Which of the following compounds, when dissolved in a 0.01-M solution of HClO_ has a solubility greater than in pure water: AgBr, BaFz, Ca_(PO \(_{4}\) ) \(2,\) ZnS, PbI \(_{2}\) ? Explain your answer.

Short Answer

Expert verified
BaF2 has a solubility greater than in pure water when dissolved in a 0.01-M solution of HClO due to the formation of weak acid HF, which shifts the equilibrium to increase solubility.

Step by step solution

01

- Understanding the Common Ion Effect

Recognize that the common ion effect influences the solubility of ionic compounds in a solution that contains one of the ions present in the compound. The presence of a common ion decreases the solubility of the compound.
02

- Identifying if HClO contains a common ion with the given compounds

Analyze whether the compounds contain the same ions as the ions in HClO. The ion ClO- from HClO does not appear in any of the given compounds, which means that HClO does not contain a common ion with any of the compounds listed.
03

- Determine the compound where HClO can improve solubility

Consider that a 0.01-M HClO solution, a weak acid, can slightly ionize to give H+ ions. These H+ ions can react with anions of the slightly soluble salts to form a weak acid or a gas, and this removal of anions can drive the solubility equilibrium forward, increasing solubility. Among the given compounds, only the anions from AgBr (Br-), BaF2 (F-), and PbI2 (I-) can form a weak acid (HBr, HF, and HI respectively) upon reacting with H+. The BaF2 stands out because HF is known to be a relatively weak acid, so the formation of HF from F- ions would most significantly improve the solubility of BaF2 in the presence of excess H+ ions.
04

- Conclusion

The solubility of BaF2 will be greater in a 0.01-M solution of HClO than in pure water due to the formation of weak acid HF when fluoride ions react with the H+ ions from the partial ionization of HClO. The other compounds either do not react with the H+ ions or form a stronger acid, which does not shift the solubility equilibrium significantly.

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Key Concepts

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

Solubility of Ionic Compounds
Solubility refers to the ability of a substance to dissolve in a solvent, such as water. Ionic compounds dissolve when water molecules surround and stabilize the ions, allowing them to separate from the solid structure of the compound. The solubility of an ionic compound is influenced by various factors including temperature, the nature of the solvent, and the presence of other ions in the solution.

Common Ion Effect and Solubility

The common ion effect is a key factor that impacts the solubility of ionic compounds. When an ionic compound is added to a solution that already contains one of the ions present in that compound, the presence of the common ion suppresses the compound's solubility. This phenomenon occurs because the addition of the common ion shifts the dissolution equilibrium according to Le Chatelier's principle, favoring the undissolved form of the compound.

For example, if a solution already contains Cl- ions and you add NaCl to it, the solubility of NaCl will be lower than it would be in pure water, as the solution is already saturated with Cl- ions.
Chemical Equilibrium
Chemical equilibrium is a state where the rate of the forward reaction equals the rate of the reverse reaction; therefore, the concentrations of reactants and the concentrations of products remain constant over time. This doesn't mean the reactions have stopped, but rather they are proceeding at equal rates in both directions.

Le Chatelier's Principle in Equilibrium

In the context of solubility, when an ionic compound is dissolved in water, it establishes a dynamic equilibrium between the solid (undissolved) form and its ions in solution. Le Chatelier's principle predicts that if the concentration of one of the ions is increased (as in the case of a common ion effect), the equilibrium will shift to favor the formation of the undissolved compound, thus reducing its solubility in the solution.

Practical understanding of chemical equilibrium and the factors that affect it, such as concentration changes, temperature, and pressure, is crucial for predicting how different environments might influence the behavior of chemical reactions, including the dissolving of ionic compounds.
Weak Acid Behavior
Weak acids do not fully dissociate in water, unlike strong acids which dissociate completely. This partial dissociation leads to a dynamic equilibrium between the undissociated acid and the ions it produces.

