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Chapter 16: Problem 21

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

Short Answer

Expert verified
In short, silver chloride (AgCl) dissolves readily in 2M ammonia solution (NH₃) due to the formation of a soluble complex ion, \([Ag(NH_3)_2]^{+}\), which increases its solubility. However, in a 2M ammonium nitrate solution (NH₄NO₃), the ammonium ions compete with silver ions for the ammonia, reducing the concentration of free ammonia and decreasing the solubility of AgCl.

Step by step solution

01

Write the solubility equilibrium for silver chloride in water

First, we need to write the solubility equilibrium for silver chloride dissolving in water. Silver chloride is a sparingly soluble salt that, when it dissolves in water, releases silver ions (Ag⁺) and chloride ions (Cl⁻) as follows: \[AgCl_{(s)} \rightleftharpoons Ag^{+}_{(aq)} + Cl^{-}_{(aq)}\]
02

Explain the interaction between silver chloride and ammonia in solution

In the presence of ammonia, the solubility of silver chloride increases due to the formation of a soluble complex ion with ammonia. The complex ion that forms is: \([Ag(NH_3)_2]^{+}\). The reaction between silver ions and ammonia is represented as: \[Ag^{+}_{(aq)} + 2NH_3_{(aq)} \rightleftharpoons [Ag(NH_{3})_{2}]^{+}_{(aq)}\] By Le Chatelier's principle, adding more ammonia to the solution drives this equilibrium to the formation of more complex ions, increasing the solubility of AgCl.
03

Explain the interaction between silver chloride and ammonium nitrate in solution

Ammonium nitrate is a salt of ammonium cation (NH₄⁺) and nitrate anion (NO₃⁻). When it dissolves in water, it releases ammonium ions and nitrate ions as follows: \[NH_4NO_3_{(s)} \rightarrow NH_4^{+}_{(aq)} + NO_3^{-}_{(aq)}\] Ammonium ions in the solution can compete with silver ions for the ammonia. The reaction between ammonium ions and ammonia is represented as: \[NH^{+}_4_{(aq)} + NH_3_{(aq)} \rightleftharpoons [NH_4(NH_{3})]^{+}_{(aq)}\] This reaction decreases the concentration of free ammonia in the solution and, by Le Chatelier's principle, shifts the silver-ammonia equilibrium back to the left, decreasing the solubility of AgCl.
04

Sum up the explanation

In summary, silver chloride dissolves readily in 2M ammonia solution due to the formation of a soluble complex ion, \([Ag(NH_3)_2]^{+}\), which increases its solubility. On the other hand, when AgCl is in a 2M ammonium nitrate solution, the ammonium ions in the solution compete with silver ions for the ammonia, decreasing the concentration of free ammonia and consequently reducing the solubility of AgCl.

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Most popular questions from this chapter

\(\mathrm{Ag}_{2} \mathrm{S}(s)\) has a larger molar solubility than CuS even though \(\mathrm{Ag}_{2} \mathrm{S}\) has the smaller \(K_{\mathrm{sp}}\) value. Explain how this is possible.

a. Calculate the molar solubility of \(\mathrm{AgBr}\) in pure water. \(K_{\mathrm{sp}}\) for \(\mathrm{AgBr}\) is \(5.0 \times 10^{-13}\) . b. Calculate the molar solubility of \(\mathrm{AgBr}\) in \(3.0M\) \(\mathrm{NH}_{3} .\) 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 \(\mathrm{AgBr}\) 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?

Calculate the solubility (in moles per liter) of \(\mathrm{Fe}(\mathrm{OH})_{3}\) $\left(K_{\mathrm{sp}}=4 \times 10^{-38}\right)$ in each of the following. a. water b. a solution buffered at pH \(=5.0\) c. a solution buffered at pH\(=11.0\)

The equilibrium constant for the following reaction is \(1.0 \times 10^{23} :\) $$\mathrm{Cr}^{3+}(a q)+\mathrm{H}_{2} \mathrm{EDTA}^{2-}(a q) \rightleftharpoons \mathrm{CrEDTA}^{-}(a q)+2 \mathrm{H}^{+}(a q)$$ EDTA is used as a complexing agent in chemical analysis. Solutions of EDTA, usually containing the disodium salt $\mathrm{Na}_{2} \mathrm{H}_{2} \mathrm{EDTA}$ , are used to treat heavy metal poisoning. Calculate \(\left[\mathrm{Cr}^{3+}\right]\) at equilibrium in a solution originally \(0.0010 M\) in \(\mathrm{Cr}^{3+}\) and \(0.050 M\) in $\mathrm{H}_{2} \mathrm{EDTA}^{2-}\( and buffered at \)\mathrm{pH}=6.00.$

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 \(\mathrm{AgI?}\)
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