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Chapter 15: Problem 61

In liquid \(\mathrm{NH}_{3}\) (a) \(\mathrm{CH}_{3} \mathrm{COOH}\) behaves as strong acid (b) \(\mathrm{NaNH}_{2}\) is a base (c) \(\mathrm{NH}_{4} \mathrm{Cl}\) is an acid (d) All the above facts are true

Short Answer

Expert verified
All given statements are true (d).

Step by step solution

01

Understand the context of liquid NH3

In liquid ammonia (NH3), the solvent acts similarly to water in aqueous solutions. We look for similar properties, such as acids and bases, understanding that liquid NH3 is a polar solvent.
02

Examine CH3COOH in NH3

Acetic acid (CH3COOH) in liquid NH3 is expected to act more strongly as it donates a proton and ammonia's proton affinity enhances acetic acid's acidic properties.
03

Consider NaNH2 in NH3

Sodium amide (NaNH2) in liquid ammonia acts as a base. It can accept protons (NH2-) and form NH3. Sodium amide in NH3 is comparable to NaOH in water.
04

Analyze NH4Cl in NH3

Ammonium chloride (NH4Cl) in liquid ammonia acts as an acid because the NH4+ ion can donate a proton to ammonia, forming NH3.
05

Evaluate all given statements

Both (a) CH3COOH behaves as a strong acid, (b) NaNH2 is a base, and (c) NH4Cl is an acid in liquid NH3 hold true. This confirms that answer (d) all the above statements are indeed correct.

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Key Concepts

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

Acidic and Basic Properties
Understanding acidic and basic properties in liquid ammonia (NH鈧�) involves recognizing how substances behave when they are dissolved in this solvent. In liquid NH鈧�, similar to how water functions in aqueous solutions, substances can either donate or accept protons. This transfer of protons determines their acidic or basic nature.
  • When a substance donates a proton in NH鈧�, it behaves as an acid.
  • Conversely, if it accepts a proton, it acts as a base.
For example, acetic acid (CH鈧僀OOH) behaves as a strong acid in liquid NH鈧� because it readily donates protons, while ammonium chloride (NH鈧凜l) acts as an acid because its ammonium ions (NH鈧勨伜) donate protons to the ammonia molecules. Meanwhile, sodium amide (NaNH鈧�) acts as a base because it can accept protons to form NH鈧�.
Polar Solvent
Liquid ammonia ( H鈧� ext) is classified as a polar solvent, similar to water. This means that its molecules have a partial positive charge on one end and a partial negative charge on the other. This polarity allows it to dissolve other polar substances efficiently. The polarity of liquid NH鈧� makes it an effective medium for facilitating reactions that involve the transfer of protons. As a polar solvent, NH鈧� can stabilize ions formed during these reactions:
  • The positive end of NH鈧� can attract negatively charged ions.
  • The negative end can attract positively charged ions.
This property is what allows substances like H鈧僀OOH ext) and NaNH鈧� ext) to exhibit exaggerated acidic or basic properties compared to their behavior in less polar or non-polar environments.
Proton Transfer Reactions
Proton transfer reactions are at the heart of understanding acid-base behavior in liquid ammonia. These reactions involve moving protons from one molecule to another. In liquid NH鈧�:
  • When acetic acid ( H鈧僀OOH ext) dissolves, it donates protons to the ammonia molecules, creating acetate ions and protons, thereby displaying its acidic behavior.
  • Sodium amide ( ANH鈧� ext) accepts protons from the surrounding H鈧� ext) molecules, leading to the production of additional ammonia and highlighting its basic nature.
  • Similarly, ammonium chloride ( H鈧凜l ext) donates protons from the H鈧勨伜 ext) ion, illustrating its acidic property.
These transfers are facilitated by the polar nature of ammonia, which efficiently stabilizes the resulting ions, making proton transfer reactions more pronounced.

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Most popular questions from this chapter

Structure of \(\mathrm{H}_{2} \mathrm{O}_{2}\) is (a) planar (b) linear (c) non planar (d) three dimensional

The percentage by weight of hydrogen in \(\mathrm{H}_{2} \mathrm{O}_{2}\) is (a) \(6.5\) (b) \(5.88\) (c) 25 (d) 50

Which one of the following pairs of reactant does not form oxygen when they react with each other? (a) \(\mathrm{Cl}_{2}, \mathrm{NaOH}\) solution (cold, dilute) (b) \(\mathrm{F}_{2}, \mathrm{NaOH}\) solution (hot, concentrated) (c) \(\mathrm{F}_{2}, \mathrm{H}_{2} \mathrm{O}\) (d) \(\mathrm{CaOCl}_{2}, \mathrm{H}_{2} \mathrm{SO}_{4}\) (dilute)

Which of the following reactions shows the correct sequence of the Ostwald process in the manufacture of nitric acid? (a) \(4 \mathrm{NH}_{3}+5 \mathrm{O}_{2} \stackrel{750^{\circ} \mathrm{C}-900^{\circ} \mathrm{C}, \text { catalyst }}{\longrightarrow} 4 \mathrm{NO}+6 \mathrm{H}_{2} \mathrm{O}\) \(\mathrm{NO} \stackrel{\text { heat } \mathrm{O}_{2}}{\longrightarrow} \mathrm{NO}_{2} \stackrel{\mathrm{H}_{2} \mathrm{O}}{\mathrm{O}} \mathrm{HNO}_{3}\) (b) \(\mathrm{S}+\mathrm{O}_{2} \longrightarrow \mathrm{SO}_{2} \stackrel{\mathrm{O}_{2}}{\longrightarrow} \mathrm{SO}_{3} \frac{3}{+\mathrm{HNO}_{3}}\) \(\longrightarrow \mathrm{NaNO}_{3}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{NaHSO}_{4}\) (d) both (a) and (b)

\(\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s})\) is heated in a test tube. Vapours are brought in contact with red litmus paper, which changes to blue and then to red. It is because of (a) formation of \(\mathrm{NH}_{3}\) and \(\mathrm{HCl}\) (b) formation of \(\mathrm{NH}_{4} \mathrm{OH}\) and \(\mathrm{HCl}\) (c) greater diffusion of \(\mathrm{HCl}\) than \(\mathrm{NH}_{3}\) (d) greater diffusion of \(\mathrm{NH}_{3}\) than \(\mathrm{HCl}\)
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