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Chapter 17: Problem 85

Both \(\mathrm{Ag}^{+}\) and \(\mathrm{Zn}^{2+}\) form complex ions with \(\mathrm{NH}_{3}\). Write balanced equations for the reactions. However, \(\mathrm{Zn}(\mathrm{OH})_{2}\) is soluble in \(6 \mathrm{M} \mathrm{NaOH},\) and \(\mathrm{AgOH}\) is not. Explain.

Short Answer

Expert verified
The equations for the reactions are \(\mathrm{Ag}^{+} + 2\mathrm{NH}_{3} \rightarrow \mathrm{[Ag(NH_{3})_{2}]}^{+} \) and \(\mathrm{Zn}^{2+} + 4\mathrm{NH}_{3} \rightarrow \mathrm{[Zn(NH_{3})_{4}]}^{2+} \). \(\mathrm{Zn(OH)}_{2}\) is soluble in \(6 \mathrm{M} \mathrm{NaOH}\), and \(\mathrm{AgOH}\) is not because \(\mathrm{Zn(OH)}_{2}\) can form an soluble complex ion \(\mathrm{[Zn(OH)_{4}]}^{2-} \) with excess \(\mathrm{OH}^{-}\) ions whereas \(\mathrm{AgOH}\) cannot.

Step by step solution

01

Writing Balanced Equations

The reaction of \(\mathrm{Ag}^{+}\) and \(\mathrm{Zn}^{2+}\) with \(\mathrm{NH}_{3}\) can be written as follows: \[\mathrm{Ag}^{+} + 2\mathrm{NH}_{3} \rightarrow \mathrm{[Ag(NH_{3})_{2}]}^{+} \]\[\mathrm{Zn}^{2+} + 4\mathrm{NH}_{3} \rightarrow \mathrm{[Zn(NH_{3})_{4}]}^{2+} \]
02

Understanding solubility rules

The hydroxides of transition metals and post-transition metals are generally insoluble, but certain hydroxides become soluble in the presence of excess ions. A precipitate of \(\mathrm{Zn(OH)}_{2}\) dissolves in excess \(\mathrm{OH}^{-}\) ions to form soluble complex ions. However, \(\mathrm{AgOH}\) does not dissolve because the formation of soluble complex ions with \(\mathrm{OH}^{-}\) ions is not possible.
03

Writing Solubility Equations

The solubility of \(\mathrm{Zn(OH)}_{2}\) can be written as: \[\mathrm{Zn(OH)}_{2} + 2\mathrm{OH}^{-} \rightarrow \mathrm{[Zn(OH)_{4}]}^{2-} \] This shows that \(\mathrm{Zn(OH)}_{2}\) dissolves in the presence of excess \(\mathrm{OH}^{-}\) ions. Since no such reaction exists for \(\mathrm{AgOH}\), it remains insoluble even in the presence of \(\mathrm{OH}^{-}\) ions.

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