R3.1.11 Conjugate Acid-Base pair (Higher Level Only)

Conjugate Acid-Base Pair: The Relationship Between $K_{a}$ , $K_{b}$ , and $K_{w}$

Acid Ionization Constant ( $K_{a}$ )

When a weak acid $H A$ dissociates in water, the equilibrium can be expressed as:

$H A (a q) ⇌ H^{+} (a q) + A^{-} (a q)$

The equilibrium constant for this reaction, $K_{a}$ , is defined as:

$K_{a} = \frac{[H^{+}] [A^{-}]}{[H A]}$

Here:

$[H^{+}]$ : Concentration of hydrogen ions (protons)
$[A^{-}]$ : Concentration of the conjugate base
$[H A]$ : Concentration of the undissociated acid

Tip

For strong acids, $K_{a}$ is very large because the acid dissociates almost completely. For weak acids, $K_{a}$ is small, indicating partial dissociation.

Base Ionization Constant ( $K_{b}$ )

The conjugate base $A^{-}$ of the weak acid $H A$ can react with water to form $O H^{-}$ ions:

$A^{-} (a q) + H_{2} O (l) ⇋ O H^{-} (a q) + H A (a q)$

The equilibrium constant for this reaction, $K_{b}$ , is defined as:

$K_{b} = \frac{[O H^{-}] [H A]}{[A^{-}]}$

Note

The stronger the conjugate base, the larger the $K_{b}$ . Conversely, weak conjugate bases have smaller $K_{b}$ values.

The Ionic Product of Water ( $K_{w}$ )

Water itself undergoes a small degree of ionization:

$H_{2} O (l) ⇋ H^{+} (a q) + O H^{-} (a q)$

The equilibrium constant for this reaction, known as the ionic product of water, is:

$K_{w} = [H^{+}] [O H^{-}]$

At 298 K, $K_{w} = 1.0 \times 10^{- 14}$ .

Common Mistake

Many students forget that $K_{w}$ changes with temperature. Always check the temperature before using $K_{w} = 1.0 \times 10^{- 14}$ .

Deriving $K_{a} \times K_{b} = K_{w}$

Let’s combine the ionization reactions for $H A$ and $A^{-}$ :
1. Acid ionization: $H A ⇋ H^{+} + A^{-}$
2. Base ionization: $A^{-} + H_{2} O ⇋ O H^{-} + H A$
Adding these reactions cancels $H A$ and $A^{-}$ , leaving:

$H_{2} O ⇋ H^{+} + O H^{-}$

The equilibrium constant for this overall reaction is $K_{w}$ . According to the rules of equilibrium constants:

$K_{w} = K_{a} \times K_{b}$

Example

For acetic acid ( $K_{a} = 1.8 \times 10^{- 5}$ ) and its conjugate base, acetate ( $K_{b} = 5.6 \times 10^{- 10}$ ): $K_{a} \times K_{b} = (1.8 \times 10^{- 5}) (5.6 \times 10^{- 10}) = 1.0 \times 10^{- 14} = K_{w}$

Applications of $K_{a} \times K_{b} = K_{w}$

1. Calculating $K_{b}$ from $K_{a}$

Suppose you know $K_{a}$ for a weak acid and need to find $K_{b}$ for its conjugate base. Simply rearrange the relationship:

$K_{b} = \frac{K_{w}}{K_{a}}$

Example

For formic acid ( $K_{a} = 1.8 \times 10^{- 4}$ ), the $K_{b}$ of its conjugate base is: $K_{b} = \frac{1.0 \times 10^{- 14}}{1.8 \times 10^{- 4}} = 5.6 \times 10^{- 11}$

2. Calculating $K_{a}$ from $K_{b}$

Similarly, if $K_{b}$ is known, you can calculate $K_{a}$ :

$K_{a} = \frac{K_{w}}{K_{b}}$

Self review

Given $K_{b} = 2.5 \times 10^{- 5}$ , what is $K_{a}$ for the conjugate acid?

3. Solving Acid–Base Equilibrium Problems

When analyzing weak acid–base equilibria, knowing $K_{a}$ , $K_{b}$ , and $K_{w}$ allows you to:

Predict the pH of a solution
Determine the extent of ionization
Compare the strengths of acids and bases

Tip

Remember: Strong acids have weak conjugate bases, and weak acids have strong conjugate bases. $K_{a}$ and $K_{b}$ are inversely related.

Reflection and Further Questions

Self review

What happens to $K_{a}$ and $K_{b}$ if the temperature of the system increases?
Why does $K_{w}$ depend on temperature, and how might this affect calculations in non-standard conditions?
How can the relationship $K_{a} \times K_{b} = K_{w}$ be used to evaluate the safety or effectiveness of chemical processes?

Theory of Knowledge

How does the relationship between $K_{a}$ , $K_{b}$ , and $K_{w}$ illustrate the interconnectedness of chemical systems?
Can this principle be applied to other areas of knowledge?

You've read 2/2 free chapters this week.

Upgrade to PLUS or PRO to unlock all notes, for every subject.

Introduction to Conjugate Acid-Base Pairs

A conjugate acid-base pair consists of two species that differ by a single proton ( $H^+$ ). When an acid donates a proton, it becomes its conjugate base, and when a base accepts a proton, it becomes its conjugate acid.
The concept of conjugate acid-base pairs was first introduced by Johannes Brønsted and Thomas Lowry in 1923, forming the foundation of the Brønsted-Lowry acid-base theory.

Analogy

Think of a conjugate acid-base pair like a reversible transformation between two forms of a character in a story—like Dr. Jekyll and Mr. Hyde. They are the same entity, just in different states.

Example

When hydrochloric acid (

HCl

) donates a proton to water, it forms its conjugate base, chloride ion (

Cl^-

), while water becomes its conjugate acid, hydronium ion (

H_3O^+

HCl + H_2O \rightarrow Cl^- + H_3O^+

Note

The strength of an acid is inversely related to the strength of its conjugate base. A strong acid has a weak conjugate base, and vice versa.

Definition

Conjugate Acid-Base Pair

A pair of chemical species that differ by a single proton ( $H^+$ ), where one is an acid and the other is its conjugate base.

R3.1.11 Conjugate Acid-Base pair (Higher Level Only)

Conjugate Acid-Base Pair: The Relationship Between Ka, Kb, and Kw

Acid Ionization Constant (Ka)

Base Ionization Constant (Kb)

The Ionic Product of Water (Kw)

Deriving Ka×Kb=Kw

Applications of Ka×Kb=Kw

1. Calculating Kb from Ka

2. Calculating Ka from Kb

3. Solving Acid–Base Equilibrium Problems

Reflection and Further Questions

You've read 2/2 free chapters this week.

Introduction to Conjugate Acid-Base Pairs

Conjugate Acid-Base Pair: The Relationship Between $K_{a}$ , $K_{b}$ , and $K_{w}$

Acid Ionization Constant ( $K_{a}$ )

Base Ionization Constant ( $K_{b}$ )

The Ionic Product of Water ( $K_{w}$ )

Deriving $K_{a} \times K_{b} = K_{w}$

Applications of $K_{a} \times K_{b} = K_{w}$

1. Calculating $K_{b}$ from $K_{a}$

2. Calculating $K_{a}$ from $K_{b}$