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Conjugate base of hydrazoic acid is: (a) \(\mathrm{HN}_{3}^{-}\) (b) \(\mathrm{N}_{2}^{-}\) (c) Azide ion (d) Nitride ion

Short Answer

Expert verified
The conjugate base of hydrazoic acid is the azide ion (c).

Step by step solution

01

Understand the Concept of Conjugate Bases

A conjugate base is formed when an acid donates a proton (H鈦). The molecule remaining after the loss of this hydrogen ion is the conjugate base of the acid.
02

Identify Hydrazoic Acid

The chemical formula for hydrazoic acid is \( ext{HN}_3\). It consists of one hydrogen atom and the azide group \( ext{N}_3^鈭抃).
03

Determine the Result of Proton Donation

When hydrazoic acid \( ext{HN}_3\) donates a proton, it loses one hydrogen ion (H鈦). This results in the formation of \( ext{N}_3^-\), which is the azide ion.
04

Identify the Correct Answer

Comparing the options, the correct description of the conjugate base is the azide ion, denoted by \( ext{N}_3^-\). This matches option (c).

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Key Concepts

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

Hydrazoic Acid
Hydrazoic acid, with the chemical formula \(\mathrm{HN}_3\), is a fascinating compound with unique characteristics. It consists of one hydrogen atom connected to an azide ion \(\mathrm{N}_3^-\). This acid is known for its volatile and explosive nature in its concentrated form. It exists primarily in dilute solutions and is recognized for its ability to donate a proton.Hydrazoic acid belongs to a class of compounds known as acids, which have the common property of releasing hydrogen ions \((\mathrm{H}^+)\) in solutions. This ability to donate protons makes them active participants in various chemical reactions. Hydrazoic acid is relatively weak compared to strong acids like hydrochloric acid \((\mathrm{HCl})\) or sulfuric acid \((\mathrm{H}_2\mathrm{SO}_4)\), which means it partially dissociates in water.
Proton Donation
The concept of proton donation is a core idea in chemistry, particularly in acid-base reactions. When an acid undergoes a chemical process, it donates a proton \((\mathrm{H}^+)\) to another substance, usually a base. This is sometimes called the acid's dissociation.- During proton donation, the acid loses a positive hydrogen ion.- The remaining part of the acid becomes its conjugate base.For hydrazoic acid, the proton donation process can be expressed simply. When \(\mathrm{HN}_3\) loses its hydrogen ion, it produces \(\mathrm{N}_3^-\). It's this reaction that forms the basis for identifying the conjugate base of hydrazoic acid as the azide ion. Understanding proton donation is essential for deciphering the behavior of acids and understating their interactions in chemical reactions.
Azide Ion
The azide ion \((\mathrm{N}_3^-)\) is a linear ion composed of three nitrogen atoms. It forms the conjugate base of hydrazoic acid once a proton is donated. The azide ion is known for its resonance structures and can be quite reactive.Key properties and uses of the azide ion include:- Structure: It is a linear ion with equal bond lengths between the nitrogen atoms, stabilized by resonance.- Reactivity: It is known for reactions that generate nitrogen gas \((\mathrm{N}_2)\), which is a driving force for many chemical processes.- Applications: Azide ions are used in the synthesis of organic and inorganic azides, explosives, and propellants. They also play a role in pharmaceuticals, being part of azide groups in some medications.Understanding the azide ion, its formation, and its properties further illuminates the nature of hydrazoic acid and similar compounds in chemistry.

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Most popular questions from this chapter

\(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}+22.4 \mathrm{kcal}\) formation of \(\mathrm{NH}_{3}\) by above reaction shows: (a) Cyanamide process (b) Serpeck's process (c) Haber process (d) None of these

One mole of a compound AB reacts with one mole of compound CD according to the equation: \(\mathrm{AB}+\mathrm{CD} \rightarrow \mathrm{AD}+\mathrm{CB}\) When equilibrium had been established it was found that \(3 / 4\) mole each of reactant \(\mathrm{AB}\) and \(\mathrm{CD}\) has been converted to AD and CB. There is no change in volume. The equilibrium constant of the reaction is? (a) 9 (b) \(\frac{1}{9}\) (c) \(9 / 16\) (d) \(16 / 9\)

At constant temperature, the equilibrium constant \(\left(\mathrm{K}_{\mathrm{p}}\right.\) ) for the decomposition reaction: \(\mathrm{N}_{2} \mathrm{O}_{4} \rightleftharpoons 2 \mathrm{NO}_{2}\) is expressed by \(\mathrm{K}_{\mathrm{p}}=\left(4 \mathrm{x}^{2} \mathrm{P}\right) /\left(1-\mathrm{x}^{2}\right)\), where \(\mathrm{P}=\) pressure, \(\mathrm{x}=\) extent of decomposition. Which one of the following statements is true? (a) \(\mathrm{K}_{\mathrm{p}}\) increases with increase of \(\mathrm{P}\) (b) \(\mathrm{K}_{\mathrm{p}}\) increases with increase of \(\mathrm{x}\) (c) \(K_{p}\) increases with decrease of \(x\) (d) \(K_{p}\) remains constant with change in \(P\) and \(x\)

The value of \(\mathrm{K}_{\mathrm{p}}\) in the reaction: \(\mathrm{MgCO}_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{MgO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g})\) is (a) \(\mathrm{K}_{\mathrm{p}}=\mathrm{P}\left(\mathrm{CO}_{2}\right)\) (b) \(\mathrm{K}_{\mathrm{p}}=\frac{\mathrm{P}\left(\mathrm{MgCO}_{3}\right)}{\mathrm{P}\left(\mathrm{CO}_{2}\right) \times \mathrm{P}(\mathrm{MgO})}\)

For a reversible reaction, the concentration of the reactants are doubled, then the equilibrium constant: (a) Becomes one-fourth (b) Is doubled (c) Is halved (d) Remains same

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