Acid & Base Dissociation Constants (HL) | HL IB Chemistry Revision Notes 2025

Acid & Base Dissociation Constants

Weak acids

A weak acid is an acid that partially (or incompletely) dissociates in aqueous solutions
- For example, most carboxylic acids (e.g. ethanoic acid), HCN (hydrocyanic acid), H₂S (hydrogen sulfide) and H₂CO₃ (carbonic acid)
In general, the following equilibrium is established:

HA (aq) + H₂O (l) ⇌ A^- (aq) + H₃O⁺ (aq)

HA (aq) ⇌ A^- (aq) + H⁺ (aq)

At equilibrium, the majority of HA molecules remain unreacted
The position of the equilibrium is more towards the left and an equilibrium is established
As this is an equilibrium, we can write an equilibrium constant expression for the reaction
This constant is called the acid dissociation constant, K_a

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

Carboxylic acids are weak acids
- For example, propanoic acid, CH₃CH₂COOH (aq), dissociates according to the following equation which leads to the K_a expression for propanoic acid:

CH₃CH₂COOH (aq) + H₂O (l) ⇌ CH₃CH₂COO^– (aq) + H₃O⁺ (aq)

CH₃CH₂COOH (aq) ⇌ CH₃CH₂COO^– (aq) + H⁺ (aq)

- The acid dissociation constant expressions for propanoic acid:
  - $K_{a} = \frac{[{CH}_{3} {CH}_{2} {COO}^{-}] [H^{+}]}{[{CH}_{3} {CH}_{2} COOH]}$

Values of K_aare very small
- For example, K_a for propanoic acid = 1.34 x 10^-5
- When writing the equilibrium expression for weak acids, we assume that the concentration of H⁺(aq) due to the ionisation of water is negligible

Weak bases

A weak base will also ionise in water and we can represent this with the base dissociation constant, K_b
In general, the equilibrium established is:

B (aq) + H₂O (l) ⇌ BH⁺ (aq) + OH^- (aq)

The base dissociation constant expression is:

$K_{b} = \frac{[{BH}^{+}] [{OH}^{-}]}{[B]}$

Amines are weak bases
- For example, 1-phenylmethanamine, C₆H₅CH₂NH₂ (aq), dissociates according to the following equation which leads to the K_a expression for 1-phenylmethanamine:

C₆H₅CH₂NH₂ (aq) + H₂O (l) ⇌ C₆H₅CH₂NH₃⁺ (aq) + OH^- (aq)

- Base dissociation constant expression for 1-phenylmethanamine
  - $K_{b} = \frac{[{CH}_{5} {CH}_{2} {NH}_{3}^{+}] [{OH}^{-}]}{[C_{6} H_{5} {CH}_{2} {NH}_{2}]}$

pK_aand pK_b

The range of values of K_aand K_bis very wide
For weak acids, the values themselves are very small numbers

Table of K_avalues

Acid	K_a	*pK_a*
Methanoic acid, HCOOH	1.77 x 10^–4	3.75
Ethanoic acid, CH₃COOH	1.74 x 10^–5	4.75
Benzoic acid, C₆H₅COOH	6.46 x 10^–5	4.18
Carbonic acid, H₂CO₃	4.30 x 10^-5	6.36

For this reason, it is easier to work with another term called pK_afor acids or pK_bfor bases
In order to convert the values we need to apply the following calculations:

pK_a= -logK_aK_a= 10^-pK_a

pK_b= -logK_bK_b= 10^-pK_b

The range of pK_avalues for most weak acids lies between 3 and 7

Relative Strengths of Acids and Bases

The larger the K_a value, the stronger the acid
The larger the pK_a value, the weaker the acid
The larger the K_b value, the stronger the base
The larger the pK_b value, the weaker the base

Diagram showing the relationship between strong and weak acids / bases

Diagram showing the relative strengths of acids and bases in terms of pKa and pKb

pK_aand pK_btell us the relative strengths of acids and bases

In all aqueous solutions, an equilibrium exists in water where a few water molecules dissociate into protons and hydroxide ions
We can derive an equilibrium constant for the reaction:

H₂O (l) ⇌ H⁺(aq) + OH^- (aq)

The concentration of water is constant, so the expression for K_w is:

K_w = [H⁺][OH^-]

This is a specific equilibrium constant called the ion product for water
The product of the two ion concentrations is 1.00 x 10^-14at 298 K

For conjugate acid-base pairs, K_a and K_b are related to K_w

K_a K_b = K_w

The conjugate base of ethanoic acid is the ethanoate ion, CH₃COO^–(aq)

CH₃COOH (aq) ⇌ CH₃COO^– (aq) + H⁺ (aq)

acid conjugate base

We can then put this into the K_a expression
- $K_{a} = \frac{[{CH}_{3} {COO}^{-}] [H^{+}]}{[{CH}_{3} COOH]}$

The ethanoate ion will react with water according to the following equation

CH₃COO^– (aq) + H₂O (l) ⇌ CH₃COOH (aq) + OH^- (aq)

We can then put this into the K_b expression
- $K_{b} = \frac{[{CH}_{3} COOH] [{OH}^{-}]}{[CH 3 {COO}^{-}]}$

Now, these two expressions can be combined, which corresponds to
- K_a K_b = K_w
- K_a K_b = 10^-14
Or we could say that
- pK_a + pK_b = pK_w
- pK_a + pK_b = 14
- This makes the numbers much more easy to deal with as using K_a K_b = 10^-14will give very small numbers

Combining the K_a and K_b expressions:
- $K_{a} K_{b} = \frac{[{CH}_{3} {COO}^{-}] [H^{+}]}{[{CH}_{3} COOH]} \times \frac{[{CH}_{3} COOH] [{OH}^{-}]}{[{CH}_{3} {COO}^{-}]} K_{a} K_{b} = [H^{+}] [{OH}^{-}] = K_{w}$

Or rearranging these:
- $K_{a} = \frac{K_{w}}{K_{b}}$
- $K_{b} = \frac{K_{w}}{K_{a}}$

Acid & Base Dissociation Constants (HL) (HL IB Chemistry)

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