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Chapter 16: Problem 71

For each reaction draw a Lewis structure for each species and indicate which is the acid and which is the base: (a) \(\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{H}_{2} \mathrm{CO}_{3}\) (b) \(\mathrm{H}_{2} \mathrm{O}+\mathrm{BF}_{3} \longrightarrow \mathrm{H}_{2} \mathrm{OBF}_{3}\) (c) \(\mathrm{O}^{2-}+\mathrm{H}_{2} \mathrm{O} \longrightarrow 2 \mathrm{OH}^{-}\) (d) \(\mathrm{S}^{2-}+\mathrm{SO}_{3} \longrightarrow \mathrm{S}_{2} \mathrm{O}_{3}^{2-}\)

Short Answer

Expert verified
In reaction (a), \(\mathrm{CO}_{2}\) is base and \(\mathrm{H}_{2} \mathrm{O}\) is acid .In reaction (b), \(\mathrm{BF}_{3}\) is the acid and \(\mathrm{H}_{2} \mathrm{O}\) is the base. In reaction (c), \(\mathrm{O}^{2-}\) is the base and \(\mathrm{H}_{2} \mathrm{O}\) is the acid. In reaction (d), \(\mathrm{SO}_{3}\) is the acid and \(\mathrm{S}^{2-}\) is the base.

Step by step solution

01

Draw Lewis Structures and Identify Acid and Base for Reaction (a)

Firstly, one has to draw Lewis structures for \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\). In \(\mathrm{CO}_{2}\), carbon is in the middle with double bonds to each oxygen. For water, the oxygen has two lone pairs and is bonded to two hydrogen atoms. In the balanced reaction, it is seen that \(\mathrm{CO}_{2}\) gained two hydrogen ions. This means \(\mathrm{H}_{2} \mathrm{O}\) donated hydrogen ions, therefore it is the acid and \(\mathrm{CO}_{2}\) is the base.
02

Draw Lewis Structures and Identify Acid and Base for Reaction (b)

Draw Lewis structures for \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{BF}_{3}\). In \(\mathrm{BF}_{3}\), Boron is in the center with three bonding to three fluorine atoms. For water, the oxygen has two lone pairs and is bonded to two hydrogen atoms. In the reaction, \(\mathrm{BF}_{3}\) accepts a lone pair of electrons from \(\mathrm{H}_{2} \mathrm{O}\), making it the acid and \(\mathrm{H}_{2} \mathrm{O}\) the base.
03

Draw Lewis Structures and Identify Acid and Base for Reaction (c)

Draw Lewis structures for \(\mathrm{O}^{2-}\) and \(\mathrm{H}_{2} \mathrm{O}\). The oxide anion has six electrons on the outer shell and accepts two hydrogen ions become water molecules. It is therefore the base and \(\mathrm{H}_{2} \mathrm{O}\) is the acid.
04

Draw Lewis Structures and Identify Acid and Base for Reaction (d)

Draw Lewis structures for \(\mathrm{S}^{2-}\) and \(\mathrm{SO}_{3}\). In this reaction, sulphur ion gains oxygen atoms from \(\mathrm{SO}_{3}\), making \(\mathrm{SO}_{3}\) the acid (donating oxygen) and \(\mathrm{S}^{2-}\) the base (accepting oxygen).

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Most popular questions from this chapter

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) \(\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{O}+\mathrm{BF}_{3} \longrightarrow\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{OBF}_{3}\)

A 28.2 L volume of \(\mathrm{HCl}(\mathrm{g}),\) measured at \(742 \mathrm{mmHg}\) and \(25.0^{\circ} \mathrm{C},\) is dissolved in water. What volume of \(\mathrm{NH}_{3}(\mathrm{g}),\) measured at \(762 \mathrm{mmHg}\) and \(21.0^{\circ} \mathrm{C},\) must be absorbed by the same solution to neutralize the HCl?

What is the pH of the solution obtained by mixing \(24.80 \mathrm{mL}\) of \(0.248 \mathrm{M} \mathrm{HNO}_{3}\) and \(15.40 \mathrm{mL}\) of \(0.394 \mathrm{M}\) \(\space\) \(KOH?\)

Here is a way to test the validity of the statement made on page 719 in conjunction with the three key ideas governing the ionization of polyprotic acids. Determine the \(\mathrm{pH}\) of \(0.100 \mathrm{M}\) succinic acid in two ways: first by assuming that \(\mathrm{H}_{3} \mathrm{O}^{+}\) is produced only in the first ionization step, and then by allowing for the possibility that some \(\mathrm{H}_{3} \mathrm{O}^{+}\) is also produced in the second ionization step. Compare the results, and discuss the significance of your finding. $$\begin{array}{c} \mathrm{H}_{2} \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{4}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{HC}_{4} \mathrm{H}_{4} \mathrm{O}_{4}^{-} \\ &K_{\mathrm{a}_{1}}=6.2 \times 10^{-5} \\ \mathrm{HC}_{4} \mathrm{H}_{4} \mathrm{O}_{4}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{4}^{2-} \\ &K_{\mathrm{a}_{2}}=2.3 \times 10^{-6} \end{array}$$

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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