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Chapter 17: Problem 20

Dissolving ammonium bromide in water gives an acidic solution. Write a balanced equation showing how this can occur.

Short Answer

Expert verified
Ammonium bromide dissociates into \( NH_4^+ \) and \( Br^- \), then \( NH_4^+ \) reacts with water to form \( H_3O^+ \), making the solution acidic.

Step by step solution

01

Identify Chemical Formulas

The formula for ammonium bromide is \( NH_4Br \). When it dissolves in water, it dissociates into its ions, ammonium \( NH_4^+ \) and bromide \( Br^- \). This is the first reaction we investigate.
02

Dissociation in Water

When \( NH_4Br \) dissolves in water, it dissociates as follows: \[ NH_4Br (s) \rightarrow NH_4^+ (aq) + Br^- (aq) \]. This shows the separation of ammonium bromide into ammonium and bromide ions in aqueous solution.
03

Reaction of Ammonium Ion

The ammonium ion \( NH_4^+ \) can react with water. The equation for this is: \[ NH_4^+ (aq) + H_2O (l) \rightarrow NH_3 (aq) + H_3O^+ (aq) \]. This reaction shows that \( NH_4^+ \) donates a proton to water, forming hydronium ions \( H_3O^+ \) which makes the solution acidic.

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Key Concepts

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

Acidic Solutions
When a substance dissolves in water and increases the concentration of hydrogen ions ( H^+ ), the resulting mixture is known as an acidic solution. This happens when a compound releases or facilitates the formation of H^+ in the solution.

Important characteristics of acidic solutions include:

  • A sour taste.
  • The ability to conduct electricity due to the presence of ions.
  • A pH value less than 7.

For the case of ammonium bromide dissolving in water, H_4^+ ions are responsible for the solution's acidic nature. The presence of H_3O^+ (hydronium ions) in the solution is a direct indication of its acidic quality. Hydronium ions increase the H^+ concentration significantly, lowering the pH and affirming the acidic property of the solution.

Ion Dissociation
Ion dissociation is the process through which compounds split into ions when dissolved in a solvent like water. It is a crucial step that leads to various chemical reactions in aqueous solutions.

During this process:

  • The ionic compound separates into its constituent ions.
  • The ions become surrounded by water molecules, aiding in their dispersion through the solution.

For example, when ammonium bromide ( NH_4Br ) dissolves in water, it dissociates into ammonium ( NH_4^+ ) and bromide ( Br^- ) ions. This dissociation is key as it provides free ions that can participate in further reactions, such as the ones involving the formation of hydronium ions. Without dissociation, the chemical species wouldn't be able to interact readily within the solution.

Hydronium Ion Formation
The formation of hydronium ions (H_3O^+) is a hallmark of creating an acidic solution. This process involves the transfer of protons (H^+) from one molecule to another, specifically from an acid to water.

Let's explore how this occurs with ammonium ions:

  • The ammonium ion (NH_4^+) acts as a proton donor.
  • It transfers a proton to a nearby water molecule (H_2O), which accepts it.
  • This donation transforms the water molecule into a hydronium ion (H_3O^+).

The equation representing this interaction is: \[ NH_4^+ (aq) + H_2O (l) \rightarrow NH_3 (aq) + H_3O^+ (aq) \]

Formation of hydronium ions indicates an increase in acidity due to the rise in H^+ ion concentration. This process is crucial in understanding why solutions like that of ammonium bromide in water become acidic.

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Most popular questions from this chapter

A \(0.10 \mathrm{M}\) solution of chloroacetic acid, \(\mathrm{ClCH}_{2} \mathrm{CO}_{2} \mathrm{H},\) has a pH of \(1.95 .\) Calculate \(K_{\mathrm{a}}\) for the acid.

Hydrazine, \(\mathrm{N}_{2} \mathrm{H}_{4},\) can interact with water in two steps. $$\begin{aligned}\mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightleftarrows \mathrm{N}_{2} \mathrm{H}_{5}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \\\K_{\mathrm{bl}}=8.5 \times 10^{-7} \end{aligned}$$ $$\begin{aligned}\mathrm{N}_{2} \mathrm{H}_{5}^{+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) & \mathrm{N}_{2} \mathrm{H}_{6}^{2+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \\\& K_{\mathrm{b} 2}=8.9 \times 10^{-16} \end{aligned}$$ (a) What is the concentration of \(\mathrm{OH}^{-}, \mathrm{N}_{2} \mathrm{H}_{5}^{+},\) and \(\mathrm{N}_{2} \mathrm{H}_{6}^{2+}\) in a \(0.010 \mathrm{M}\) aqueous solution of hydrazine? (b) What is the \(\mathrm{pH}\) of the \(0.010 \mathrm{M}\) solution of hydrazine?

Which is the stronger of the following two acids? (a) benzoic acid, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}, \mathrm{p} K_{\mathrm{a}}=4.20\) (b) 2 -chlorobenzoic acid, \(\mathrm{ClC}_{6} \mathrm{H}_{4} \mathrm{CO}_{2} \mathrm{H}, \mathrm{p} K_{\mathrm{a}}=2.90\)

A hydrogen atom in the organic base pyridine, \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\) can be substituted by various atoms or groups to give \(\mathrm{XC}_{5} \mathrm{H}_{4} \mathrm{N},\) where \(\mathrm{X}\) is an atom such as \(\mathrm{Cl}\) or a group such as \(\mathrm{CH}_{3}\). The following table gives \(K_{\mathrm{a}}\) values for the conjugate acids of a variety of substituted pyridines. $$\begin{array}{ll}\text { Atom or Group } X & K_{\mathrm{a}} \text { of Conjugate Acid } \\ \hline \mathrm{NO}_{2} & 5.9 \times 10^{-2} \\\\\mathrm{Cl} & 1.5 \times 10^{-4} \\\\\mathrm{H} & 6.8 \times 10^{-6} \\\\\mathrm{CH}_{3} & 1.0 \times 10^{-6}\end{array}$$ (a) Suppose each conjugate acid is dissolved in sufficient water to give a \(0.050 \mathrm{M}\) solution. Which solution would have the highest pH? The lowest pH? (b) Which of the substituted pyridines is the strongest Bronsted base? Which is the weakest Br枚nsted base?

Given the following solutions: (a) \(0.1 \mathrm{M} \mathrm{NH}_{3}\) (b) \(0.1 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}\) (c) \(0.1 \mathrm{M} \mathrm{NaCl}\) (d) \(0.1 \mathrm{M} \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\) (e) \(0.1 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\) (f) \(0.1 \mathrm{M} \mathrm{NaCH}_{3} \mathrm{CO}_{2}\) (g) \(0.1 \mathrm{M} \mathrm{NH}_{4} \mathrm{CH}_{3} \mathrm{CO}_{2}\) (i) Which of the solutions are acidic? (ii) Which of the solutions are basic? (iii) Which of the solutions is most acidic?
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