What is the conjugate acid for NH2 in water?

The conjugate acid of $N{{H}_{2}}^{-}$ is:A. $N{{H}_{3}}$ B. $N{{H}^{2-}}$ C. $N{{H}_{4}}^{+}$ D. ${{N}_{3}}^{-}$

Answer

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Hint: $N{{H}_{2}}^{-}$ is referred to as amides and the IUPAC name sanctioned for this anion is Azanide. A conjugate acid is formed when an acid donates its proton to base. This theory is known as acid base theory.

Complete step by step answer:
Bronsted-lowry theory is basically an acid base reaction theory which has given a fundamental concept that when a base reacts with an acid , the acid forms its conjugate base and the base forms its conjugate acid.
In this theory, they have defined acids and bases by the way they react with each other. Their definitions were expressed in terms of equilibrium expression.
\[HA+B\rightleftharpoons {{A}^{-}}+H{{B}^{+}}\]
Where, $HA$ is an acid and ${{A}^{-}}$ is conjugate base
$B$ is the base $H{{B}^{+}}$ is a conjugate acid.
In this reaction, an acid loses its electron to become its conjugate base and the base accepts the electron to become its conjugate acid.
$N{{H}_{2}}^{-}+{{H}^{+}}\to N{{H}_{3}}$
Where, $N{{H}_{2}}^{-}$ is a base, ${{H}^{+}}$ is a proton and $N{{H}_{3}}$ is a conjugate acid.
Here, the base that is $N{{H}_{2}}^{-}$ accepts a proton that gives conjugate acid that is $N{{H}_{3}}$ .

So, the correct answer is Option A .

Additional information:
Ammonia is a colourless gas that is composed of nitrogen and hydrogen. It is a stable binary hydride compound. It is used in agriculture as fertilizers. It is also used in the purification of water supplies.
Conjugate acid is formed when an acid donates a proton or when it loses a hydrogen ion.

Note: When a base accepts a proton it gives conjugate acid as a product and when acid donates a proton it gives conjugate base as a product. In this question, $N{{H}_{2}}^{-}$ is a base that accepts protons to give ammonia as a conjugate acid.

Conjugate acid	Formula	Structure	pKa	Conjugate base	Name
Methane	CH4		50-60		Methyl
Ethene	CH2CH2		45		Ethenyl
Ammonia	NH3		36		Amide
Acetylene (ethyne)	C2H2		26		Acetylide
Ethanol	C2H5OH		16		Ethoxide
Water	HOH		15.7		Hydroxide
Ammonium	NH4+	H�NH3	9.8	: NH3	Ammonia
Hydroflouric acid	HF		4		Flouride

Use this table to predict which conjugate base will favorably react with which conjugate acids. Pay attention to the pKa values shown. As a general rule, the conjugate base of any acid will react with, and remove, the proton (H+ ion) from any conjugate acid that is stronger than the conjugate acid from which the conjugate base you are looking at was derived from. 

In the chart above, the conjugate base will remove a proton from every conjugate acid that is below it in the table.

Examples:

• The Amide ion will remove (quantitatively) a proton from Acetylene, Ethanol, Water, and Hydroflouric acid, these cells are shown in color above, so that the amide ion will remove a proton from the acids below it in this table (smaller pKa values, stronger acids).
• The Hydroxide ion will remove (quantitatively) a proton only from Hydroflouric acid and the Ammonium ion.
• The Ethenyl ion will remove (quantitatively) a proton from Ammonia, Acetylene, Ethanol, Water, and Hydroflouric acid.
• The Methyl carbanion will remove (quantitatively) a proton from all acids (from Ethene, Ammonia, Acetylene, Ethanol, Water, and Hydroflouric Acid)

When I write � quantitatively � I mean that there is enough difference in the pKa values (of the conjugate acids) to remove greater than 99% of the protons from the acids below them. A difference of 2 pKa values will allow the conjugate base coming from the weaker acid will remove >99% of protons from the stronger acid.

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