Hydrolysis of Anions from Weak Acids

When weak acids release H+ ions in water, the anions they form might react with other ions present in the solution. This reaction can lead to the formation of a new weak acid or a gas, a process known as hydrolysis. In a solution with additional H+ ions from a source like HClO, anions that come from the dissolution of slightly soluble salts can react with these extra H+ ions to form a new weak acid. This process can effectively 'remove' the anions from the equilibrium equation, thereby increasing the solubility of the salt.

For instance, F- ions from BaF2 can interact with excess H+ ions to form HF, a weak acid. This removal of F- ions from the solution drives the dissolution equilibrium of BaF2 forward, increasing its solubility in the process—a remarkable example of how weak acid behavior is closely linked with the solubility of ionic compounds in the presence of common ions.

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Most popular questions from this chapter

A saturated solution of a slightly soluble electrolyte in contact with some of the solid electrolyte is said to be a system in equilibrium. Explain. Why is such a system called a heterogeneous equilibrium?

The carbonate ion concentration is gradually increased in a solution containing divalent cations of magnesium, calcium, strontium, barium, and manganese. Which of the following carbonates will form first? Which of the following will form last? Explain. \(\begin{array}{ll}\text { (a) } \mathrm{MgCO}_{3} & K_{\mathrm{sp}}=3.5 \times 10^{-8}\end{array}\) (b) \(\mathrm{CaCO}_{3} \quad K_{\mathrm{sp}}=4.2 \times 10^{-7}\) (c) \(\operatorname{Sr} \mathrm{CO}_{3} \quad K_{\mathrm{sp}}=3.9 \times 10^{-9}\) \(\begin{array}{ll}\text { (d) } \mathrm{BaCO}_{3} & K_{\mathrm{sp}}=4.4 \times 10^{-5}\end{array}\) (e) \(\mathrm{MnCO}_{3} \quad K_{\mathrm{sp}}=5.1 \times 10^{-9}\)

The Handbook of Chemistry and Physics (http://openstaxcollege.org/l/16Handbook) gives solubilities of the following compounds in grams per \(100 \mathrm{mL}\) of water. Because these compounds are only slightly soluble, assume that the volume does not change on dissolution and calculate the solubility product for each. (a) \(\mathrm{BaSiF}_{6}, 0.026 \mathrm{g} / 100 \mathrm{mL}\) (contains \(\mathrm{SiF}_{6}^{2-}\) ions) (b) \(\operatorname{Ce}\left(\mathrm{IO}_{3}\right)_{4}, 1.5 \times 10^{-2} \mathrm{g} / 100 \mathrm{mL}\) (c) \(\mathrm{Gd}_{2}\left(\mathrm{SO}_{4}\right)_{3}, 3.98 \mathrm{g} / 100 \mathrm{mL}\) (d) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{PtBr}_{6}, 0.59 \mathrm{g} / 100 \mathrm{mL}\) (contains \(\mathrm{PtBr}_{6}^{2-}\) ions)

Write the Lewis structures of the reactants and product of each of the following equations, and identify the Lewis acid and the Lewis base in each: (a) \(\mathrm{CO}_{2}+\mathrm{OH}^{-} \longrightarrow \mathrm{HCO}_{3}^{-}\) (b) \(\mathrm{B}(\mathrm{OH})_{3}+\mathrm{OH}^{-} \longrightarrow \mathrm{B}(\mathrm{OH})_{4}^{-}\) (c) \(\mathrm{I}^{-}+\mathrm{I}_{2} \longrightarrow \mathrm{I}_{3}^{-}\) (d) \(\mathrm{AlCl}_{3}+\mathrm{Cl}^{-} \longrightarrow \mathrm{AlCl}_{4}^{-}\) (use Al-Cl single bonds) (e) \(\mathrm{O}^{2-}+\mathrm{SO}_{3} \longrightarrow \mathrm{SO}_{4}^{2-}\)

Assuming that no equilibria other than dissolution are involved, calculate the molar solubility of each of the following from its solubility product: (a) \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\) (b) \(\mathrm{PbBr}_{2}\) (c) AgI (d) \(\mathrm{CaC}_{2} \mathrm{O}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\)

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