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Chapter 16: Problem 3

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) $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}+\mathrm{OH}^{-} \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}+\mathrm{H}_{2} \mathrm{O}$

Short Answer

Expert verified
(a) \(\mathrm{HOBr}\) (acid), \(\mathrm{H}_{2} \mathrm{O}\) (base), \(\mathrm{OBr}^{-}\) (base), \(\mathrm{H}_{3} \mathrm{O}^{+}\) (acid); (b) \(\mathrm{HSO}_{4}^{-}\) (acid), \(\mathrm{H}_{2} \mathrm{O}\) (base), \(\mathrm{SO}_{4}^{2-}\) (base), \(\mathrm{H}_{3} \mathrm{O}^{+}\) (acid); (c) \(\mathrm{HS}^{-}\) (base), \(\mathrm{H}_{2} \mathrm{O}\) (acid), \(\mathrm{H}_{2} \mathrm{S}\) (acid), \(\mathrm{OH}^{-}\) (base); (d) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\) (acid), \(\mathrm{OH}^{-}\) (base), \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\) (base), \(\mathrm{H}_{2} \mathrm{O}\) (acid).

Step by step solution

01

Identify acids and bases for (a)

In the forward direction, \(\mathrm{HOBr}\) acts as the acid because it loses a proton to become \(\mathrm{OBr}^{-}\). \(\mathrm{H}_{2} \mathrm{O}\) is the base as it gains a proton from \(\mathrm{HOBr}\) to turn into \(\mathrm{H}_{3} \mathrm{O}^{+}\). In the reverse direction, \(\mathrm{OBr}^{-}\) gains a proton and thus acts as a base, and \(\mathrm{H}_{3} \mathrm{O}^{+}\) loses a proton to become \(\mathrm{H}_{2} \mathrm{O}\), acting as an acid.
02

Identify acids and bases for (b)

For the reaction \(\mathrm{HSO}_{4}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons\mathrm{SO}_{4}^{2-}+\mathrm{H}_{3} \mathrm{O}^{+}\), in the forward direction, \(\mathrm{HSO}_{4}^{-}\) is the acid as it donates a proton to \(\mathrm{H}_{2} \mathrm{O}\), thus forming \(\mathrm{SO}_{4}^{2-}\). \(\mathrm{H}_{2} \mathrm{O}\) is the base as it accepts this proton, turning into \(\mathrm{H}_{3} \mathrm{O}^{+}\). In the reverse direction, \(\mathrm{SO}_{4}^{2-}\) is the base and \(\mathrm{H}_{3} \mathrm{O}^{+}\) is the acid.
03

Identify acids and bases for (c)

The equation for part (c) is \(\mathrm{HS}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{2} \mathrm{S}+\mathrm{OH}^{-}\). In the forward direction, \(\mathrm{HS}^{-}\) is a base, as it gains a proton from \(\mathrm{H}_{2} \mathrm{O}\) forming \(\mathrm{H}_{2} \mathrm{S}\). \(\mathrm{H}_{2} \mathrm{O}\) is an acid, as it loses a proton. In the reverse direction, \(\mathrm{OH}^{-}\) is a base and \(\mathrm{H}_{2} \mathrm{S}\) is the acid.
04

Identify acids and bases for (d)

The reaction in part (d) is \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}+\mathrm{OH}^{-} \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}+\mathrm{H}_{2} \mathrm{O}\). Forward direction: \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\) is an acid since it donates a proton to \(\mathrm{OH}^{-}\) turning into \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\). \(\mathrm{OH}^{-}\) is a base in this case since it gains a proton. In the reverse direction: \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\) is a base, and \(\mathrm{H}_{2} \mathrm{O}\) is an acid.

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Most popular questions from this chapter

Approximately 4 metric tons of quinoline, \(\mathrm{C}_{9} \mathrm{H}_{7} \mathrm{N},\) is produced annually. The principal source of quinoline is coal tar. Quinoline is a weak base in water. A handbook gives \(K_{\mathrm{a}}=6.3 \times 10^{-10}\) for protonated quinoline, \(\mathrm{C}_{9} \mathrm{H}_{7} \mathrm{NH}^{+} .\) Write the ionization reaction for \(\mathrm{C}_{9} \mathrm{H}_{7} \mathrm{NH}^{+}\) and calculate \(\mathrm{p} \bar{K}_{\mathrm{b}}\) for quinoline.

Calculate \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) and \(\mathrm{pH}\) in saturated \(\mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{aq})\) which contains \(3.9 \mathrm{g} \mathrm{Ba}(\mathrm{OH})_{2} \cdot 8 \mathrm{H}_{2} \mathrm{O}\) per \(100 \mathrm{mL}\) of solution.

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 \(0.394 \mathrm{M}\) \(\space\) \(KOH?\)

Write the formula of the conjugate base in the reaction of each acid with water. (a) \(\mathrm{HIO}_{3} ;\) (b) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\) (c) \(\mathrm{HPO}_{4}^{2-} ;\) (d) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\)

Arrange the following 0.010 M solutions in order of increasing \(\mathrm{pH}: \mathrm{NH}_{3}(\mathrm{aq}), \mathrm{HNO}_{3}(\mathrm{aq}), \mathrm{NaNO}_{2}(\mathrm{aq})\) \(\mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq}), \quad \mathrm{NaOH}(\mathrm{aq}), \quad \mathrm{NH}_{4} \mathrm{CH}_{3} \mathrm{COO}(\mathrm{aq})\) \(\mathrm{NH}_{4} \mathrm{ClO}_{4}(\mathrm{aq})\)
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