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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 \(\mathrm{SO}_{4}{ }^{2-}\) ions.

Short Answer

Expert verified
(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 by step solution

01

(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#
02

(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#
03

(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#
04

(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#
05

(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#
06

(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#

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

(a) How many grams of ethanol, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\), should you dissolve in water to make \(1.00 \mathrm{~L}\) of vodka (which is an aqueous solution that is \(6.86 \mathrm{M}\) ethanol)? (b) Using the density of ethanol \((0.789 \mathrm{~g} / \mathrm{mL})\), calculate the volume of ethanol you need to make \(1.00 \mathrm{~L}\) of vodka.

The U.S. standard for arsenate in drinking water requires that public water supplies must contain no greater than 10 parts per billion ( \(\mathrm{ppb})\) arsenic. If this arsenic is present as arsenate, \(\mathrm{AsO}_{4}{ }^{3-}\), what mass of sodium arsenate would be present in a \(1.00\)-L sample of drinking water that just meets the standard? Parts per billion is defined on a mass basis as $$ \mathrm{ppb}=\frac{\mathrm{g} \text { solute }}{\text { g solution }} \times 10^{9} $$

(a) How many milliliters of \(0.120 \mathrm{M} \mathrm{HCl}\) are needed to completely neutralize \(50.0 \mathrm{~mL}\) of \(0.101 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) solution? (b) How many milliliters of \(0.125 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) are needed to neutralize \(0.200 \mathrm{~g}\) of \(\mathrm{NaOH}\) ? (c) If \(55.8 \mathrm{~mL}\) of a \(\mathrm{BaCl}_{2}\) solution is needed to precipitate all the sulfate ion in a \(752-\mathrm{mg}\) sample of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\), what is the molarity of the \(\mathrm{BaCl}_{2}\) solution? (d) If \(42.7 \mathrm{~mL}\) of \(0.208\) \(\mathrm{M} \mathrm{HCl}\) solution is needed to neutralize a solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\), how many grams of \(\mathrm{Ca}(\mathrm{OH})_{2}\) must be in the solution?

Using solubility guidelines, predict whether each of the following compounds is soluble or insoluble in water: (a) \(\mathrm{MgBr}_{2}\), (b) \(\mathrm{PbI}_{2}\), (c) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}\), (d) \(\mathrm{Sr}(\mathrm{OH})_{2}\), (e) \(\mathrm{ZnSO}_{4}\).

Write balanced net ionic equations for the reactions that occur in each of the following cases. Identify the spectator ion or ions in each reaction. (a) \(\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}(a q) \longrightarrow\) (b) \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{K}_{2} \mathrm{SO}_{4}(a q) \longrightarrow\) (c) \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{KOH}(a q)\)

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