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Which of the following salts will be substantially more soluble in acidic solution than in pure water: (a) \(\mathrm{ZnCO}_{3}\), (b) \(\mathrm{ZnS}\) (c) \(\mathrm{BiI}_{3}\) (d) \(\mathrm{AgCN}\), (e) \(\mathrm{Ba}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) ?

Short Answer

Expert verified
The salts that will be substantially more soluble in acidic solution than in pure water are (a) \(\mathrm{ZnCO}_{3}\), (b) \(\mathrm{ZnS}\), and (d) \(\mathrm{AgCN}\).

Step by step solution

01

Write the dissociation reactions

Write the dissociation reactions of each salt in water: (a) \(\mathrm{ZnCO}_{3}(s) \rightleftharpoons \mathrm{Zn}^{2+}(aq) + \mathrm{CO}_{3}^{2-}(aq)\) (b) \(\mathrm{ZnS}(s) \rightleftharpoons \mathrm{Zn}^{2+}(aq) + \mathrm{S}^{2-}(aq)\) (c) \(\mathrm{BiI}_{3}(s) \rightleftharpoons \mathrm{Bi}^{3+}(aq) + 3\mathrm{I}^-(aq)\) (d) \(\mathrm{AgCN}(s) \rightleftharpoons \mathrm{Ag}^+(aq) + \mathrm{CN}^-(aq)\) (e) \(\mathrm{Ba}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s) \rightleftharpoons 3\mathrm{Ba}^{2+}(aq) + 2\mathrm{PO}_{4}^{3-}(aq)\)
02

Determine possible reactions with acidic solution

Analyze the possible favorable reactions of anions with H+ ions: (a) CO₃²⁻ + 2H⁺ → H2CO3 → CO2 + H2O (b) S²⁻ + 2H⁺ → H2S (c) I⁻: No favorable reaction with H+ ions. (d) CN⁻ + H⁺ → HCN (e) PO₄³⁻ + 3H⁺ → H3PO4
03

Determine the stability of the formed products

Determine if any of the reactions in Step 2 result in the formation of a weak electrolyte (weak acid or weak base). Weak electrolytes favor solubility when reacting with acidic or basic solutions: (a) H2CO3: Weak acid, decomposes to CO2(g) and H2O, which makes the reaction favorable. (b) H2S: Weak acid, so the reaction is favorable. (c) No favorable reaction with H+ ions (d) HCN: Weak acid, so the reaction is favorable (e) H3PO4: Strong acid, does not favor solubility.
04

Determine solubility

Based on the reactions observed in Step 3, determine which salts have greater solubility in acidic solutions: (a) ZnCO₃: More soluble in acidic solution. (b) ZnS: More soluble in acidic solution. (c) BiI₃: No significant difference in solubility. (d) AgCN: More soluble in acidic solution. (e) Ba₃(PO₄)₂: No significant difference in solubility. Answer: The salts that will be substantially more soluble in acidic solution than in pure water are (a) \(\mathrm{ZnCO}_{3}\), (b) \(\mathrm{ZnS}\), and (d) \(\mathrm{AgCN}\).

