Chapter 7, Problem 6DQ

(a)

Interpretation Introduction

Interpretation: The equilibrium picture of 10 molecules of HA and HCl acid in their two separate aqueous solutions needs to be drawn.

Concept Introduction: An acid is the substance that gives H+ or ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions in its aqueous solution. On the basis of acid dissociation, acids can be classified as strong and weak acid. A strong acid is ionized completely and gives the respective ions.

  ${\text{HA}}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l) }}\stackrel{}{\to }{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+A}}^{\text{-}}{}_{\text{(aq)}}$

On the contrary, a weak acid ionized partially and reaches to equilibrium.

  ${\text{HA}}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l) }}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+A}}^{\text{-}}{}_{\text{(aq)}}$

Answer to Problem 6DQ

  

Explanation of Solution

HA is weak acid and HCl is a strong acid. For a weak acid HA, the ionization reaction can be written as:

  ${\text{HA}}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l) }}\stackrel{}{\to }{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+A}}^{\text{-}}{}_{\text{(aq)}}$

Since it is partial ionization hence some molecules of HCl and water must be present in the solution and the picture of equilibrium can be shown as:

  Whereas HCl is a strong acid so it will ionize completely as given below. Hence no HCl and water molecules must be present in the solution.

  ${\text{HCl}}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l) }}\stackrel{}{\to }{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+Cl}}^{\text{-}}{}_{\text{(aq)}}$

Hence the picture of equilibrium can be shown as:

  

(b)

Interpretation Introduction

Interpretation: The major species in the two separate aqueous solution of HA and HCl acid needs to be determined.

Concept Introduction: An acid is the substance that gives H+ or ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions in its aqueous solution. On the basis of acid dissociation, acids can be classified as strong and weak acid. A strong acid is ionized completely and gives the respective ions.

  ${\text{HA}}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l) }}\stackrel{}{\to }{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+A}}^{\text{-}}{}_{\text{(aq)}}$

On the contrary, a weak acid ionized partially and reaches to equilibrium.

  ${\text{HA}}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l) }}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+A}}^{\text{-}}{}_{\text{(aq)}}$

Answer to Problem 6DQ

  $\begin{array}{l}{\text{In HA}}_{\text{(aq)}}{\text{:HA}}_{\text{(aq)}}{\text{, H}}_{\text{2}}{\text{O}}_{\text{(l) }},{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+A}}^{\text{-}}{}_{\text{(aq)}}\\ {\text{In HCl}}_{\text{(aq)}}{\text{: H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{and Cl}}^{\text{-}}{}_{\text{(aq)}}\end{array}$

Explanation of Solution

HA is weak acid and HCl is a strong acid. For a weak acid HA, the ionization reaction can be written as:

  ${\text{HA}}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l) }}\stackrel{}{\to }{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+A}}^{\text{-}}{}_{\text{(aq)}}$

Whereas HCl is a strong acid so it will ionize completely as given below:

  ${\text{HCl}}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l) }}\stackrel{}{\to }{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+Cl}}^{\text{-}}{}_{\text{(aq)}}$

Hence the major species in both acids must be:

  $\begin{array}{l}{\text{In HA}}_{\text{(aq)}}{\text{:HA}}_{\text{(aq)}}{\text{, H}}_{\text{2}}{\text{O}}_{\text{(l) }},{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+A}}^{\text{-}}{}_{\text{(aq)}}\\ {\text{In HCl}}_{\text{(aq)}}{\text{: H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{and Cl}}^{\text{-}}{}_{\text{(aq)}}\end{array}$

(c)

Interpretation Introduction

Interpretation: The K_a value from the equilibrium picture of 10 molecules of HA and HCl acid in their two separate aqueous solution needs to be calculated.

  Concept Introduction: An acid is the substance that gives H+ or ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions in its aqueous solution. On the basis of acid dissociation, acids can be classified as strong and weak acid. A strong acid is ionized completely and gives the respective ions.

