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Chapter 16: Problem 75

The three following reactions are acid-base reactions according to the Lewis theory. Draw Lewis structures, and identify the Lewis acid and Lewis base in each reaction. (a) \(\mathrm{B}(\mathrm{OH})_{3}+\mathrm{OH}^{-} \longrightarrow\left[\mathrm{B}(\mathrm{OH})_{4}\right]^{-}\) (b) \(\mathrm{N}_{2} \mathrm{H}_{4}+\mathrm{H}_{3} \mathrm{O}^{+} \longrightarrow \mathrm{N}_{2} \mathrm{H}_{5}^{+}+\mathrm{H}_{2} \mathrm{O}\) (c) \(\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{O}+\mathrm{BF}_{3} \longrightarrow\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{OBF}_{3}\)

Short Answer

Expert verified
For reaction (a), \(\mathrm{B}(\mathrm{OH})_{3}\) is the Lewis acid and \(\mathrm{OH}^{-}\) is the Lewis base. For reaction (b), \(\mathrm{H}_{3} \mathrm{O}^{+}\) is the Lewis acid and \(\mathrm{N}_{2} \mathrm{H}_{4}\) is the Lewis base. For reaction (c), \(\mathrm{BF}_{3}\) is the Lewis acid and \(\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{O}\) is the Lewis base.

Step by step solution

01

Chemical Reaction A - Lewis Structures and Identification

Given the reaction \(\mathrm{B}(\mathrm{OH})_{3}+\mathrm{OH}^{-} \longrightarrow\left[\mathrm{B}(\mathrm{OH})_{4}\right]^{-}\), \(\mathrm{B}(\mathrm{OH})_{3}\) becomes \(\left[\mathrm{B}(\mathrm{OH})_{4}\right]^{-}\), and thus accepts an electron pair from \(\mathrm{OH}^{-}\), so it is the Lewis acid. \(\mathrm{OH}^{-}\) donates an electron pair, so it's the Lewis base.
02

Chemical Reaction B - Lewis Structures and Identification

Given the reaction \(\mathrm{N}_{2} \mathrm{H}_{4}+\mathrm{H}_{3} \mathrm{O}^{+} \longrightarrow \mathrm{N}_{2} \mathrm{H}_{5}^{+}+\mathrm{H}_{2} \mathrm{O}\), \(\mathrm{H}_{3} \mathrm{O}^{+}\) becomes \(\mathrm{H}_{2} \mathrm{O}\) by getting an electron from \(\mathrm{N}_{2} \mathrm{H}_{4}\). So, \(\mathrm{H}_{3} \mathrm{O}^{+}\) is the Lewis acid while \(\mathrm{N}_{2} \mathrm{H}_{4}\) is the Lewis base.
03

Chemical Reaction C - Lewis Structures and Identification

Given the reaction \(\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{O}+\mathrm{BF}_{3} \longrightarrow\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{OBF}_{3}\), \(\mathrm{BF}_{3}\) becomes \(\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{OBF}_{3}\) by accepting a pair of electrons. So, \(\mathrm{BF}_{3}\) is the Lewis acid, while \(\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{O}\) by donating a pair of electrons is the Lewis base.

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Most popular questions from this chapter

What is the \(\mathrm{pH}\) of the solution obtained when \(125 \mathrm{mL}\) of \(0.606 \mathrm{M} \mathrm{NaOH}\) is diluted to \(15.0 \mathrm{L}\) with water?

Caproic acid, \(\mathrm{HC}_{6} \mathrm{H}_{11} \mathrm{O}_{2},\) found in small amounts in coconut and palm oils, is used in making artificial flavors. A saturated aqueous solution of the acid contains \(11 \mathrm{g} / \mathrm{L}\) and has \(\mathrm{pH}=2.94 .\) Calculate \(K_{\mathrm{a}}\) for the acid. $$\mathrm{HC}_{6} \mathrm{H}_{11} \mathrm{O}_{2}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{C}_{6} \mathrm{H}_{11} \mathrm{O}_{2}^{-} \quad K_{\mathrm{a}}=?$$

The solubility of 1 -naphthylamine, \(\mathrm{C}_{10} \mathrm{H}_{7} \mathrm{NH}_{2}, \mathrm{a}\) substance used in the manufacture of dyes, is given in a handbook as 1 g per \(590 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\). What is the approximate \(\mathrm{pH}\) of a saturated aqueous solution of 1-naphthylamine? $$\begin{array}{r} \mathrm{C}_{10} \mathrm{H}_{7} \mathrm{NH}_{2}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{C}_{10} \mathrm{H}_{7} \mathrm{NH}_{3}^{+}+\mathrm{OH}^{-} \\ \mathrm{p} K_{\mathrm{b}}=3.92 \end{array}$$

Suppose you wanted to produce an aqueous solution of \(\mathrm{pH}=8.65\) by dissolving one of the following salts in water. Which salt would you use, and at what molarity? (a) \(\mathrm{NH}_{4} \mathrm{Cl} ;\) (b) \(\mathrm{KHSO}_{4} ;\) (c) \(\mathrm{KNO}_{2}\); (d) \(\mathrm{NaNO}_{3}\).

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