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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

Expert verified
The salts that will be substantially more soluble in an HNO鈧 solution than in pure water are (b) CuS and (c) Cd(OH)鈧. This is because the anions in these salts (S虏鈦 and OH鈦) can react with H鈧僌鈦 ions, promoting dissolution.

Step by step solution

01

Analyze the reaction of each salt in acidic solution

Determine how each salt reacts with HNO鈧 and whether dissolution can be promoted by acid-base reactions: (a) \(\mathrm{BaSO}_{4}\): The dissolution reaction for BaSO鈧 can be written as: \[ BaSO_{4} \leftrightarrows Ba^{2+} + SO^{2-}_{4} \] No formation of additional H鈧僌鈦 ions occurs (since sulfate is a weak base and not particularly reactive with H鈧僌鈦), indicating that dissolution of this salt should not be significantly affected by the presence of HNO鈧. (b) \(\mathrm{CuS}\): The dissolution reaction for CuS can be written as: \[ CuS \leftrightarrows Cu^{2+} + S^{2-} \] In this case, the sulfide ion (S虏鈦) can react with H鈧僌鈦 ions resulting in the production of HS鈦 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 HNO鈧 solution compared to water. (c) \(\mathrm{Cd}(\mathrm{OH})_{2}\): The dissolution reaction for Cd(OH)鈧 can be written as: \[ Cd(OH)_{2} \leftrightarrows Cd^{2+} + 2OH^{-} \] In this case, the hydroxide ions (OH鈦) can react with H鈧僌鈦 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)鈧 more soluble in an HNO鈧 solution compared to water. (d) \(\mathrm{PbF}_{2}\): The dissolution reaction for PbF鈧 can be written as: \[ PbF_{2} \leftrightarrows Pb^{2+} + 2 F^{-} \] As fluoride is a weak base and not very reactive towards H鈧僌鈦 ions, the dissolution of this salt should not be significantly affected by the presence of HNO鈧. (e) \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\): The dissolution reaction for Cu(NO鈧)鈧 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 H鈧僌鈦 occurs, dissolution of this salt should not be significantly affected by the presence of HNO鈧. Also, Cu(NO鈧)鈧 is already a soluble salt, so it won't become 鈥渟ubstantially more soluble鈥 in HNO3.
02

Identify which salts will be substantially more soluble in HNO鈧 solution

From the analysis in step 1, we can conclude that: (a) BaSO鈧: Not affected by the HNO鈧 solution (b) CuS: Substantially more soluble in HNO鈧 solution (c) Cd(OH)鈧: Substantially more soluble in HNO鈧 solution (d) PbF鈧: Not affected by the HNO鈧 solution (e) Cu(NO鈧)鈧: Not affected by the HNO鈧 solution Therefore, the salts that will be substantially more soluble in an HNO鈧 solution than in pure water are: (b) CuS (c) Cd(OH)鈧

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Key Concepts

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

Acid-Base Reaction
In chemistry, an acid-base reaction involves the transfer of protons (H鈦 ions) between reactants. This is crucial in determining the solubility of salts in acidic solutions. For example, acids like nitric acid (HNO鈧) can donate protons (H鈦) to other substances.
  • These protons can react with anions of salts, such as sulfide (S虏鈦) and hydroxide (OH鈦), present in salts like Copper sulfide (CuS) and Cadmium hydroxide (Cd(OH)鈧).
  • 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 (Cu虏鈦 and S虏鈦) and the undissolved solid.
  • When protons from an acid (like H鈧僌鈦) react with sulfide ions (S虏鈦), it forms bisulfide ions (HS鈦), 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鈦) 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)鈧) are substantially more soluble in acidic solutions because their anions (S虏鈦 and OH鈦) can readily react with H鈦 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.

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Most popular questions from this chapter

(a) The molar solubility of \(\mathrm{PbBr}_{2}\) at \(25^{\circ} \mathrm{C}\) is \(1.0 \times 10^{-2} \mathrm{~mol} / \mathrm{L} .\) Calculate \(K_{s p} .(\mathbf{b})\) If \(0.0490 \mathrm{~g}\) of \(\mathrm{AgIO}_{3}\) dis- solves per liter of solution, calculate the solubility-product constant. (c) Using the appropriate \(K_{s p}\) value from Appendix D, calculate the pH of a saturated solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\).

Predict whether the equivalence point of each of the following titrations is below, above, or at pH 7: (a) \(\mathrm{NaHCO}_{3}\) titrated with \(\mathrm{NaOH},(\mathbf{b}) \mathrm{NH}_{3}\) titrated with \(\mathrm{HCl},\) (c) KOH titrated with HBr.

(a) Calculate the percent ionization of \(0.250 \mathrm{M}\) lactic acid \(\left(K_{a}=1.4 \times 10^{-4}\right) .(\mathbf{b})\) Calculate the percent ionization of \(0.250 \mathrm{M}\) lactic acid in a solution containing \(0.050 \mathrm{M}\) sodium lactate.

For each of the following slightly soluble salts, write the net ionic equation, if any, for reaction with a strong acid: (a) MnS, (b) \(\mathrm{PbF}_{2}\), (c) \(\mathrm{AuCl}_{3}\) (d) \(\mathrm{Hg}_{2} \mathrm{C}_{2} \mathrm{O}_{4},\) (e) CuBr.

A buffer contains 0.20 mol of acetic acid and 0.25 mol of sodium acetate in \(2.50 \mathrm{~L}\). (a) What is the pH of this buffer? (b) What is the pH of the buffer after the addition of 0.05 mol of \(\mathrm{NaOH}\) ? (c) What is the pH of the buffer after the addition of \(0.05 \mathrm{~mol}\) of \(\mathrm{HCl}\) ?

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