Chapter 18, Problem 18.186P

(a)

Interpretation Introduction

Interpretation:

The given acids have to be ranked in the increasing order of ${\text{K}}_{\text{a}}.$

Concept Introduction:

Consider the following acid dissociation reaction.

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

Where,

    $\begin{array}{l}\text{HA = acid}\\ {\text{A}}^{\text{-}}\text{ = conjugate base of the acid HA}\end{array}$

The equilibrium constant for the reaction can be written as follows,

  $\text{K=}\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\left[{\text{A}}^{\text{-}}\right]}{\left[\text{HA}\right]\left[{\text{H}}_{\text{2}}\text{O}\right]}$

The concentration of water is taken as unity. So the new equilibrium constant, the acid dissociation constant (${\text{K}}_{\text{a}}$) takes the value as follows,

  ${\text{K×1 = K}}_{\text{a}}\text{=}\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\left[{\text{A}}^{\text{-}}\right]}{\left[\text{HA}\right]}$

For strong acids the dissociation will be more, so more ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ is produced. Hence larger value of ${\text{K}}_{\text{a}}$ is obtained.

(b)

Interpretation Introduction

Interpretation:

The given acids are ranked in the increasing order ${\text{pK}}_{\text{a}}.$

Concept Introduction:

Consider the following acid dissociation reaction.

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

Where,

  $\begin{array}{l}\text{HA = acid}\\ {\text{A}}^{\text{-}}\text{ = conjugate base of the acid HA}\end{array}$

The equilibrium constant for the reaction can be written as follows,

  $\text{K=}\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\left[{\text{A}}^{\text{-}}\right]}{\left[\text{HA}\right]\left[{\text{H}}_{\text{2}}\text{O}\right]}$

The concentration of water is taken as unity. So the new equilibrium constant, the acid dissociation constant (${\text{K}}_{\text{a}}$) takes the value as follows,

  ${\text{K×1 = K}}_{\text{a}}\text{=}\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\left[{\text{A}}^{\text{-}}\right]}{\left[\text{HA}\right]}$

For strong acids the dissociation will be more, so more ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ is produced. Hence larger value of ${\text{K}}_{\text{a}}$ is obtained.

${\text{pK}}_{\text{a}}$ value is also used to express the strength of acids. The expression ${\text{pK}}_{\text{a}}$ is given below.

  ${\text{pK}}_{\text{a}}{\text{ = - log}}_{\text{10}}{\text{K}}_{\text{a}}$

As the acid strength increases, ${\text{K}}_{\text{a}}$ increases hence ${\text{pK}}_{\text{a}}$ decreases.

(c)

Interpretation Introduction

Interpretation:

The conjugate base of the given acids has to be ranked in the increasing order of ${\text{pK}}_{\text{b}}.$

Concept Introduction:

Consider the following acid dissociation reaction.

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

Where,

  $\begin{array}{l}\text{HA = acid}\\ {\text{A}}^{\text{-}}\text{ = conjugate base of the acid HA}\end{array}$

There is relationship between ${\text{K}}_{\text{a}}$ and ${\text{K}}_{\text{b}}$ of HA and ${\text{A}}^{\text{-}}.$ This is obtained by adding the dissociation reaction of both HA and A- in water as follows,

  $\frac{\begin{array}{l}\cancel{\text{HA}}{\text{ + H}}_{\text{2}}\text{O }\stackrel{}{\rightleftarrows }{\text{ H}}_{\text{3}}{\text{O}}^{\text{+}}\text{+ }\cancel{{\text{A}}^{\text{-}}}\\ \cancel{{\text{A}}^{\text{-}}}{\text{ + H}}_{\text{2}}\text{O }\stackrel{}{\rightleftarrows }\cancel{\text{HA }}{\text{+ OH}}^{\text{-}}\end{array}}{{\text{2 H}}_{\text{2}}\text{O }\stackrel{}{\rightleftarrows }{\text{ H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{ + OH}}^{\text{-}}}$

Sum of these two reactions is equal to the self-ionization of water. Hence equilibrium constant for the ionization of water can be written as the product of the equilibrium constant for these two reaction as follows,

  ${\text{K}}_{\text{w}}{\text{ = K}}_{\text{a}}{\text{ (of HA) × K}}_{\text{b}}{\text{ (of A}}^{\text{-}}\text{)}$

From the equation it is clear that ${\text{K}}_{\text{a}}$ and ${\text{K}}_{\text{b}}$ have inverse relation. Hence the value of  ${\text{pK}}_{\text{a}}$ and ${\text{pK}}_{\text{b}}$ have inverse relation.

(d)

Interpretation Introduction

Interpretation:

The percentage dissociation of $\text{HX}$ has to be calculated.

Concept Introduction:

Consider the following acid dissociation reaction.

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

Where,

  $\begin{array}{l}\text{HA = acid}\\ {\text{A}}^{\text{-}}\text{ = conjugate base of the acid HA}\end{array}$

The percentage dissociation of a particular acid can be calculated as follows,

  $\text{percentage dissociation = }\frac{\text{amount of acid dissociated}}{\text{total amount of acid present initially}}\text{×100 \%}$

(e)

Interpretation Introduction

Interpretation:

The highest $\text{HY}$ and lowest $\text{pH}$ solution containing equimolar amounts of sodium salts of acids

Concept Introduction:

Consider the following reaction of sodium salt of acid in water.

  ${\text{NaA + H}}_{\text{2}}\text{O }\stackrel{}{\rightleftarrows }{\text{ Na}}^{\text{+}}{\text{ + A}}^{\text{-}}{\text{ + H}}^{\text{+}}{\text{ + OH}}^{\text{-}}$

The pH of the solution depends on the strength of $\text{HA}$. $\text{NaOH}$ will always exist as ions since it is a strong base. But depending on the strength of acid the dissociation of $\text{HA}$ varies.

The $\text{HY}$ value of base can be calculated from $\left[{\text{OH}}^{\text{-}}\right]$ as follows,

  ${\text{pOH = - log}}_{\text{10}}\left[{\text{OH}}^{\text{-}}\right]$