Chapter 16, Problem 16.124QP

Interpretation Introduction

Interpretation:

In a 0.080 M ${\text{NH}}_{\text{3}}$ solution, what percent of the ${\text{NH}}_{\text{3}}$ is present as ${\text{NH}}_{\text{4}}{}^{\text{+}}$ has to be calculated.

Concept Information:

Strong base and weak base:

Strong base dissociates into its constituent ions fully.  It produces more of hydroxide ions while dissolved in water.  Weak bases partially dissociates into its constituent ions.

According to Bronsted-Lowry, strong base is a good proton acceptor whereas weak base is a poor proton acceptor

Since, the ionization of a weak base is incomplete, it is treated in the same way as the ionization of a weak acid.

The ionization of a weak base $\text{B}$ is given by the below equation.

${\text{B}}_{\text{(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}\to {\text{HB}}^{\text{+}}{}_{\text{(aq)}}{\text{+OH}}^{\text{-}}{}_{\text{(aq)}}$

The equilibrium expression for the ionization of weak base $\text{B}$ will be,

${\text{K}}_b=\frac{[{\text{HB}}^{\text{+}}][{\text{OH}}^{\text{-}}]}{[\text{B}]}$

Where,

$K_b$ is base ionization constant,

$[{\text{OH}}^{-}\text{]}$ is concentration of hydroxide ion

$[{\text{HB}}^{\text{+}}\text{]}$  is concentration of conjugate acid

$[\text{B]}$ is concentration of the base

Percent ionization:

A quantitative measure of the degree of ionization is percent ionization.

For a weak, base $\text{HB}$ percent ionization can be calculated as follows,

$\text{percent}\text{ionization}=\frac{{\text{[OH}}^{\text{-}}\text{]}}{\text{[HB]}}\times 100\%$

To Calculate: The percent of the ${\text{NH}}_{\text{3}}$ is present as ${\text{NH}}_{\text{4}}{}^{\text{+}}$ in the given solution