Chapter 16.8, Problem 16PPA

(a)

Interpretation Introduction

Interpretation:

The ${\text{K}}_a$ and $K_b$ for each of the given ions has to be calculated.

Concept Information:

Acid ionization constant ${\text{K}}_{\text{a}}$:

The equilibrium expression for the reaction ${\text{HA}}_{(aq)}\stackrel{}{\to }{\text{H}}^{+}{}_{(aq)}+{\text{A}}^{\text{-}}{}_{(aq)}$ is given below.

$K_a=\frac{[{\text{H}}^{\text{+}}{\text{][A}}^{\text{-}}\text{]}}{[\text{HA]}}$

Where $K_a$ is acid ionization constant, $[{\text{H}}^{\text{+}}\text{]}$  is concentration of hydrogen ion, $[{\text{A}}^{\text{-}}\text{]}$  is concentration of acid anion, $[\text{HA]}$ is concentration of the acid

Base  ionization constant $K_b$

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

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

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

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

Relationship between $K_a$ and $K_b$

$K_a\times K_b=K_w$

Where $K_w$ is autoionization of water.

$\begin{array}{l}\text{The autoionization of water}:\\ K_w=[{\text{H}}^{\text{+}}\text{]}[{\text{OH}}^{\text{-}}]\\ =1.0\times {10}^{-14}\end{array}$

To Calculate: The $K_b$ of the benzoate ion ${\text{(C}}_{\text{6}}{\text{H}}_5{\text{COO}}^{\text{-}})$

(b)

Interpretation Introduction

Interpretation:

The ${\text{K}}_a$ and $K_b$ for each of the given ions has to be calculated.

Concept Information:

Acid ionization constant ${\text{K}}_{\text{a}}$:

The equilibrium expression for the reaction ${\text{HA}}_{(aq)}\stackrel{}{\to }{\text{H}}^{+}{}_{(aq)}+{\text{A}}^{\text{-}}{}_{(aq)}$ is given below.

$K_a=\frac{[{\text{H}}^{\text{+}}{\text{][A}}^{\text{-}}\text{]}}{[\text{HA]}}$

Where $K_a$ is acid ionization constant, $[{\text{H}}^{\text{+}}\text{]}$  is concentration of hydrogen ion, $[{\text{A}}^{\text{-}}\text{]}$  is concentration of acid anion, $[\text{HA]}$ is concentration of the acid

Base  ionization constant $K_b$

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

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

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

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

Relationship between $K_a$ and $K_b$

$K_a\times K_b=K_w$

Where $K_w$ is autoionization of water.

$\begin{array}{l}\text{The autoionization of water}:\\ K_w=[{\text{H}}^{\text{+}}\text{]}[{\text{OH}}^{\text{-}}]\\ =1.0\times {10}^{-14}\end{array}$

To Calculate: The $K_b$ of the chloroacetate ion ${\text{(CH}}_{\text{2}}{\text{ClCOO}}^{\text{-}})$

(c)

Interpretation Introduction

Interpretation:

The ${\text{K}}_a$ and $K_b$ for each of the given ions has to be calculated.

Concept Information:

Acid ionization constant ${\text{K}}_{\text{a}}$:

The equilibrium expression for the reaction ${\text{HA}}_{(aq)}\stackrel{}{\to }{\text{H}}^{+}{}_{(aq)}+{\text{A}}^{\text{-}}{}_{(aq)}$ is given below.

$K_a=\frac{[{\text{H}}^{\text{+}}{\text{][A}}^{\text{-}}\text{]}}{[\text{HA]}}$

Where $K_a$ is acid ionization constant, $[{\text{H}}^{\text{+}}\text{]}$  is concentration of hydrogen ion, $[{\text{A}}^{\text{-}}\text{]}$  is concentration of acid anion, $[\text{HA]}$ is concentration of the acid

Base  ionization constant $K_b$

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

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

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

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

Relationship between $K_a$ and $K_b$

$K_a\times K_b=K_w$

Where $K_w$ is autoionization of water.

$\begin{array}{l}\text{The autoionization of water}:\\ K_w=[{\text{H}}^{\text{+}}\text{]}[{\text{OH}}^{\text{-}}]\\ =1.0\times {10}^{-14}\end{array}$

To Calculate: The $K_a$ of the ethylammonium ion ${\text{(C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{NH}}_3{}^{\text{+}})$