Problem 34 - Chemistry: The Central Science

Question

State whether each of the following statements is true or false. Justify your answer in each case. (a) \(\mathrm{NH}_{3}\) contains no \(\mathrm{OH}^{-}\)ions, and yet its aqueous solutions are basic. (b) HF is a strong acid. (c) Although sulfuric acid is a strong electrolyte, an aqueous solution of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) contains more \(\mathrm{HSO}_{4}^{-}\)ions than \(\mathrm{SO}_{4}{\underline{\phantom{xx}}}^{2-}\) ions.

Step-by-Step Solution

Answer

(a) True. \(\mathrm{NH}_{3}\) contains no \(\mathrm{OH}^{-}\) ions, but when dissolved in water, it behaves as a weak base, accepting a proton from a water molecule to form \(\mathrm{NH}_{4}^{+}\) and \(\mathrm{OH}^{-}\) ions, making the solution basic. (b) False. HF is a weak acid, not a strong one. When it dissolves in water, it only partially dissociates into \(\mathrm{H}^+\) and \( \mathrm{F}^-\) ions. (c) True. In an aqueous solution of \(\mathrm{H}_{2}\mathrm{SO}_{4}\), there will be more \(\mathrm{HSO}_{4}^-\) ions than \(\mathrm{SO}_{4}^{2-}\) ions due to its dissociation chemistry, despite it being a strong electrolyte.

Step 1: (Statement a: Analysis)

For Statement (a), we need to determine if it's true that \(\mathrm{NH}_3\) contains no \(\mathrm{OH}^-\) ions, yet its aqueous solutions are basic. To assess the statement, we must analyze the behavior of \(\mathrm{NH}_3\) in water and its resulting pH. #tag_end#

Step 2: (Statement a: Resolution)

Statement (a) is true. \(\mathrm{NH}_{3}\) itself contains no \(\mathrm{OH}^{-}\) ions, as its formula indicates. However, when dissolved in water, it behaves as a weak base, accepting a proton from a water molecule to form \(\mathrm{NH}_{4}^{+}\) and \(\mathrm{OH}^{-}\) ions. The reaction can be described as follows: \[\mathrm{NH}_{3}(aq) + \mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{NH}_{4}^{+}(aq) + \mathrm{OH}^{-}(aq)\] As a result, the \(\mathrm{OH}^{-}\) ions produced increase the pH of the solution, making it basic. #tag_end#

Step 3: (Statement b: Analysis)

For Statement (b), we need to determine whether HF is a strong acid or not. A strong acid completely dissociates in water, while a weak acid only partially dissociates. We can evaluate this statement by considering the degree of dissociation of HF in water. #tag_end#

Step 4: (Statement b: Resolution)

Statement (b) is false. HF is a weak acid, not a strong one. When it dissolves in water, it only partially dissociates into \(\mathrm{H}^+\) and \( \mathrm{F}^-\) ions. Its equilibrium reaction can be written as: \[ \mathrm{HF}(aq) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{F}^{-}(aq)\] Since the reaction does not proceed to completion and there is only a partial dissociation of HF, it is considered a weak acid. #tag_end#

Step 5: (Statement c: Analysis)

For Statement (c), we need to determine whether an aqueous solution of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) contains more \(\mathrm{HSO}_{4}^-\) ions than \(\mathrm{SO}_{4}^{2-}\) ions, despite sulfuric acid being a strong electrolyte. To evaluate this statement, we should examine the dissociation chemistry of sulfuric acid in water. #tag_end#

Step 6: (Statement c: Resolution)

