Question

Conjugate base of $N{H}_3$ is:
A.$N{H_4}^{+}$
B.$N{H_2}^{+}$
C.$N{H_2}^{-}$
D.$N_2$

Answer 1

Hint: We can define Bronsted-Lowry acid as proton donor and comprises a hydrogen atom. It may be a neutral molecule or may contain a net positive or negative charge.
We can define a conjugate base as the product formed by a loss of proton from an acid. The conjugate base of the acid A will be ${\text{A}}^{\text{ - }}\text{.}$

Complete step by step answer:
Based on the Bronsted-Lowry Theory, a substance, which releases a proton, is an acid and a base is a substance, which accepts a proton.
Let us consider an example of hydrochloric acid reacting with ammonia to form ammonium ions and chloride ions.
We can write the chemical reaction as,
$HCl\left(aq\right)+N{H}_3\left(aq\right)\stackrel{}{\to }N{H_4}^{+}\left(aq\right)+C{l}^{-}\left(aq\right)$
In the above equation, we can see that hydrochloric acid has donated a proton to ammonia, and ammonia has accepted a proton. Therefore, we can say hydrochloric acid is Bronsted-Lowry acid (proton donor), and ammonia as Bronsted-Lowry base (proton acceptor). Here, ammonium ion is a conjugate acid of ammonia and chloride ions are conjugate bases of hydrochloric acid.
We can define a conjugate base as the product formed by loss of proton from an acid. The conjugate base of the acid A will be $A^{-}.$
A conjugate acid is the product formed by gain of a proton by a base. The conjugate acid of the base B will be $H{B}^{+}\text{.}$
Now let us identify the conjugate base of $N{H}_3$.
Ammonia $\left(N{H}_3\right)$ loses a proton and acts as an acid. We can write the chemical equation as,
$$N{H}_3\stackrel{-H^{+}}{\to }N{H_2}^{-}$$
Ammonia loses a proton and forms its conjugate base $N{H_2}^{-}$.
Ammonia is the acid and $N{H_2}^{-}$ is the conjugate base of $N{H}_3$.
We can give the structure of ammonia as,

The conjugate base of ammonia will have the structure as,

$\therefore$Option (C) is correct.

Note:
We know that acid loses a proton and forms a conjugate base. Base accepts a proton and forms conjugate acid.
Example 1: Let us consider the reaction given below,
$$N{H}_3\left(g\right)+HI\left(g\right)\stackrel{}{\to }{I}^{-}\left(aq\right)+N{H_4}^{+}\left(aq\right)$$
Hydrogen iodide loses its proton to form iodide. Ammonia gains a proton to form ammonium ion.
The acid in the reaction is $HI$
The conjugate base of the acid is $I^{-}$
The base in the reaction is $N{H}_3$
The conjugate acid of the base is $N{H_4}^{+}$
Example 2: Let us consider the reaction given below,
$HCOOH\left(l\right)+H_{2}O\left(l\right)\stackrel{}{\to }HCO{O}^{-}\left(aq\right)+H_3{O}^{+}\left(aq\right)$
Formic acid loses a proton to form formate ion. Water gains a proton and forms hydronium ion.
The acid in the reaction is $HCOOH$
The conjugate base of the acid is $HCO{O}^{-}$
The base in the reaction is $H_{2}O$
The conjugate acid of the base is $H_3{O}^{+}$
Example 3: Let us consider the reaction given below,
$HS{O}_4{}_{\left(aq\right)}^{-}+H_2{O}_{(l)}\stackrel{}{\to }{H}_{2}S{O_4}_{\left(aq\right)}+O{H^{-}}_{\left(aq\right)}$
$HS{O_4}^{-}$ gains a proton and forms sulphuric acid. Water loses a proton to form hydronium ion.
The acid in the reaction is $H_{2}O$
The conjugate base of the acid is $O{H}^{-}$
The base in the reaction is $HS{O_4}^{-}$
The conjugate acid of the base is $H_{2}S{O}_4$