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) \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2} ?\)

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.

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