Understanding Ka and Acid Dissociation Revision Notes | A Level AQA Chemistry

1. Why Weak Acids Behave Differently

Picture two beakers, each containing the same concentration of acid. In the first beaker — hydrochloric acid — every single HCl molecule has split apart into H⁺ and Cl⁻ ions. In the second beaker — ethanoic acid at the same concentration — the vast majority of molecules are still intact, with only a small fraction having dissociated. The pH values are measurably different, even though the concentrations are identical.

That difference is the heart of this topic. Strong acids dissociate completely; weak acids dissociate only slightly in aqueous solution. The word slightly is doing real work here — it means a dynamic equilibrium is established, and that equilibrium lies heavily to the left, favouring the undissociated acid.

The general dissociation equation for a weak acid HA is written with a reversible arrow to show the equilibrium:

Common examples of weak acids include ethanoic acid (CH₃COOH), carbonic acid (H₂CO₃), and hydrofluoric acid (HF). Each of these establishes an equilibrium in water, with the equilibrium position lying well to the left — meaning the concentration of undissociated acid molecules far exceeds the concentration of ions.

2. Constructing the Ka Expression

Because a weak acid dissociation is an equilibrium, you can apply the same equilibrium constant logic you used for Kc. The key difference is that water is the solvent — present in such large excess that its concentration is treated as constant and is omitted from the expression.

Starting from the equilibrium $\text{HA}(aq) \rightleftharpoons \text{H}^+(aq) + \text{A}^-(aq)$, you write the equilibrium constant expression in the usual way: products over reactants, each raised to the power of its stoichiometric coefficient. Because all coefficients here are 1, no powers are needed.

For a specific example, ethanoic acid dissociates as: $\text{CH}_3\text{COOH}(aq) \rightleftharpoons \text{H}^+(aq) + \text{CH}_3\text{COO}^-(aq)$. The Ka expression becomes $K_a = \frac{[\text{H}^+][\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]}$. Notice the undissociated acid sits on the denominator — a large denominator (lots of undissociated acid) gives a small Ka, which is exactly what you expect for a weak acid.

The units of Ka are derived from the concentration terms. Since there are two concentration terms on the numerator and one on the denominator, the units work out as mol dm⁻³. You should state units in any calculation unless the question specifies otherwise.

3. Ka as a Measure of Acid Strength

Ka is not an arbitrary number — it tells you directly how far the dissociation equilibrium lies to the right. A larger Ka means the numerator (ion concentrations) is large relative to the denominator (undissociated acid), so a greater proportion of the acid has dissociated. A smaller Ka means the equilibrium lies further to the left, with most molecules remaining intact.

This is where Ka becomes a precise, quantitative measure of acid strength within the category of weak acids. Two weak acids can be compared directly using their Ka values: the acid with the higher Ka is the stronger of the two, because it produces a higher concentration of H⁺ ions at equilibrium for the same initial concentration.

Scanning the table, methanoic acid has a Ka roughly ten times larger than ethanoic acid. Because [H⁺] ≈ √(Ka × c), a 10-fold increase in Ka gives approximately √10 ≈ 3.2 times more H⁺ ions at the same concentration — a measurably lower pH. Note that this is a square-root relationship, not a linear one: doubling Ka does not double [H⁺]. The Ka values for weak acids are typically in the range $10^{-3}$ to $10^{-10}$ mol dm⁻³, which is many orders of magnitude smaller than the effective Ka of a strong acid (which would be extremely large, reflecting complete dissociation).

4. Calculating Ka from Concentration Data

Exam questions frequently give you an initial concentration of a weak acid and either the equilibrium H⁺ concentration or the pH, then ask you to calculate Ka. The method follows a consistent pattern every time, and working through it carefully avoids the most common errors.

There is one assumption built into most A-level Ka calculations: because the acid is weak and dissociates only slightly, the equilibrium concentration of HA is approximated as equal to the initial concentration. This is the approximation that [HA]equilibrium ≈ [HA]initial. It is valid when the degree of dissociation is small (typically less than 5%), and AQA expects you to state this assumption when you use it.

Worked example: Calculating Ka for a Weak Acid

A 0.100 mol dm⁻³ solution of ethanoic acid has an equilibrium H⁺ concentration of 1.32 × 10⁻³ mol dm⁻³. Calculate Ka for ethanoic acid and state its units.

