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Chapter 16: Problem 87

Like water, ammonia undergoes autoionization in liquid ammonia: $$\mathrm{NH}_{3}+\mathrm{NH}_{3} \rightleftharpoons \mathrm{NH}_{4}^{+}+\mathrm{NH}_{2}^{-}$$ (a) Identify the Bronsted acids and Brønsted bases in this reaction. (b) What species correspond to \(\mathrm{H}^{+}\) and \(\mathrm{OH}^{-},\) and what is the condition for a neutral solution?

Short Answer

Expert verified
In the given autoionization of ammonia: Bronsted acid is NH3 giving out a proton to form NH4+ (the conjugate acid). Bronsted base is also NH3 that accepts a proton to form NH2- (the conjugate base). The species corresponding to H+ is NH4+ and to OH- is NH2-. The condition for a neutral solution is that the concentrations of NH4+ and NH2- are equal, thus securing dynamic equilibrium.

Step by step solution

01

Identify the Bronsted Acids and Bases

In the given equilibrium reaction, one molecule of ammonia (\(\mathrm{NH}_{3}\)) is donating a proton (H\(^{+}\)); so it is acting as a Bronsted acid. The resulting ion, the ammonium ion (\(\mathrm{NH}_{4}^{+}\)), is the conjugate acid. The other molecule of ammonia is accepting a proton (H\(^+\)); hence it acts as a Bronsted base. The resulting ion, the amide ion (\(\mathrm{NH}_{2}^{-}\)), is the conjugate base.
02

Identify the Species Corresponding to H+ and OH-

The species corresponding to \(\mathrm{H^{+}}\) (Hydronium ion) is \(\mathrm{NH_{4}^{+}}\) (Ammonium ion) as it is the ion produced after NH3 (the Bronsted acid) donates a proton. The species corresponding to \(\mathrm{OH^{-}}\) (Hydroxide ion) is \(\mathrm{NH_{2}^{-}}\) (Amide ion) as the Bronsted base (NH3) accepts a proton to become NH2-.
03

Condition for a Neutral Solution

A solution of ammonia will be neutral when the concentrations of the ammonium ions (\(\mathrm{NH_{4}^{+}}\)) and the amide ions (\(\mathrm{NH_{2}^{-}}\)) are equal. This is because in a neutral solution, the amount of acidic species (ammonium ions) must be equal to the amount of basic species (amide ions). This condition represents a state of dynamic equilibrium in the solution.

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Most popular questions from this chapter

Identify the acid-base conjugate pairs in each of these reactions: (a) \(\mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{HCN} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COOH}+\mathrm{CN}^{-}\) (b) \(\mathrm{HCO}_{3}^{-}+\mathrm{HCO}_{3}^{-} \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}_{3}+\mathrm{CO}_{3}^{2-}\) (c) \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}+\mathrm{NH}_{3} \rightleftharpoons \mathrm{HPO}_{4}^{2-}+\mathrm{NH}_{4}^{+}\) (d) \(\mathrm{HClO}+\mathrm{CH}_{3} \mathrm{NH}_{2} \rightleftharpoons \mathrm{CH}_{3} \mathrm{NH}_{3}^{+}+\mathrm{ClO}^{-}\) (e) \(\mathrm{CO}_{3}^{2-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{HCO}_{3}^{-}+\mathrm{OH}^{-}\) (f) \(\mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COOH}+\mathrm{OH}^{-}\)

What are the names and formulas of the conjugate bases of these acids: (a) \(\mathrm{HNO}_{2},\) (b) \(\mathrm{H}_{2} \mathrm{SO}_{4},\) (c) \(\mathrm{H}_{2} \mathrm{~S},\) (d) HCN, (e) HCOOH (formic acid)?

Write an equation relating \(\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) in solution at \(25^{\circ} \mathrm{C}\).

Teeth enamel is largely hydroxyapatite \(\left[\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH}\right]\). When it dissolves in water (a process called demineralization), it dissociates as follows: $$\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH} \longrightarrow 5 \mathrm{Ca}^{2+}+3 \mathrm{PO}_{4}^{3-}+\mathrm{OH}^{-}$$ The reverse process, called remineralization, is the body's natural defense against tooth decay. Acids produced from food remove the \(\mathrm{OH}^{-}\) ions and thereby weaken the enamel layer. Most toothpastes contain a flouride compound such as \(\mathrm{NaF}\) or \(\mathrm{SnF}_{2}\). What is the function of these compounds in preventing tooth decay?

Predict the direction that predominates in this reaction: $$\mathrm{F}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{HF}(a q)+\mathrm{OH}^{-}(a q)$$
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