Question

Which of the following salts will be substantially more soluble in an \(\mathrm{HNO}_{3}\) solution than in pure water: (a) \(\mathrm{BaSO}_{4}\), (b) \(\mathrm{CuS},\) (c) \(\mathrm{Cd}(\mathrm{OH})_{2}\) (d) \(\mathrm{PbF}_{2}\), (e) \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2} ?\)

Short answer

The salts that will be substantially more soluble in an HNO₃ solution than in pure water are (b) CuS and (c) Cd(OH)₂. This is because the anions in these salts (S²⁻ and OH⁻) can react with H₃O⁺ ions, promoting dissolution.

Step-by-step solution

Analyze the reaction of each salt in acidic solution

Determine how each salt reacts with HNO₃ and whether dissolution can be promoted by acid-base reactions: (a) \(\mathrm{BaSO}_{4}\): The dissolution reaction for BaSO₄ can be written as: \[ BaSO_{4} \leftrightarrows Ba^{2+} + SO^{2-}_{4} \] No formation of additional H₃O⁺ ions occurs (since sulfate is a weak base and not particularly reactive with H₃O⁺), indicating that dissolution of this salt should not be significantly affected by the presence of HNO₃. (b) \(\mathrm{CuS}\): The dissolution reaction for CuS can be written as: \[ CuS \leftrightarrows Cu^{2+} + S^{2-} \] In this case, the sulfide ion (S²⁻) can react with H₃O⁺ ions resulting in the production of HS⁻ ions. The reaction is as follows: \[ S^{2-} + H_{3}O^{+} \rightarrow HS^{-} + H_{2}O \] This reaction shifts the equilibria towards dissolution, making CuS more soluble in an HNO₃ solution compared to water. (c) \(\mathrm{Cd}(\mathrm{OH})_{2}\): The dissolution reaction for Cd(OH)₂ can be written as: \[ Cd(OH)_{2} \leftrightarrows Cd^{2+} + 2OH^{-} \] In this case, the hydroxide ions (OH⁻) can react with H₃O⁺ ions resulting in the formation of water: \[ OH^- + H_{3}O^{+} \rightarrow 2H_{2}O \] This reaction shifts the equilibria towards dissolution, making Cd(OH)₂ more soluble in an HNO₃ solution compared to water. (d) \(\mathrm{PbF}_{2}\): The dissolution reaction for PbF₂ can be written as: \[ PbF_{2} \leftrightarrows Pb^{2+} + 2 F^{-} \] As fluoride is a weak base and not very reactive towards H₃O⁺ ions, the dissolution of this salt should not be significantly affected by the presence of HNO₃. (e) \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\): The dissolution reaction for Cu(NO₃)₂ can be written as: \[ Cu(NO_{3})_{2} \leftrightarrows Cu^{2+} + 2NO_{3}^{-} \] Since Cu(2+) is the cation in HNO_3 as well and no additional formation of H₃O⁺ occurs, dissolution of this salt should not be significantly affected by the presence of HNO₃. Also, Cu(NO₃)₂ is already a soluble salt, so it won't become “substantially more soluble” in HNO3.

Identify which salts will be substantially more soluble in HNO₃ solution

From the analysis in step 1, we can conclude that: (a) BaSO₄: Not affected by the HNO₃ solution (b) CuS: Substantially more soluble in HNO₃ solution (c) Cd(OH)₂: Substantially more soluble in HNO₃ solution (d) PbF₂: Not affected by the HNO₃ solution (e) Cu(NO₃)₂: Not affected by the HNO₃ solution Therefore, the salts that will be substantially more soluble in an HNO₃ solution than in pure water are: (b) CuS (c) Cd(OH)₂

Key concepts

Acid-Base Reaction

In chemistry, an acid-base reaction involves the transfer of protons (H⁺ ions) between reactants. This is crucial in determining the solubility of salts in acidic solutions. For example, acids like nitric acid (HNO₃) can donate protons (H⁺) to other substances.

• These protons can react with anions of salts, such as sulfide (S²⁻) and hydroxide (OH⁻), present in salts like Copper sulfide (CuS) and Cadmium hydroxide (Cd(OH)₂).
• This reaction forms less negatively charged ions or neutral molecules like water, which in turn shifts the equilibrium towards more salt dissolving.

This interaction is a fundamental concept in understanding why certain salts dissolve more in acidic environments, enhancing their solubility.

Chemical Equilibrium

Chemical equilibrium is a state in which the rates of the forward and reverse reactions are equal, resulting in no net change in the concentrations of reactants and products.

• For a salt like Copper sulfide (CuS), the equilibrium exists between the dissolved ions (Cu²⁺ and S²⁻) and the undissolved solid.
• When protons from an acid (like H₃O⁺) react with sulfide ions (S²⁻), it forms bisulfide ions (HS⁻), disrupting the equilibrium.

The reaction tends to shift towards more dissolution of the salt to restore equilibrium, thus making the salt appear more soluble.

Dissolution Process

The dissolution process involves a substance, usually a solid (like a salt), dissolving into a liquid, forming a solution. Understanding this process involves looking into both the chemical interaction and physical dissolving mechanisms.

• When placed in water, a salt like Copper sulfide (CuS) dissociates into its ionic components.
• However, the presence of nitric acid (an acid) causes further chemical reactions that promote more dissolving of the salt.
• The acid's protons (H⁺) help by reacting with negatively charged ions, aiding the salt to go into solution.

This not only increases the salt's solubility but also exemplifies the intricate balance of chemical and physical processes in dissolution.

Salt Solubility in Acids

Salt solubility in acids is a concept reflecting how some salts dissolve better in the presence of acids than in pure water. This is due to interactions between the acid and the anions from the salt.

• Salts like Copper sulfide (CuS) and Cadmium hydroxide (Cd(OH)₂) are substantially more soluble in acidic solutions because their anions (S²⁻ and OH⁻) can readily react with H⁺ ions.
• Such interactions lower the concentration of the anion in the solution, prompting more salt to dissolve.

This understanding helps in predicting the solubility behavior of different salts in acidic environments, aiding in various practical applications, from industry to laboratory settings.