  ${\text{HA}}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l) }}\stackrel{}{\to }{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+A}}^{\text{-}}{}_{\text{(aq)}}$

On the contrary, a weak acid ionized partially and reaches to equilibrium.

  ${\text{HA}}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l) }}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+A}}^{\text{-}}{}_{\text{(aq)}}$

Answer to Problem 6DQ

  $\begin{array}{l}{\text{K}}_{\text{a}}\text{ for HA = 5}\\ {\text{K}}_{\text{a}}\text{for HCl }=100\end{array}$

Explanation of Solution

The K_aexpression for both acids can be written as given :

  $\begin{array}{l}\text{Ka = }\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{] [A}}^{\text{-}}{}_{\text{(aq)}}]}{[{\text{HA}}_{\text{(aq)}}]}\\ \text{Ka = }\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{] [Cl}}^{\text{-}}{}_{\text{(aq)}}]}{[{\text{HCl}}_{\text{(aq)}}]}\end{array}$

Substitute the values of number of molecules from each picture:

  $\begin{array}{l}{\text{K}}_{\text{a}}\text{ for HA = }\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{] [A}}^{\text{-}}{}_{\text{(aq)}}]}{[{\text{HA}}_{\text{(aq)}}]}\text{= }\frac{\text{[5] [5}]}{[\text{5}]}=5\\ {\text{K}}_{\text{a}}\text{for HCl = }\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{] [Cl}}^{\text{-}}{}_{\text{(aq)}}]}{[{\text{HCl}}_{\text{(aq)}}]}\frac{\text{[10] [10}]}{[\text{0}]}=100\end{array}$

(d)

Interpretation Introduction

Interpretation: The order from strongest to weakest base for ${\text{H}}_{\text{2}}\text{O}$ , ${\text{A}}^{\text{-}}$ and ${\text{Cl}}^{\text{-}}$ needs to be determined.

Concept Introduction: An acid is the substance that gives H+ or ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions in its aqueous solution. On the basis of acid dissociation, acids can be classified as strong and weak acid. A strong acid is ionized completely and gives the respective ions.

  ${\text{HA}}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l) }}\stackrel{}{\to }{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+A}}^{\text{-}}{}_{\text{(aq)}}$

On the contrary, a weak acid ionized partially and reaches to equilibrium.

  ${\text{HA}}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l) }}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+A}}^{\text{-}}{}_{\text{(aq)}}$

Answer to Problem 6DQ

  ${\text{A}}^{\text{-}}{\text{> H}}_{\text{2}}{\text{O> Cl}}^{\text{-}}$

Explanation of Solution

HA is weak acid and HCl is a strong acid. For a weak acid HA, the ionization reaction can be written as:

  ${\text{HA}}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l) }}\stackrel{}{\to }{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+A}}^{\text{-}}{}_{\text{(aq)}}$

Whereas HCl is a strong acid so it will ionize completely as given below:

  ${\text{HCl}}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l) }}\stackrel{}{\to }{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+Cl}}^{\text{-}}{}_{\text{(aq)}}$

According to the Bronsted-Lowery acid-base theory, a strong acid gives H+ ions and form weak conjugate base whereas a strong base accepts H+ ion to form weak conjugate acid of it. Since HA is weak acid and forms ${\text{A}}^{\text{-}}$ as its conjugate base therefore it must be strong conjugate base whereas HCl is a strong acid and form ${\text{Cl}}^{\text{-}}$ ion as its conjugate base therefore it must be weak conjugate base. With both acids, ${\text{H}}_{\text{2}}\text{O}$ acts as base and accepts H+ ions hence it must have intermediate value of both bases; ${\text{A}}^{\text{-}}$ and ${\text{Cl}}^{\text{-}}$ . That makes the order of strength as:

  ${\text{A}}^{\text{-}}{\text{> H}}_{\text{2}}{\text{O> Cl}}^{\text{-}}$