Statement (c) is true. Sulfuric acid (\(\mathrm{H}_{2}\mathrm{SO}_{4}\)) is a strong electrolyte and a strong acid, which means that it dissociates completely in water. However, it is also a diprotic acid, which means it can lose two protons in successive steps. The dissociation reactions of sulfuric acid can be written as: \[\mathrm{H}_{2}\mathrm{SO}_{4}(aq) \rightarrow \mathrm{H}^{+}(aq) + \mathrm{HSO}_{4}^{-}(aq)\] \[\mathrm{HSO}_{4}^{-}(aq) \rightarrow \mathrm{H}^{+}(aq) + \mathrm{SO}_{4}^{2-}(aq)\] The first reaction goes to completion, while the second reaction is an equilibrium, meaning that not all \(\mathrm{HSO}_{4}^{-}\) ions will dissociate to form \(\mathrm{SO}_{4}^{2-}\) ions. Because of this, there will be more \(\mathrm{HSO}_{4}^-\) ions than \(\mathrm{SO}_{4}^{2-}\) ions in an aqueous solution of \(\mathrm{H}_{2}\mathrm{SO}_{4}\). #tag_end#

Key Concepts

Ammonia Solutions

Ammonia (\(\mathrm{NH}_3\)) might seem like an unusual candidate for a base because it does not directly contain hydroxide ions (\(\mathrm{OH}^-\)). However, in an aqueous solution, ammonia exhibits basic properties. When ammonia is dissolved in water, it undergoes a reaction with water molecules.
This reaction results in the formation of ammonium ions (\(\mathrm{NH}_4^+\)) and hydroxide ions (\(\mathrm{OH}^-\)): \[\mathrm{NH}_{3}(aq) + \mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{NH}_{4}^{+}(aq) + \mathrm{OH}^{-}(aq)\]

• The hydroxide ions formed in the solution increase the pH, making the solution basic.
• The production of \(\mathrm{OH}^-\) ions is what gives ammonia its basic characteristics despite its initial composition.

This partial ionization is why ammonia is considered a weak base, as not all ammonia molecules engage in this transformation.

Hydrofluoric Acid

Hydrofluoric acid (HF) is often misunderstood due to its acidity and powerful reactivity, but it's classified as a weak acid when it comes to its dissociation in water. A strong acid is characterized by complete dissociation in aqueous solutions.
In contrast, HF only partially dissociates into hydrogen ions (\(\mathrm{H}^+\)) and fluoride ions (\(\mathrm{F}^-\)):\[\mathrm{HF}(aq) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{F}^{-}(aq)\]

• The equilibrium nature of this reaction means that not all HF molecules will dissociate.
• This results in a relatively high concentration of non-dissociated HF molecules remaining in the solution.

Because of this partial dissociation, HF does not fully release its protons, and thus it's considered a weak acid. Despite being a weak acid, HF can be quite dangerous and corrosive, so handle with care.

Sulfuric Acid Dissociation

Sulfuric acid (\(\mathrm{H}_{2}\mathrm{SO}_{4}\)) is a strong acid and also known as a strong electrolyte due to its ability to completely dissociate in water in its first dissociation step. Being a diprotic acid, it loses protons in two stages:

First dissociation:\[\mathrm{H}_{2}\mathrm{SO}_{4}(aq) \rightarrow \mathrm{H}^{+}(aq) + \mathrm{HSO}_{4}^{-}(aq)\]This reaction goes to completion, meaning almost all \(\mathrm{H}_{2}\mathrm{SO}_{4}\) molecules will dissociate fully to form \(\mathrm{HSO}_{4}^{-}\) ions and \(\mathrm{H}^{+}\) ions.

Second dissociation:\[\mathrm{HSO}_{4}^{-}(aq) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{SO}_{4}^{2-}(aq)\]

• The second reaction does not proceed to completion and is at equilibrium, therefore not all \(\mathrm{HSO}_{4}^{-}\) ions convert to \(\mathrm{SO}_{4}^{2-}\) ions.
• As a result, an aqueous solution of sulfuric acid contains more \(\mathrm{HSO}_{4}^{-}\) ions compared to \(\mathrm{SO}_{4}^{2-}\) ions.

Thus, the two-step dissociation impacts the ion distribution, with the presence of both dissociated and non-dissociated ions.