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,
$HCl\left( {aq} \right) + N{H_3}\left( {aq} \right)\xrightarrow{{}}N{H_4}^ + \left( {aq} \right) + C{l^ - }\left( {aq} \right)$
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,
\[N{H_3}\xrightarrow{{ - {H^ + }}}N{H_2}^ - \]
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,
\[N{H_3}\left( g \right) + HI\left( g \right)\xrightarrow{{}}{I^ - }\left( {aq} \right) + N{H_4}^ + \left( {aq} \right)\]
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,
$HCOOH\left( l \right) + {H_2}O\left( l \right)\xrightarrow{{}}HCO{O^ - }\left( {aq} \right) + {H_3}{O^ + }\left( {aq} \right)$
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,
$HS{O_4}{^ - _{\left( {aq} \right)}} + {H_2}{O_{(l)}}\xrightarrow{{}}{H_2}S{O_4}_{\left( {aq} \right)} + O{H^ - }_{\left( {aq} \right)}$
$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 ${H_2}S{O_4}$