight into how an external change can affect a system at equilibrium. This principle states that if a dynamic equilibrium is disturbed by changing the conditions, the system responds in a way to counteract the change and establish a new equilibrium.

In the case of \( \mathrm{PbCl}_{2} \) in an \( \mathrm{HCl} \) solution, the introduction of extra chloride ions from \( \mathrm{HCl} \) disturbs the equilibrium. According to Le Chatelier's Principle, the system will try to counteract this increase by shifting the equilibrium back towards the reactants, which in this case results in the precipitation of more \( \mathrm{PbCl_2} \). This shift minimizes the effect of the common ion, ultimately reducing the solubility of \( \mathrm{PbCl_2} \).

However, when \( \mathrm{HNO_3} \) is added to the \( \mathrm{PbCl_2} \) solution, it does not introduce any chloride ions and therefore, has little effect on the equilibrium. The system remains relatively unaffected by this addition, meaning the solubility does not differ significantly from what is seen in water.
Aqueous Solutions
An aqueous solution is one where water is the solvent. In such solutions, soluble ionic compounds dissociate into their respective ions. This dissociation is central to the behaviors observed in solubility equilibria.

For instance, when \( \mathrm{PbCl}_{2} \) dissolves in water, it breaks down into \( \mathrm{Pb^{2+}} \) and \( \mathrm{Cl^{-}} \) ions. This dissociation forms the foundation for further interactions like those introduced by common ions or external perturbations.

Aqueous solutions also facilitate the application of principles like Le Chatelier's, allowing us to predict how a system will respond to changes in concentration or the introduction of additional chemicals.

These concepts are especially important when considering the addition of different solutes like \( \mathrm{HCl} \) or \( \mathrm{HNO_3} \). The ease with which these acids dissociate into ions in aqueous solutions explains their differing impacts on the solubility of compounds like \( \mathrm{PbCl}_{2} \). Understanding these behaviors requires a comprehensive grasp of the solution's ionic composition and the effects of various chemical interactions.

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Most popular questions from this chapter

The following \(K_{\mathrm{sp}}\) values are found in a handbook. Write the solubility product expression to which each one applies. For example, \(K_{\mathrm{sp}}(\mathrm{AgCl})=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]=\) \(1.8 \times 10^{-10}\). (a) \(K_{\mathrm{sp}}\left(\mathrm{Cr} \mathrm{F}_{3}\right)=6.6 \times 10^{-11}\) (b) \(K_{\mathrm{sp}}\left[\mathrm{Au}_{2}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]=1 \times 10^{-10}\) (c) \(K_{\mathrm{sp}}\left[\mathrm{Cd}_{3}\left(\mathrm{PO}_{4}\right)_{2}\right]=2.1 \times 10^{-33}\) (d) \(K_{\mathrm{sp}}\left(\mathrm{Sr} \mathrm{F}_{2}\right)=2.5 \times 10^{-9}\)

Pure water is saturated with slightly soluble \(\mathrm{PbI}_{2}\) Which of the following is a correct statement concerning the lead ion concentration in the solution, and what is wrong with the others? (a) \(\left[\mathrm{Pb}^{2+}\right]=\left[\mathrm{I}^{-}\right]\); (b) \(\left[\mathrm{Pb}^{2+}\right]=K_{\mathrm{sp}}\) of \(\mathrm{PbI}_{2} ;(\mathrm{c})\left[\mathrm{Pb}^{2+}\right]=\sqrt{K_{\mathrm{sp}}}\) of \(\mathrm{PbI}_{2}\); (d) \(\left[\mathrm{Pb}^{2+}\right]=0.5\left[\mathrm{I}^{-}\right]\)

Assume that, to be visible to the unaided eye, a precipitate must weigh more than \(1 \mathrm{mg}\). If you add \(1.0 \mathrm{mL}\) of \(1.0 \mathrm{M} \mathrm{NaCl}(\mathrm{aq})\) to \(100.0 \mathrm{mL}\) of a clear saturated aqueous AgCl solution, will you be able to see \(\mathrm{AgCl}(\mathrm{s})\) precipitated as a result of the common-ion effect? Explain.

Which one of the following solutions can be used to separate the cations in an aqueous solution in which \(\left[\mathrm{Ba}^{2+}\right]=\left[\mathrm{Ca}^{2+}\right]=0.050 \mathrm{M}: 0.10 \mathrm{M} \mathrm{NaCl}(\mathrm{aq}), 0.05 \mathrm{M}\) \(\mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq}),\) \(0.001 \mathrm{M}\) \(\mathrm{NaOH}(\mathrm{aq}),\) or \(0.50 \mathrm{M}\) \(\mathrm{Na}_{2} \mathrm{CO}_{3}(\text { aq }) ?\) Explain why.

Will \(\mathrm{Al}(\mathrm{OH})_{3}(\mathrm{s})\) precipitate from a buffer solution that is \(0.45 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}\) and \(0.35 \mathrm{M} \mathrm{NaCH}_{3} \mathrm{COO}\) and also \(0.275 \mathrm{M}\) in \(\mathrm{Al}^{3+}(\mathrm{aq}) ?\) For \(\mathrm{Al}(\mathrm{OH})_{3}, K_{\mathrm{sp}}=\) \(1.3 \times 10^{-33} ;\) for \(\mathrm{CH}_{3} \mathrm{COOH}, K_{\mathrm{a}}=1.8 \times 10^{-5}\).

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