Chapter 16, Problem 16.89QP

(a)

Interpretation Introduction

Interpretation:

Like water, liquid ammonia undergoes autoionization.

• The Bronsted acid and Bronsted base in this reaction has to be identified.

Concept Information:

Bronsted acid-base reaction:

Bronsted definition deals with the transfer of protons.

Bronsted's definition is based on the  chemical reaction that occur when both acids and bases are added with each other. In Bronsted's theory acid donates proton , while base accepts proton from acid resulting in the formation of water.

Example: Consider the following reaction.

    $\text{HCl}\text{+}{\text{NH}}_{\text{3}}\to {\text{NH}}_{\text{4}}{}^{\text{+}}\text{+}{\text{Cl}}^{\text{-}}$

Hydrogen chloride donates a proton, and hence it is a Bronsted acid. Ammonia accepts a proton, and hence it is a Bronsted base.

Autoionization of water:

The equation of equilibrium for autoionization of water is,

  ${\text{H}}_{\text{2}}\text{O}\to {\text{H}}^{\text{+}}\text{+}{\text{OH}}^{\text{-}}$

  ${\text{K}}_{\text{w}}\text{=}{\text{[H}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}$

The equilibrium expression for water at $\text{25}{}^{\text{o}}\text{C}$ is,

  ${\text{[H}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]= 1}\times {\text{10}}^{\text{-14}}$

Therefore,

  $\begin{array}{l}{\text{K}}_{\text{w}}\text{=}{\text{[H}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}\\ \text{=1}\times {\text{10}}^{\text{-14}}\end{array}$

(b)

Interpretation Introduction

Interpretation:

Like water, liquid ammonia undergoes autoionization.

• The species correspond to ${\text{H}}^{\text{+}}$ and ${\text{OH}}^{\text{-}}$ has to be given
• The condition for neutral solution has to be given

Concept Information:

Bronsted acid-base reaction:

Bronsted definition deals with the transfer of protons.

Bronsted's definition is based on the  chemical reaction that occur when both acids and bases are added with each other. In Bronsted's theory acid donates proton , while base accepts proton from acid resulting in the formation of water.

Example: Consider the following reaction.

    $\text{HCl}\text{+}{\text{NH}}_{\text{3}}\to {\text{NH}}_{\text{4}}{}^{\text{+}}\text{+}{\text{Cl}}^{\text{-}}$

Hydrogen chloride donates a proton, and hence it is a Bronsted acid. Ammonia accepts a proton, and hence it is a Bronsted base.

Autoionization of water:

The equation of equilibrium for autoionization of water is,

  ${\text{H}}_{\text{2}}\text{O}\to {\text{H}}^{\text{+}}\text{+}{\text{OH}}^{\text{-}}$

  ${\text{K}}_{\text{w}}\text{=}{\text{[H}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}$

The equilibrium expression for water at $\text{25}{}^{\text{o}}\text{C}$ is,

  ${\text{[H}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]= 1}\times {\text{10}}^{\text{-14}}$

Therefore,

  $\begin{array}{l}{\text{K}}_{\text{w}}\text{=}{\text{[H}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}\\ \text{=1}\times {\text{10}}^{\text{-14}}\end{array}$