Weak Acids and Ka

Learning Objectives (OCR Spec 5.1.3)

By the end of this lesson you should be able to:

• Define a weak acid and explain partial dissociation
• Write the Ka expression for a weak monobasic acid
• State the meaning and units of Ka
• Define pKa as -log10(Ka) and convert between Ka and pKa
• Compare the strengths of weak acids using Ka and pKa
• Recognise that strong acids do not need Ka (it is effectively infinite)

What is a Weak Acid?

A weak acid is one that only partially dissociates in aqueous solution. For a general weak monobasic acid HA:

HA(aq) <=> H+(aq) + A-(aq)

The double arrow reflects that this is a dynamic equilibrium - most HA molecules are still intact at any given time, and only a small fraction exist as H+ and A-. Typical weak acids dissociate 1-10% at ordinary concentrations.

Examples of weak acids at OCR A-Level include:

Acid	Formula	Ka (298 K) / mol dm-3	pKa
Hydrofluoric	HF	5.6 x 10^-4	3.25
Methanoic	HCOOH	1.6 x 10^-4	3.80
Ethanoic	CH3COOH	1.74 x 10^-5	4.76
Propanoic	CH3CH2COOH	1.3 x 10^-5	4.89
Benzoic	C6H5COOH	6.3 x 10^-5	4.20
Hydrogen cyanide	HCN	4.9 x 10^-10	9.31
Phenol	C6H5OH	1.3 x 10^-10	9.89

You are not required to memorise these but you must be able to use them.

The Acid Dissociation Constant Ka

Applying the equilibrium law to HA(aq) <=> H+(aq) + A-(aq):

Ka = [H+][A-] / [HA]

where each concentration is the equilibrium concentration. Ka is called the acid dissociation constant (or acid ionisation constant). For a given acid at a given temperature Ka has a fixed value.

Units of Ka: (mol dm-3)(mol dm-3) / (mol dm-3) = mol dm-3.

The larger the Ka, the further the equilibrium lies to the right, and the stronger the acid. A Ka of 10^-2 indicates a relatively strong weak acid; a Ka of 10^-10 indicates a very weak acid.

Why water does not appear

The equation H2O is the solvent and its concentration is essentially constant, so it is absorbed into the equilibrium constant (compare with Kw). Had we written the Bronsted-Lowry form HA + H2O <=> H3O+ + A- and used [H2O] in the denominator, we would end up with the same Ka after absorbing the constant [H2O].

pKa

Because Ka values span many orders of magnitude, it is convenient to take a logarithm:

pKa = -log10(Ka)

and conversely:

Ka = 10^-pKa

Worked Example 1: Ethanoic acid has Ka = 1.74 x 10^-5 mol dm-3. Calculate pKa.

pKa = -log10(1.74 x 10^-5) = 4.76 (2 dp)

Worked Example 2: Phenol has pKa = 9.89. Calculate Ka.

Ka = 10^-9.89 = 1.29 x 10^-10 mol dm-3

Notice the inverse relationship: a large Ka corresponds to a small pKa, and vice versa. A stronger acid has a larger Ka but a smaller pKa.

Comparing Weak Acid Strengths

Use this rule: the stronger the acid, the larger Ka and the smaller pKa.

Acid	Ka	pKa	Strength
HF	5.6 x 10^-4	3.25	Strongest (of these)
HCOOH	1.6 x 10^-4	3.80
C6H5COOH	6.3 x 10^-5	4.20
CH3COOH	1.74 x 10^-5	4.76
HCN	4.9 x 10^-10	9.31
C6H5OH	1.3 x 10^-10	9.89	Weakest

All of these are "weak" compared to strong acids - even HF with Ka of ~10^-3.3 is millions of times weaker than HCl.

The Ka of a Strong Acid

You may ask why we never quote Ka for strong acids. In principle HCl <=> H+ + Cl- does have a Ka - it is simply enormous, perhaps 10^6 or greater. In practice all HCl dissociates, so [HA] at the denominator approaches zero and Ka is undefined / infinite for practical purposes. We write HCl with a single arrow -> and use pH = -log10(c) directly.

Writing Ka Expressions

The Ka expression always goes: products over reactants, each to the power of the stoichiometric coefficient.

Ethanoic acid

CH3COOH(aq) <=> H+(aq) + CH3COO-(aq) Ka = [H+][CH3COO-] / [CH3COOH]

Methanoic acid

HCOOH(aq) <=> H+(aq) + HCOO-(aq) Ka = [H+][HCOO-] / [HCOOH]

Ammonium ion (a weak acid in water)

NH4+(aq) <=> H+(aq) + NH3(aq) Ka = [H+][NH3] / [NH4+]

Hydrogen cyanide

HCN(aq) <=> H+(aq) + CN-(aq) Ka = [H+][CN-] / [HCN]

Every time, only H+ and the conjugate base appear in the numerator; undissociated HA is in the denominator.

Ka, Kb and Kw