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The conjugate acid of \(\mathrm{NH}_{2}^{-}\) is: (a) \(\mathrm{NH}_{4}^{+}\) (b) \(\mathrm{NH}_{3}\) (c) \(\mathrm{NH}_{2} \mathrm{OH}\) (d) \(\mathrm{N}_{2} \mathrm{H}_{4}\)

Short Answer

Expert verified
The conjugate acid of \( \text{NH}_2^- \) is \( \text{NH}_3 \), option (b).

Step by step solution

01

Identify the Conjugate Acid-Base Pair

Given the base \( \text{NH}_2^- \), find its conjugate acid by adding a proton (H⁺). This is because the conjugate acid of a base is formed by adding a hydrogen ion to it.
02

Add a Proton to the Base

Add a hydrogen ion (\( \text{H}^+ \)) to \( \text{NH}_2^- \). This is done by combining \( \text{NH}_2^- \) with \( \text{H}^+ \), resulting in \( \text{NH}_3 \). This is the neutral form of the molecule after accepting a proton.
03

Verify the Result

Compare \( \text{NH}_3 \) with the given options: - (a) \( \text{NH}_4^+ \) - (b) \( \text{NH}_3 \) - (c) \( \text{NH}_2 \text{OH} \) - (d) \( \text{N}_2 \text{H}_4 \) Identify \( \text{NH}_3 \) as option (b), which matches our result from Step 2.

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Key Concepts

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

Ammonia (NH3)
Ammonia, or \( \text{NH}_3 \), is a simple molecule made up of nitrogen and hydrogen. It is a common compound found in nature and is particularly known for its distinctive odor. Ammonia has several uses, both in industrial applications and household cleaning products.
In chemistry, it is notable for acting as a base, meaning it can accept a proton (\( \text{H}^+ \)). This property makes ammonia a central player in acid-base reactions.
Ammonia's ability to act as a base is due to the unshared pair of electrons on the nitrogen atom. This pair allows \( \text{NH}_3 \) to attract and bond with protons (\( \text{H}^+ \)), forming \( \text{NH}_4^+ \). This process highlights ammonia's role in different chemical processes and its importance in understanding how substances interact in solution.
Proton Transfer
Proton transfer is a fundamental concept in acid-base chemistry, referring to the movement of a proton (\( \text{H}^+ \)) from one molecule to another. This transfer is a key aspect of acid-base reactions and helps define the roles of substances as acids or bases.
When a base, such as \( \text{NH}_2^- \), accepts a proton, it becomes its conjugate acid, \( \text{NH}_3 \). This transformation is called protonation and is part of identifying conjugate acid-base pairs.
Proton transfers are essential in many biochemical processes and reactions in nature. They also significantly impact the pH levels of solutions because adding or removing protons can change the acidity or basicity, altering the chemical environment.
Acid-Base Reaction
An acid-base reaction involves the transfer of protons between participants. This transfer helps establish which substances are acids (proton donors) and which are bases (proton acceptors). These reactions are a core topic in chemistry, explaining how different species interact and transform.
In the context of our exercise, \( \text{NH}_3 \) acts as the conjugate acid formed when \( \text{NH}_2^- \) accepts a proton. This example illustrates the nature of acid-base reactions, where proton transfers create new substances.
Understanding acid-base reactions includes recognizing the importance of conjugate acid-base pairs and the role they play in chemical equilibrium. These concepts are essential for predicting the outcome of reactions and understanding the behavior of substances in a chemical solution. By mastering this concept, students can accurately anticipate how different chemical entities will react in various environments.

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Most popular questions from this chapter

In which of the following reactions, the concentration of product is higher than the concentration of reactant at equilibrium? = equilibrium constant): (a) \(\mathrm{A} \rightleftharpoons \mathrm{B} ; \mathrm{K}=0.001\) (b) \(\mathrm{M} \rightleftharpoons \mathrm{N} ; \mathrm{K}=10\) (c) \(\mathrm{X} \rightleftharpoons \mathrm{Y} ; \mathrm{K}=0.005\) (d) \(\mathrm{R} \rightleftharpoons \mathrm{P} ; \mathrm{K}=0.01\)

Which of these reactions are having homogeneous equilibrium: (a) \(\mathrm{N}_{2} \mathrm{O}_{4} \rightleftharpoons 2 \mathrm{NO}_{2}\) \(' \mathrm{~g}^{\prime} \quad{ }^{6} \mathrm{~g}^{\prime}\) (b) \(\mathrm{H}_{2}+\mathrm{Cl}_{2} \rightleftharpoons 2 \mathrm{HCl}\) \(' \mathrm{~g}^{\prime} \quad{ }^{2} \mathrm{~g}^{\prime} \quad{ }^{\mathrm{s}} \mathrm{g}^{\prime}\) (c) \(\mathrm{CaCO}_{3} \rightleftharpoons \mathrm{CaO}+\mathrm{CO}_{2}\) 's' \(\quad{ }^{\prime} \mathrm{s}^{\prime} \quad{ }^{\prime} \mathrm{g}\) ' (d) Both (a) and (b)

One of the following equilibria is not affected by change in volume of the flask: (a) \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g}) \mathrm{Cl}_{2}(\mathrm{~g})\) (b) \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\) (c) \(\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\) (d) \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\)

A chemical reaction is catalysed by a catalyst \(\mathrm{X}\). Hence, \(\mathrm{X}:\) (a) Increases activation energy of the reaction (b) Does not affect equilibrium constant of the reaction (c) Decreases rate constant of the reaction (d) Reduces enthalpy of the reaction

The ratio of \(\mathrm{K}_{\mathrm{p}} / \mathrm{K}_{\mathrm{c}}\) for the reaction: \(\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{~g})\) is (a) 1 (b) RT (c) \((\mathrm{RT})^{1 / 2}\) (d) \((\mathrm{RT})^{-1 / 2}\)

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