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Chapter 16: Problem 80

Use Lewis structures to diagram the following reaction in the manner of reaction (16.19) $$2 \mathrm{NH}_{3}+\mathrm{Ag}^{+} \longrightarrow\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}$$ Identify the Lewis acid and Lewis base.

Short Answer

Expert verified
The Lewis acid is the Silver ion (Ag+) and the Lewis base is the ammonia (NH3). The Lewis acid (Ag+) accepts an electron pair from the Lewis base (NH3) to form a complex ion [Ag(NH3)2]+

Step by step solution

01

Determine the Lewis Structures

Firstly, Lewis structures should be drawn for the reactants and products. The nitrogen atom in ammonia (NH3) has a lone pair of electrons that can be donated, and Ag+ is a positively charged ion that can accept a pair of electrons to achieve a stable electron configuration.
02

Identify the Lewis Acid

A Lewis acid is defined as an atom, ion or molecule that accepts an electron pair to form a covalent bond. In this case, the silver ion \(Ag^{+}\) acts as a Lewis acid because it accepts a pair of electrons from ammonia.
03

Identify the Lewis Base

A Lewis base is an atom, ion or molecule that donates an electron pair to form a covalent bond. In this case, the nitrogen in the ammonia (NH3) is the Lewis base, because it donates its lone pair of electrons.
04

Visualize the Reaction

The reaction is visualized as two ammonia (NH3) molecules donates their lone pair of electrons to the silver ion (Ag+) to form a coordination compound. The Lewis base is the electron-donating species (NH3), and the Lewis acid is the electron-accepting species (Ag+).

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Most popular questions from this chapter

What is the (a) degree of ionization and (b) percent ionization of propionic acid in a solution that is \(0.45 \mathrm{M}\) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H} ?\) $$\begin{aligned} &\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CO}_{2}^{-}\\\ &&\mathrm{p} K_{\mathrm{a}}=4.89 \end{aligned}$$

\(3.00 \mathrm{mol}\) of calcium chlorite is dissolved in enough water to produce 2.50 L of solution. \(K_{\mathrm{a}}=2.9 \times 10^{-8}\) for \(\mathrm{HClO}\), and \(K_{\mathrm{a}}=1.1 \times 10^{-2}\) for \(\mathrm{HClO}_{2}\). Compute the pH of the solution.

For each of the following, identify the acids and bases involved in both the forward and reverse directions. (a) $\mathrm{HOBr}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{OBr}^{-}+\mathrm{H}_{3} \mathrm{O}^{+}$ (b) $\mathrm{HSO}_{4}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{SO}_{4}^{2-}+\mathrm{H}_{3} \mathrm{O}^{+}$ (c) $\mathrm{HS}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{2} \mathrm{S}+\mathrm{OH}^{-}$ (d) $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}+\mathrm{OH}^{-} \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}+\mathrm{H}_{2} \mathrm{O}$

Codeine, \(\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{O}_{3} \mathrm{N},\) is an opiate, has analgesic and antidiarrheal properties, and is widely used. In water, codeine is a weak base. A handbook gives \(\mathrm{p} K_{\mathrm{a}}=6.05\) for protonated codeine, \(\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{O}_{3} \mathrm{NH}^{+} .\) Write the reaction for \(\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{O}_{3} \mathrm{NH}^{+}\) and calculate \(\mathrm{p} K_{\mathrm{b}}\) for codeine.

Explain why \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) in a strong acid solution doubles as the total acid concentration doubles, whereas in a weak acid solution, \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) increases only by about a factor of \(\sqrt{2}\)
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