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Chapter 15: Problem 66

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 \(K_{\mathrm{sp}}\) values for \(\mathrm{AgCl}, \mathrm{AgBr},\) and \(\mathrm{AgI} ?\)

Short Answer

Expert verified
In this exercise, we analyzed a series of chemical reactions involving silver halides. The reactions and their balanced chemical equations are as follows: 1. \(AgNO_3 + NaCl \rightarrow AgCl_{(s)} + NaNO_3\) 2. \(AgCl_{(s)} + 2NH_3 \rightarrow [Ag(NH_3)_2]^+ + Cl^-\) 3. \([Ag(NH_3)_2]^+ + Cl^- + KBr \rightarrow AgBr_{(s)} + 2NH_3 + KCl\) 4. \(AgBr_{(s)} + 2Na_2S_2O_3 \rightarrow [Ag(S_2O_3)_2]^{3-} + 2NaBr\) 5. \([Ag(S_2O_3)_2]^{3-} + 2NaBr + 2KI \rightarrow AgI_{(s)} + 2K_2S_2O_3 + 2NaBr \) Based on these reactions, we concluded that the solubility product constants (\(K_{sp}\)) for the silver halides follow this order: \(K_{sp} (AgCl) > K_{sp} (AgBr) > K_{sp} (AgI)\).

Step by step solution

01

Formation of Silver Chloride Precipitate

When sodium chloride is added to a solution of silver nitrate, a white precipitate of silver chloride is formed. The balanced chemical equation for this reaction is: \(AgNO_3 + NaCl \rightarrow AgCl_{(s)} + NaNO_3\)
02

Dissolution of Silver Chloride Precipitate in Ammonia

When ammonia is added to the mixture containing the silver chloride precipitate, the precipitate dissolves. This occurs because ammonia forms a complex ion with silver ions. The balanced chemical equation for this reaction is: \(AgCl_{(s)} + 2NH_3 \rightarrow [Ag(NH_3)_2]^+ + Cl^-\)
03

Formation of Silver Bromide Precipitate

When potassium bromide is added to the solution, a pale yellow precipitate of silver bromide appears. The balanced chemical equation for this reaction is: \[ [Ag(NH_3)_2]^+ + Cl^- + KBr \rightarrow AgBr_{(s)} + 2NH_3 + KCl\]
04

Dissolution of Silver Bromide Precipitate in Sodium Thiosulfate

When a solution of sodium thiosulfate is added, the yellow silver bromide precipitate dissolves. The balanced chemical equation for this reaction is: \(AgBr_{(s)} + 2Na_2S_2O_3 \rightarrow [Ag(S_2O_3)_2]^{3-} + 2NaBr\)
05

Formation of Silver Iodide Precipitate

Finally, when potassium iodide is added to the solution, a yellow precipitate of silver iodide forms. The balanced chemical equation for this reaction is: \[ [Ag(S_2O_3)_2]^{3-} + 2NaBr + 2KI \rightarrow AgI_{(s)} + 2K_2S_2O_3 + 2NaBr \]
06

Final Analysis: Comparing \(K_{sp}\) Values

As we observed, AgCl, AgBr, and AgI form precipitates under different conditions, and their solubility varies in the presence of various complexing agents. This gives us an idea about the relative solubility product constants (\(K_{sp}\)) for each silver halide. Since AgCl precipitated first and then dissolved when ammonia was added, we can infer that its \(K_{sp}\) is the highest among these three silver halides. AgBr precipitated after the addition of ammonia and dissolved in sodium thiosulfate, indicating that its \(K_{sp}\) is lower than that of AgCl but higher than that of AgI. AgI precipitated last and did not dissolve in any of the complexing agents, suggesting that its \(K_{sp}\) is the lowest among the three. In conclusion, we can arrange the silver halides in the order of their decreasing solubility product constant values as follows: \(K_{sp} (AgCl) > K_{sp} (AgBr) > K_{sp} (AgI)\)

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Most popular questions from this chapter

a. Calculate the molar solubility of AgBr in pure water. \(K_{\mathrm{sp}}\) for AgBr is \(5.0 \times 10^{-13}\). b. Calculate the molar solubility of \(\mathrm{AgBr}\) in \(3.0\) \(M\) \(\mathrm{NH}_{3} .\) The overall formation constant for \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\) is \(1.7 \times 10^{7}\) that is, $$\mathrm{Ag}^{+}(a q)+2 \mathrm{NH}_{3}(a q) \longrightarrow \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}(a q) \quad K=1.7 \times 10^{7}$$ c. Compare the calculated solubilities from parts a and b. Explain any differences. d. What mass of AgBr will dissolve in \(250.0 \mathrm{mL}\) of \(3.0 M \mathrm{NH}_{3} ?\) e. What effect does adding \(\mathrm{HNO}_{3}\) have on the solubilities calculated in parts a and b?

Cream of tartar, a common ingredient in cooking, is the common name for potassium bitartrate (abbreviated KBT, molar mass \(=188.2 \mathrm{g} / \mathrm{mol}\) ). Historically, KBT was a crystalline solid that formed on the casks of wine barrels during the fermentation process. Calculate the maximum mass of KBT that can dissolve in \(250.0 \mathrm{mL}\) of solution to make a saturated solution. The \(K_{\mathrm{sp}}\) value for \(\mathrm{KBT}\) is \(3.8 \times 10^{-4}\).

A mixture contains \(1.0 \times 10^{-3} M\) Cu \(^{2+}\) and \(1.0 \times 10^{-3} M\) \(\mathrm{Mn}^{2+}\) and is saturated with \(0.10 M \mathrm{H}_{2} \mathrm{S} .\) Determine a \(\mathrm{pH}\) where CuS precipitates but MnS does not precipitate. \(K_{\mathrm{sp}}\) for \(\mathrm{CuS}=8.5 \times 10^{-45}\) and \(K_{\mathrm{sp}}\) for \(\mathrm{MnS}=2.3 \times 10^{-13}\).

What mass of \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) must be added to \(1.0 \mathrm{L}\) of a \(1.0-M \mathrm{HF}\) solution to begin precipitation of \(\mathrm{CaF}_{2}(s) ?\) For \(\mathrm{CaF}_{2}, K_{\mathrm{sp}}=\) \(4.0 \times 10^{-11}\) and \(K_{\mathrm{a}}\) for \(\mathrm{HF}=7.2 \times 10^{-4} .\) Assume no volume change on addition of \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}(s)\).

The overall formation constant for \(\mathrm{HgI}_{4}^{2-}\) is \(1.0 \times 10^{30}\) That is, $$ 1.0 \times 10^{30}=\frac{\left[\mathrm{HgI}_{4}^{2-}\right]}{\left[\mathrm{Hg}^{2+}\right]\left[\mathrm{I}^{-}\right]^{4}} $$ What is the concentration of \(\mathrm{Hg}^{2+}\) in \(500.0 \mathrm{mL}\) of a solution that was originally \(0.010\) \(M\) \(\mathrm{Hg}^{2+}\) and \(0.78\) \(M\) \(\mathrm{I}^{-} ?\) The reaction is $$\mathrm{Hg}^{2+}(a q)+4 \mathrm{I}^{-}(a q) \rightleftharpoons \mathrm{HgI}_{4}^{2-}(a q)$$
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