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Chapter 16: 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 because it reacts with the acid to form soluble silver ions (Ag+) and release hydrogen sulfide gas (H2S), increasing its solubility. In contrast, silver chloride (AgCl) does not react with the acidic solution and hence, cannot be dissolved by the acid. This occurs because sulfur has a greater affinity for hydrogen than chloride, making it more likely to react with the acidic solution.

Step by step solution

01

Recall the solubility rules and reactions of silver sulfide and silver chloride in water

Silver sulfide (Ag2S) and silver chloride (AgCl) are slightly soluble in water due to the formation of their respective complexes. However, in an acidic solution, some compounds react with the acid, changing their solubility. We'll explore these reactions to determine which compound is more likely to dissolve.
02

Write the chemical reactions of Ag2S and AgCl with acids

In acidic solutions, silver ion (Ag+) reacts with the anions of the compounds. The chemical reactions can be represented as follows: For silver sulfide: \[ Ag2S(s) + 2H^+(aq) \rightarrow 2Ag^+(aq) + H2S(g) \] For silver chloride: \[ AgCl(s) + H^+(aq) \nrightarrow Ag^+(aq) + HCl(aq) \]
03

Analyze the reactions between the compounds and the acid

In the case of silver sulfide (Ag2S), it reacts with acid to form soluble silver ions (Ag+) and releases hydrogen sulfide gas (H2S). This reaction increases the solubility of silver sulfide. However, for silver chloride (AgCl), there is no reaction between the compound and the acid, meaning it cannot be dissolved by the acid.
04

Compare the solubility and determine the answer

Since silver sulfide (Ag2S) reacts with acid and increases its solubility while silver chloride (AgCl) does not react, we can conclude that silver sulfide (Ag2S) is more likely to dissolve in an acidic solution. The reason behind this is that sulfur has a greater affinity for hydrogen than chloride, making it more likely to react with the acidic solution and form the soluble species silver ion (Ag+) and hydrogen sulfide gas (H2S).

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Most popular questions from this chapter

Calculate the solubility of each of the following compounds in moles per liter. Ignore any acid-base properties. a. \(\mathrm{Ag}_{3} \mathrm{PO}_{4}, K_{\mathrm{sp}}=1.8 \times 10^{-18}\) b. \(\mathrm{CaCO}_{3}, K_{\mathrm{sp}}=8.7 \times 10^{-9}\) c. \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}, K_{\mathrm{sp}}=1.1 \times 10^{-18}\left(\mathrm{Hg}_{2}^{2+}\right.\) is the cation in solution.)

As sodium chloride solution is added to a solution of silver nitrate, a white precipitate forms. Ammonia is added to the mixture and the precipitate dissolves. When potassium bromide solution is then added, a pale yellow precipitate appears. When a solution of sodium thiosulfate is added, the yellow precipitate dissolves. Finally, potassium iodide is added to the solution and a yellow precipitate forms. Write equations for all the changes mentioned above. What conclusions can you draw concerning the sizes of the \(K_{\text {sp }}\) values for \(\mathrm{AgCl}, \mathrm{AgBr}\), and \(\mathrm{AgI}\) ?

The copper(I) ion forms a chloride salt that has \(K_{\text {sp }}=1.2 \times\) \(10^{-6}\). 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 \(0.10 M\)

Calculate the molar solubility of \(\mathrm{Co}(\mathrm{OH})_{3}, K_{\mathrm{sp}}=2.5 \times 10^{-43}\)

Calculate the solubility of solid \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\left(K_{\mathrm{sp}}=1.3 \times 10^{-32}\right)\) in a \(0.20-M \mathrm{Na}_{3} \mathrm{PO}_{4}\) solution.
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