1. $$\text{CH}_3\text{COOH}(aq) \rightleftharpoons \text{H}^+(aq) + \text{CH}_3\text{COO}^-(aq)$$ — Always start by writing the equilibrium with a reversible arrow. This confirms which species appear in the Ka expression.
2. $$K_a = \frac{[\text{H}^+][\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]}$$ — Products in the numerator, undissociated acid in the denominator. Water is omitted.
3. $[\text{H}^+] = 1.32 \times 10^{-3}$ mol dm$^{-3}$ $[\text{CH}_3\text{COO}^-] = 1.32 \times 10^{-3}$ mol dm$^{-3}$ (equal to $[\text{H}^+]$ by stoichiometry) $[\text{CH}_3\text{COOH}] \approx 0.100$ mol dm$^{-3}$ (approximation: initial concentration used) — For every H⁺ produced, one CH₃COO⁻ is also produced — they are formed in a 1:1 ratio. The approximation that [HA]eq ≈ [HA]initial is valid here because 1.32 × 10⁻³ is much smaller than 0.100 (about 1.3% dissociation). State this assumption explicitly in an exam.
4. $$K_a = \frac{(1.32 \times 10^{-3}) \times (1.32 \times 10^{-3})}{0.100}$$ — Substitute the three equilibrium concentrations directly. Keep the powers of ten explicit — do not collapse them.
5. $(1.32 \times 10^{-3}) \times (1.32 \times 10^{-3}) = 1.32 \times 1.32 \times 10^{-3} \times 10^{-3}$ $1.32 \times 1.32 = 1.7424$ $10^{-3} \times 10^{-3} = 10^{-6}$ Numerator $= 1.7424 \times 10^{-6}$ mol$^2$ dm$^{-6}$ — Multiply the coefficients and add the exponents separately. Never skip this step — collapsing powers of ten is a common source of error.
6. $$K_a = \frac{1.7424 \times 10^{-6}}{0.100} = 1.74 \times 10^{-5} \text{ mol dm}^{-3}$$ — Dividing mol² dm⁻⁶ by mol dm⁻³ gives mol dm⁻³. The answer rounds to 3 significant figures, matching the data given. This agrees with the accepted Ka for ethanoic acid.

Answer: $K_a = 1.74 \times 10^{-5}$ mol dm⁻³

Notice how the approximation made the calculation much more straightforward — instead of solving a quadratic equation, you substituted the initial concentration directly for [HA]. AQA mark schemes award a mark for stating this assumption, so always write it down explicitly.

5. Comparing Weak and Strong Acids

Now that Ka is established, it is worth stepping back to compare weak and strong acids side by side — because AQA questions frequently ask you to compare them, and the comparison needs to be precise rather than vague.

One comparison that catches students out is conductivity. Because a weak acid produces far fewer ions than a strong acid at the same concentration, its electrical conductivity is lower. This is a direct, measurable consequence of partial dissociation and is a useful piece of evidence for distinguishing weak from strong acids experimentally.

Another distinction worth noting: if you dilute a weak acid, the degree of dissociation increases (the equilibrium shifts right), even though Ka itself stays constant. When water is added, all concentrations fall, causing the reaction quotient Q to drop below Ka. The system then shifts right to restore Ka — and because the right-hand side of HA ⇌ H⁺ + A⁻ has two moles of solute particles compared to one on the left, the shift towards more particles is consistent with Le Chatelier's principle. Ka does not change because temperature has not changed.

6. Exam Technique and Mark-Winning Precision

Ka questions appear across all three A-level papers and reward students who are precise with language and method. The marks are often straightforward to collect — but they are easy to drop through imprecise wording or skipped steps.

When a question asks you to construct a Ka expression, write the full equilibrium equation first, then derive the expression from it. Examiners want to see the logical chain. If you write the Ka expression without the equilibrium equation, you risk writing it for the wrong species — and you lose the method mark.

For calculate questions, the mark scheme typically awards: one mark for the correct Ka expression, one mark for correct substitution with the approximation stated, and one mark for the correct numerical answer with units. Missing units costs the final accuracy mark. Writing the approximation statement — "[HA]eq ≈ [HA]initial because the acid is weak and dissociation is slight" — secures a mark that many students leave on the table.