Question

Using the Bronsted-Lowry model, write an equation to show why each of the following species produces a basic aqueous solution. (a) \(\mathrm{NH}_{3}\) (b) \(\mathrm{NO}_{2}^{-}\) (c) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\) (d) \(\mathrm{CO}_{3}^{2-}\) (e) \(\mathrm{F}^{-}\) (f) \(\mathrm{HCO}_{3}^{-}\)

Short answer

Question: Explain why each of the following species produces a basic aqueous solution using the Bronsted-Lowry model: (a) \(\mathrm{NH}_{3}\), (b) \(\mathrm{NO}_{2}^{-}\), (c) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\), (d) \(\mathrm{CO}_{3}^{2-}\), (e) \(\mathrm{F}^{-}\), and (f) \(\mathrm{HCO}_{3}^{-}\). Answer: In the Bronsted-Lowry model, a base accepts a proton (\(\mathrm{H^{+}}\)) from a proton donor, such as water. The following reactions demonstrate how each of the given species acts as a base by accepting a proton from water and producing hydroxide ions (\(\mathrm{OH}^{-}\)): (a) \(\mathrm{NH}_{3}(aq) + \mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{NH}_{4}^{+}(aq) + \mathrm{OH}^{-}(aq)\) (b) \(\mathrm{NO}_{2}^{-}(aq) + \mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{HNO}_{2}(aq) + \mathrm{OH}^{-}(aq)\) (c) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}(aq) + \mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}(aq) + \mathrm{OH}^{-}(aq)\) (d) \(\mathrm{CO}_{3}^{2-}(aq) + \mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{HCO}_{3}^{-}(aq) + \mathrm{OH}^{-}(aq)\) (e) \(\mathrm{F}^{-}(aq) + \mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{HF}(aq) + \mathrm{OH}^{-}(aq)\) (f) \(\mathrm{HCO}_{3}^{-}(aq) + \mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2}\mathrm{CO}_{3}(aq) + \mathrm{OH}^{-}(aq)\)

Step-by-step solution

(a) \(\mathrm{NH}_{3}\) as a base in aqueous solution

To show why \(\mathrm{NH}_{3}\) produces a basic aqueous solution, we need to write a reaction between \(\mathrm{NH}_{3}\) and water. In this reaction, \(\mathrm{NH}_{3}\) should act as a proton acceptor and water should act as a proton donor, forming \(\mathrm{NH}_{4}^{+}\) and \(\mathrm{OH}^{-}\): \(\mathrm{NH}_{3}(aq) + \mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{NH}_{4}^{+}(aq) + \mathrm{OH}^{-}(aq)\)

(b) \(\mathrm{NO}_{2}^{-}\) as a base in aqueous solution

In a reaction between \(\mathrm{NO}_{2}^{-}\) and water, \(\mathrm{NO}_{2}^{-}\) will act as a proton acceptor and water will act as a proton donor. The products will be \(\mathrm{HNO}_{2}\) and \(\mathrm{OH}^{-}\): \(\mathrm{NO}_{2}^{-}(aq) + \mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{HNO}_{2}(aq) + \mathrm{OH}^{-}(aq)\)

(c) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\) as a base in aqueous solution

Like ammonia, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\), also known as aniline, can act as a proton acceptor in its reaction with water. Water will act as a proton donor, forming the corresponding conjugate acid and hydroxide ion: \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}(aq) + \mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}(aq) + \mathrm{OH}^{-}(aq)\)

(d) \(\mathrm{CO}_{3}^{2-}\) as a base in aqueous solution

The carbonate ion (\(\mathrm{CO}_{3}^{2-}\)) can act as a proton acceptor in its reaction with water. Water will act as a proton donor, forming hydrogen carbonate and hydroxide ion: \(\mathrm{CO}_{3}^{2-}(aq) + \mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{HCO}_{3}^{-}(aq) + \mathrm{OH}^{-}(aq)\)

(e) \(\mathrm{F}^{-}\) as a base in aqueous solution

The fluoride ion (\(\mathrm{F}^{-}\)) can act as a proton acceptor in its reaction with water. Water will act as a proton donor, forming a hydrofluoric acid and hydroxide ion: \(\mathrm{F}^{-}(aq) + \mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{HF}(aq) + \mathrm{OH}^{-}(aq)\)

(f) \(\mathrm{HCO}_{3}^{-}\) as a base in aqueous solution

The hydrogen carbonate ion (\(\mathrm{HCO}_{3}^{-}\)) can act as a proton acceptor in its reaction with water. Water will act as a proton donor, forming carbonic acid and hydroxide ion: \(\mathrm{HCO}_{3}^{-}(aq) + \mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2}\mathrm{CO}_{3}(aq) + \mathrm{OH}^{-}(aq)\) In each of these reactions, the species in question accepts a proton from water and produces hydroxide ions, which is the characteristic of a base according to the Bronsted-Lowry model.