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Key Concepts

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

Dissociation Reactions
When a salt dissolves in water, it separates or dissociates into its respective ions. This is a crucial step for determining solubility changes in different solutions. For instance, when
  • \( \mathrm{ZnCO}_{3}\) dissolves, it splits into \( \mathrm{Zn}^{2+}(aq)\) and \( \mathrm{CO}_{3}^{2-}(aq)\).
  • Similarly, \( \mathrm{ZnS}\) dissociates into \( \mathrm{Zn}^{2+}(aq)\) and \( \mathrm{S}^{2-}(aq)\).
Each component ion can further react with other available substances in the solution. For example, the carbonate ion \( \mathrm{CO}_{3}^{2-}(aq) \) can react with hydrogen ions \( \mathrm{H}^{+}\) from an acidic solution to form \( \mathrm{CO}_2 \) and \( \mathrm{H}_2\mathrm{O} \), increasing the solubility of \( \mathrm{ZnCO}_{3} \) in this new environment. This is because removing one of the dissociated ions \( \mathrm{CO}_{3}^{2-}(aq) \) from the solution shifts the equilibrium, driving more \( \mathrm{ZnCO}_{3} \) to dissolve.
Weak Electrolytes
Weak electrolytes are substances that partially dissociate into ions in solution. They are essential in analyzing solubility in different environments. When a salt in water forms a weak electrolyte like a weak acid or base through a reaction, its solubility often increases as the system shifts to dissolve more salt to balance the concentrations.For example:
  • \( \mathrm{H}_2\mathrm{CO}_3 \) formed from \( \mathrm{CO}_{3}^{2-} \) in acidic solutions is a weak acid, which further decomposes into \( \mathrm{CO}_2 \) and \( \mathrm{H}_2\mathrm{O} \), facilitating additional dissolution.
  • \( \mathrm{HCN} \) formed from \( \mathrm{CN}^- \) also remains mostly undissociated, enhancing the solubility of \( \mathrm{AgCN} \).
Weak electrolytes thus play a central role in determining whether a salt becomes more soluble in an acidic solution compared to pure water.
Solubility Rules
Solubility rules are guidelines used to predict the solubility of compounds under different conditions. They help determine which salts are likely to dissolve more in an acidic solution than in water. The main idea is that certain anions react with \( \mathrm{H}^{+} \) ions to produce weak electrolytes, encouraging further dissolution of the salt.For salts such as:
  • \( \mathrm{ZnCO}_{3} \), the \( \mathrm{CO}_{3}^{2-} \) reacts with \( \mathrm{H}^{+} \) to form \( \mathrm{H}_2\mathrm{CO}_3 \), which breaks down to \( \mathrm{CO}_2 \) and \( \mathrm{H}_2\mathrm{O} \).
  • \( \mathrm{ZnS} \), the sulfide ion \( \mathrm{S}^{2-} \) reacts with \( \mathrm{H}^{+} \) forming \( \mathrm{H}_2\mathrm{S} \), a weak acid.
  • \( \mathrm{AgCN} \), the \( \mathrm{CN}^{-} \) ion forms \( \mathrm{HCN} \) when it reacts, which has limited dissociation, pushing more of the salt to dissolve.
Using these rules allows us to identify that some salts, like \( \mathrm{ZnCO}_{3} \), \( \mathrm{ZnS} \), and \( \mathrm{AgCN} \), show increased solubility in acidic conditions due to the reactions forming weak acids or bases.

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Most popular questions from this chapter

How many milliliters of \(0.105 \mathrm{M} \mathrm{HCl}\) are needed to titrate each of the following solutions to the equivalence point: (a) \(45.0 \mathrm{~mL}\) of \(0.0950 \mathrm{M} \mathrm{NaOH}\), (b) \(22.5 \mathrm{~mL}\) of \(0.118 \mathrm{M} \mathrm{NH}_{3}\), (c) \(125.0 \mathrm{~mL}\) of a solution that contains \(1.35 \mathrm{~g}\) of \(\mathrm{NaOH}\) per liter?

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

(a) Consider the equilibrium \(\mathrm{B}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\) \(\mathrm{HB}^{+}(a q)+\mathrm{OH}^{-}(a q) .\) Using Le Châtelier's principle, explain the effect of the presence of a salt of \(\mathrm{HB}^{+}\) on the ionization of B. (b) Give an example of a salt that can decrease the ionization of \(\mathrm{NH}_{3}\) in solution.

(a) A 0.1044-g sample of an unknown monoprotic acid requires \(22.10 \mathrm{~mL}\) of \(0.0500 \mathrm{M} \mathrm{NaOH}\) to reach the end point. What is the molecular weight of the unknown? (b) As the acid is titrated, the pH of the solution after the addition of \(11.05 \mathrm{~mL}\) of the base is \(4.89 .\) What is the \(K_{a}\) for the acid? (c) Using Appendix D, suggest the identity of the acid. Do both the molecular weight and \(K_{a}\) value agree with your choice?

Calculate the molar solubility of \(\mathrm{Fe}(\mathrm{OH})_{2}\) when buffered at \(\mathrm{pH}(\mathrm{a}) 8.0\), (b) \(10.0,(\mathrm{c}) 12.0 .\)

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