Chapter 14, Problem 1E

Write equations that show NH₃ as both a conjugate acid and a conjugate base.

Interpretation Introduction

Interpretation:

The equations should be written which show ${\text{NH}}_3$ as conjugate acid and conjugate base.

Concept Introduction:

The compounds which are differ by the existence of one proton that is H+ is said to be conjugate acid-base pairs. Every acid has its conjugate base; it is formed when the proton is donated whereas every base has its conjugate acid, it is formed when the proton is accepted.

Answer to Problem 1E

The equations which show ${\text{NH}}_3$ as conjugate acid and conjugate base is:

  ${\text{NH}}_{\text{2}}^{\text{-}}{\text{+H}}^{\text{+}}\to {\text{NH}}_{\text{3}}$ (Conjugate acid)${\text{NH}}_4^{+}{\text{(aq)+OH}}^{-}(\text{aq})\to {\text{NH}}_{\text{3}}({\text{aq)+H}}_{\text{2}}\text{O(l)}$ (Conjugate base)

Explanation of Solution

The chemical equation which shows ${\text{NH}}_3$ as conjugate acid is given by:

  ${\text{NH}}_{\text{2}}^{\text{-}}{\text{+H}}^{\text{+}}\to {\text{NH}}_{\text{3}}$

In above reaction, ${\text{NH}}_{\text{2}}^{\text{-}}$ acts as a base and ${\text{NH}}_3$ acts as a conjugate acid because it is formed by accepting one proton that is ${\text{H}}^{\text{+}}$ .

The chemical equation which shows ${\text{NH}}_3$ as conjugate base is given by:

  ${\text{NH}}_4^{+}{\text{(aq)+OH}}^{-}(\text{aq})\to {\text{NH}}_{\text{3}}({\text{aq)+H}}_{\text{2}}\text{O(l)}$

In above reaction, ${\text{NH}}_4^{+}$ acts as an acid and ${\text{NH}}_3$ acts as a conjugate base because it is formed by donating proton, resulting formation of water.