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Chapter 15: Problem 7

Which is more likely to dissolve in an acidic solution, silver sulfide or silver chloride? Why?

Short Answer

Expert verified
Silver sulfide (Ag2S) is more likely to dissolve in an acidic solution than silver chloride (AgCl) because the excess H+ ions in the acidic solution react with the S^(2-) ions produced by the dissolution of silver sulfide, which shifts the dissolution equilibrium to the right and increases the solubility of silver sulfide. In contrast, the solubility of silver chloride is not affected by the presence of H+ ions since they do not react with Cl- ions, so its solubility remains largely unchanged in the acidic solution.

Step by step solution

01

Identify the relevant chemical reactions

First, we need to identify the chemical reactions that silver sulfide and silver chloride undergo when in contact with an acidic solution. Silver sulfide and silver chloride are both slightly soluble in water and can undergo dissolution reactions to form their respective ions. The general dissolution reactions for these compounds can be written as: For silver sulfide (Ag2S): \[ Ag_2S(s) \rightleftharpoons 2Ag^+(aq) + S^{2-}(aq) \] For silver chloride (AgCl): \[ AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq) \] The acidic solution contains an excess of H+ ions, which can react with species present in the solution.
02

Consider solubility equilibria in acidic conditions

Now, let's evaluate the solubility of the two compounds in an acidic solution. Since the acidic solution contains a high concentration of H+ ions, we should consider reactions between H+ ions and other ions generated during the dissolution of silver sulfide and silver chloride. The H+ ions can react with S^(2-) ions to form hydrogen sulfide (H2S) as shown below: \[ H^+(aq) + S^{2-}(aq) \rightarrow HS^-(aq) \] \[ H^+(aq) + HS^-(aq) \rightarrow H_2S(aq) \] However, the H+ ions do not form stable compounds with Cl- ions.
03

Determine which compound is more likely to dissolve

As we can see from the reactions, in an acidic solution, the H+ ions will react with the S^(2-) ions produced by the dissolution of silver sulfide, leading to the formation of hydrogen sulfide (H2S). This reaction will pull the dissolution equilibrium of silver sulfide to the right, increasing its solubility in the acidic solution. On the other hand, the Cl- ions produced by the dissolution of silver chloride do not react with the H+ ions, and thus the solubility of silver chloride remains largely unchanged in the acidic solution.
04

Compare the solubility and provide the reason

Based on the information in steps 2 and 3, we can conclude that silver sulfide (Ag2S) is more likely to dissolve in an acidic solution than silver chloride (AgCl). The reason for this is that the excess H+ ions in the acidic solution react with the S^(2-) ions produced by the dissolution of silver sulfide, which shifts the dissolution equilibrium to the right and increases the solubility of silver sulfide. In contrast, the dissolution equilibrium of silver chloride is not affected by the presence of H+ ions since they do not react with Cl- ions, so its solubility remains largely unchanged in the acidic solution.

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Most popular questions from this chapter

Write balanced equations for the dissolution reactions and the corresponding solubility product expressions for each of the following solids. a. \(\mathrm{AgC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) b. \(\mathrm{Al}(\mathrm{OH})_{3}\) c. \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\)

The solubility of \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}(s)\) in a \(7.2 \times 10^{-2}-M \mathrm{KIO}_{3}\) solution is \(6.0 \times 10^{-9} \mathrm{mol} / \mathrm{L} .\) Calculate the \(K_{\mathrm{sp}}\) value for \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}(s)\).

The copper(I) ion forms a chloride salt that has \(K_{\mathrm{sp}}=1.2 \times\) \(10^{-6} .\) Copper(I) also forms a complex ion with \(\mathrm{Cl}^{-}:\) $$\mathrm{Cu}^{+}(a q)+2 \mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{CuCl}_{2}^{-}(a q) \quad K=8.7 \times 10^{4}$$ a. Calculate the solubility of copper(I) chloride in pure water. (Ignore \(\mathrm{CuCl}_{2}^{-}\) formation for part a.) b. Calculate the solubility of copper(I) chloride in \(0.10 M\) NaCl.

Silver chloride dissolves readily in \(2 M \mathrm{NH}_{3}\) but is quite insoluble in \(2 M \mathrm{NH}_{4} \mathrm{NO}_{3} .\) Explain.

If a solution contains either \(\mathrm{Pb}^{2+}(a q)\) or \(\mathrm{Ag}^{+}(a q),\) how can temperature be manipulated to help identify the ion in solution?
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