Chapter 8, Problem 8.55AP

(a)

Interpretation Introduction

Interpretation:

The acid, base, conjugate acid and conjugate base of the reaction has to be labeled.

The given equation is,

  $\text{HI}\left(\text{g}\right){\text{+NH}}_{\text{3}}\left(\text{g}\right)\stackrel{}{\to }{\text{NH}}_{\text{4}}{}^{\text{+}}\left(\text{aq}\right){\text{+I}}^{\text{-}}\left(\text{aq}\right)$

Concept Introduction:

Bronsted-Lowry Acids:  A Bronsted-Lowry acid is proton donor and contains a hydrogen atom.  It may be a neutral molecule or may contain a net positive or negative charge.

Bronsted-Lowry Bases:  A Bronsted-Lowry base is a proton acceptor.  A base should contain a lone pair of electrons, which donates to form a new bond.  It can be neutral or can contain a negative charge.

Conjugate acid:  A conjugate acid is the product formed by a gain of a proton by a base.  The conjugate acid of the base B will be ${\text{HB}}^{\text{+}}\text{.}$

Conjugate base:  A conjugate base is the product formed by a loss of proton from an acid.  The conjugate base of the acid A will be ${\text{A}}^{\text{-}}\text{.}$

Answer to Problem 8.55AP

The acid in the reaction is $\text{HI}\text{.}$

The conjugate base of the acid is ${\text{I}}^{\text{-}}\text{.}$

The base in the reaction is ${\text{NH}}_{\text{3}}\text{.}$

The conjugate acid of the base is ${\text{NH}}_4{}^{\text{+}}\text{.}$

Explanation of Solution

Acid loses a proton and forms conjugate base.

Base accepts a proton and forms conjugate acid.

The given reaction is,

  $\text{HI}\left(\text{g}\right){\text{+NH}}_{\text{3}}\left(\text{g}\right)\stackrel{}{\to }{\text{NH}}_{\text{4}}{}^{\text{+}}\left(\text{aq}\right){\text{+I}}^{\text{-}}\left(\text{aq}\right)$

Hydroiodic acid loses its proton to form iodide.  Ammonia gains a proton to form ammonium ion.

  $\begin{array}{l}\text{HI}\left(\text{g}\right){\text{+NH}}_{\text{3}}\left(\text{g}\right)\stackrel{}{\to }{\text{NH}}_{\text{4}}{}^{\text{+}}\left(\text{aq}\right){\text{+I}}^{\text{-}}\left(\text{aq}\right)\\ \text{Acid}\text{Base}\text{Conjugate}\text{Conjugate}\\ \text{Acid}\text{Base}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The acid, base, conjugate acid and conjugate base of the reaction has to be labeled.

The given equation is,

  $\text{HCOOH}\left(\text{l}\right){\text{+H}}_{\text{2}}\text{O}\left(\text{l}\right)\stackrel{}{\to }{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\left(\text{aq}\right){\text{+HCOO}}^{\text{-}}\left(\text{aq}\right)$

Concept Introduction:

Refer to part (a).

Answer to Problem 8.55AP

The acid in the reaction is $\text{HCOOH}\text{.}$

The conjugate base of the acid is ${\text{HCOO}}^{\text{-}}\text{.}$

The base in the reaction is ${\text{H}}_2\text{O}\text{.}$

The conjugate acid of the base is ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\text{.}$

Explanation of Solution

The given reaction is,

  $\text{HCOOH}\left(\text{l}\right){\text{+H}}_{\text{2}}\text{O}\left(\text{l}\right)\stackrel{}{\to }{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\left(\text{aq}\right){\text{+HCOO}}^{\text{-}}\left(\text{aq}\right)$

Formic acid loses a proton to form formate ion.  Water gains a proton and forms hydronium ion.

  $\begin{array}{l}\text{HCOOH}\left(\text{l}\right){\text{+H}}_{\text{2}}\text{O}\left(\text{l}\right)\stackrel{}{\to }{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\left(\text{aq}\right){\text{+HCOO}}^{\text{-}}\left(\text{aq}\right)\\ \text{Acid}\text{Base}\text{Conjugate}\text{Conjugate}\\ \text{Acid}\text{Base}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The acid, base, conjugate acid and conjugate base of the reaction has to be labeled.

The given equation is,

  ${\text{HSO}}_{\text{4}}{}^{\text{-}}\left(\text{aq}\right){\text{+H}}_{\text{2}}\text{O}\left(\text{l}\right)\stackrel{}{\to }{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(\text{aq}\right){\text{+OH}}^{\text{-}}\left(\text{aq}\right)$

Concept Introduction:

Refer to part (a).

Answer to Problem 8.55AP

The acid in the reaction is ${\text{H}}_{\text{2}}\text{O}\text{.}$

The conjugate base of the acid is ${\text{OH}}^{\text{-}}\text{.}$

The base in the reaction is ${\text{HSO}}_{\text{4}}{}^{\text{-}}\text{.}$

The conjugate acid of the base is ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{.}$

Explanation of Solution

The given reaction is,

  ${\text{HSO}}_{\text{4}}{}^{\text{-}}\left(\text{aq}\right){\text{+H}}_{\text{2}}\text{O}\left(\text{l}\right)\stackrel{}{\to }{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(\text{aq}\right){\text{+OH}}^{\text{-}}\left(\text{aq}\right)$

Hydrogen sulphate $\left({\text{HSO}}_{\text{4}}{}^{\text{-}}\right)$ gains a proton and forms sulphuric acid.  Water loses a proton to form hydronium ion.

  $\begin{array}{l}{\text{HSO}}_{\text{4}}{}^{\text{-}}\left(\text{aq}\right){\text{+H}}_{\text{2}}\text{O}\left(\text{l}\right)\stackrel{}{\to }{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(\text{aq}\right){\text{+OH}}^{\text{-}}\left(\text{aq}\right)\\ \text{Base}\text{Water}\text{Conjugate}\text{Conjugate}\\ \text{Acid}\text{Base}\end{array}$