Chapter 8, Problem 90P

Interpretation Introduction

Interpretation:

If 0.10 mol of ${\text{NH}}_{\text{3}}$ is added to 0.50 mol of HCl and make 1.0 L solution, what happens to the ${\text{NH}}_{\text{3}}$ should be explained and if any ${\text{NH}}_{\text{3}}$ remains in the solution should be determined.

Concept Introduction:

The acid and base reacts to form salt and water, this is known as neutralization reaction. Here, acid forms conjugate base and base forms conjugate acid. In the given problem, ${\text{NH}}_{\text{3}}$ is base, it can accept proton from HCl which is an acid to form ${\text{NH}}_{\text{3}}{}^{+}$ or conjugate acid. Similarly, HCl forms ${\text{Cl}}^{-}$ as a conjugate base.

Finally, ${\text{NH}}_{\text{3}}{}^{+}$ and ${\text{Cl}}^{-}$ reacts to form ${\text{NH}}_{\text{4}}\text{Cl}$, the reaction is represented as follows:

${\text{NH}}_{\text{3}}+\text{HCl}\underset{}{\overset{}{\rightleftarrows }}{\text{NH}}_{\text{4}}\text{Cl}$.