Chapter 15, Problem 15.59QP

Write the equation relating K_a for a weak acid and K_b for its conjugate base. Use NH₃ and its conjugate acid to derive the relationship between K_a and K_b.

Interpretation Introduction

Interpretation:

• The equation relating $K_a$  for a weak acid and $K_b$ for its conjugate base has to be written.
• Using ${\text{NH}}_{\text{3}}$ and its conjugate acid ${\text{NH}}^{\text{-}}{}_{\text{4}}$, the relationship between $K_a$ and $K_b$ has to be derived.

Concept Information:

When a Bronsted acid donates a proton, what remains of the acid is known as a conjugate base; when a Bronsted base accepts a proton, the newly formed protonated species is known as a conjugate acid.  This can be given by the below equation as shown in Figure 1.

Figure 1

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

Acids ionize in water. Strong acids ionize completely whereas weak acids ionize to some limited extent.

The ionization of a weak acid $\text{HA}$ can be given as follows,

${\text{HA}}_{\text{(aq)}}\to {\text{H}}^{\text{+}}{}_{\text{(aq)}}{\text{+A}}^{\text{-}}{}_{\text{(aq)}}$

The equilibrium expression for the above reaction 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 $K_b$

Strong base dissociates into its constituent ions fully. It produces more of hydroxide ions while dissolved in water. Weak bases partially dissociates into its constituent ions.

Since, the ionization of a weak base is incomplete, it is treated in the same way as the ionization of a weak acid.

The ionization of a weak base $\text{B}$ is given by the below equation.

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

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

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

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

To Write: The equation relating $K_a$  for a weak acid and $K_b$ for its conjugate base

Answer to Problem 15.59QP

The equation relating $K_a$  for a weak acid and $K_b$ for its conjugate base is $K_a\times K_b=K_w$

Relationship between $K_a$ and  $K_b$ using ${\text{NH}}_{\text{3}}$ and its conjugate acid ${\text{NH}}^{\text{-}}{}_{\text{4}}$ is,

$\begin{array}{l}{K}_w=K_a\times K_b,\\ K_w=\frac{[{\text{NH}}_{\text{4}}{}^{\text{+}}][{\text{OH}}^{\text{-}}]}{[{\text{NH}}_{\text{3}}]}\times \frac{[{\text{NH}}_{\text{3}}{\text{][H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{[{\text{NH}}_{\text{4}}{}^{+}\text{]}}\\ =[{\text{OH}}^{\text{-}}]{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\end{array}$

Explanation of Solution

A simple relationship between the ionization constant of a weak acid ($K_a$) and the ionization constant of its conjugate base ($K_b$) can be derived as follows,

Consider, weak acid $\text{HA}$

$\begin{array}{l}{\text{HA}}_{\text{(aq)}}\to {\text{H}}^{\text{+}}{}_{\text{(aq)}}\text{+}{\text{A}}^{\text{-}}{}_{\text{(aq)}}\\ \text{conjugate}\\ \text{base}\end{array}$

The acid ionization constant will be

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

The conjugate base  reacts with water as follows,

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

The base ionization constant will be

${\text{K}}_b=\frac{[\text{HA}][{\text{OH}}^{\text{-}}]}{[{\text{A}}^{\text{-}}]}$

The product of the two ionization constants will be $K_w$ (autoionization of water)

$\begin{array}{l}{K}_a\times K_b=\frac{[{\text{H}}^{\text{+}}{\text{][A}}^{\text{-}}\text{]}}{[\text{HA]}}\times \frac{[\text{HA}][{\text{OH}}^{\text{-}}]}{[{\text{A}}^{\text{-}}]}\\ \text{On cancelling the common terms, we get}\\ K_a\times K_b=[{\text{H}}^{\text{+}}\text{]}[{\text{OH}}^{\text{-}}]\\ \text{The autoionization of water}{K}_w=[{\text{H}}^{\text{+}}\text{]}[{\text{OH}}^{\text{-}}]\\ \text{Therefore,}\\ K_a\times K_b=K_w\end{array}$

This is the simple relation relating $K_a$ and $K_b$ of weak acid.

To derive: Using ${\text{NH}}_{\text{3}}$ and its conjugate acid ${\text{NH}}^{\text{-}}{}_{\text{4}}$, the relationship between $K_a$ and $K_b$

Using ${\text{NH}}_{\text{3}}$ and its conjugate acid ${\text{NH}}^{\text{+}}{}_4$ to derive the above relationship gives:

$\begin{array}{l}{\text{NH}}_{\text{3}}{}_{(aq)}+{\text{H}}_{\text{2}}\text{O}\to {\text{NH}}_{\text{4}}{{}^{\text{+}}}_{(aq)}+{\text{OH}}^{\text{-}}{}_{(aq)}\\ {\text{K}}_b=\frac{[{\text{NH}}_{\text{4}}{}^{\text{+}}][{\text{OH}}^{\text{-}}]}{[{\text{NH}}_{\text{3}}]}\\ {\text{NH}}_{\text{4}}{{}^{+}}_{(aq)}+{\text{H}}_{\text{2}}\text{O}\to {\text{NH}}_{\text{3}}{}_{(aq)}+{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{(aq)}\\ K_a=\frac{[{\text{NH}}_{\text{3}}{\text{][H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{[{\text{NH}}_{\text{4}}{}^{+}\text{]}}\\ \text{Since},K_w=K_a\times K_b,\text{we}\text{have}:\\ K_w=\frac{[{\text{NH}}_{\text{4}}{}^{\text{+}}][{\text{OH}}^{\text{-}}]}{[{\text{NH}}_{\text{3}}]}\times \frac{[{\text{NH}}_{\text{3}}{\text{][H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{[{\text{NH}}_{\text{4}}{}^{+}\text{]}}\\ =[{\text{OH}}^{\text{-}}]{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\end{array}$

Conclusion

àThe equation relating $K_a$  for a weak acid and $K_b$ for its conjugate base was written.

àUsing ${\text{NH}}_{\text{3}}$ and its conjugate acid ${\text{NH}}^{\text{-}}{}_{\text{4}}$, the relationship between $K_a$ and $K_b$ was derived.