QCE Chemistry - Unit 3 - Chemical equilibrium systems

Dissociation Constants | QCE Chemistry

Learn how Ka and Kb describe weak acid and weak base equilibria in QCE Chemistry.

Updated 2026-05-18 - 4 min read

Dissociation constants are equilibrium constants for weak acid and weak base reactions in water. They help compare acid and base strength using numbers rather than just labels.

Acid dissociation constant, Ka

For a weak acid:

$ \mathrm{HA(aq)} + \mathrm{H_2O(l)} \rightleftharpoons \mathrm{H_3O^+(aq)} + \mathrm{A^-(aq)} $

$ K_a = \frac{[\mathrm{H_3O^+}][\mathrm{A^-}]}{[\mathrm{HA}]} $

Water is omitted because it is a pure liquid.

A larger $K_a$ means the acid ionises to a greater extent, so the acid is stronger.

Base dissociation constant, Kb

For a weak base:

$ \mathrm{B(aq)} + \mathrm{H_2O(l)} \rightleftharpoons \mathrm{BH^+(aq)} + \mathrm{OH^-(aq)} $

$ K_b = \frac{[\mathrm{BH^+}][\mathrm{OH^-}]}{[\mathrm{B}]} $

A larger $K_b$ means the base accepts protons more readily, so the base is stronger.

pKa and pKb

$ pK_a = -\log_{10}(K_a) $

$ pK_b = -\log_{10}(K_b) $

Because of the negative logarithm:

• larger $K_a$ means smaller $pK_a$
• larger $K_b$ means smaller $pK_b$
• lower $pK_a$ means stronger acid
• lower $pK_b$ means stronger base

This feels backwards at first, so always connect $pK_a$ back to $K_a$.

Conjugate strength

Strong acids have very weak conjugate bases. Weak acids have conjugate bases that are stronger by comparison.

For conjugate acid-base pairs at $25^\circ\mathrm{C}$:

$ K_a \times K_b = K_w $

This means if an acid has a large $K_a$, its conjugate base must have a small $K_b$.

So if you know $K_a$ for a weak acid, you can calculate $K_b$ for its conjugate base:

$ K_b = \frac{K_w}{K_a} $

The same relationship works in reverse. This is useful when a data table gives one value but the question asks about the conjugate partner.

Setting up weak acid calculations

Weak acid questions are equilibrium questions. If the initial acid concentration is $c$ and $x$ ionises, the equilibrium concentrations are often:

$ [\mathrm{HA}] = c - x,\quad [\mathrm{H_3O^+}] = x,\quad [\mathrm{A^-}] = x $

Then:

$ K_a = \frac{x^2}{c-x} $

If $K_a$ is very small, the shortcut $c-x \approx c$ may be valid. State the approximation when you use it, because it is an assumption, not a new formula.

Worked example

Calculation example

For ethanoic acid:

$ \mathrm{CH_3COOH(aq)} + \mathrm{H_2O(l)} \rightleftharpoons \mathrm{H_3O^+(aq)} + \mathrm{CH_3COO^-(aq)} $

If $K_a = 1.8 \times 10^{-5}$, then:

$ pK_a = -\log_{10}(1.8 \times 10^{-5}) = 4.74 $

The small $K_a$ confirms that ethanoic acid only partially ionises in water.

Exam traps

Other traps:

• including water in $K_a$ or $K_b$
• confusing weak with dilute
• comparing $K_a$ values without checking they refer to acids
• forgetting that $K_a$ and $K_b$ depend on temperature

Quick check

Sources

• QCAA Chemistry subject page
• QCAA Chemistry 2025 syllabus
• OpenStax Chemistry 2e: Relative Strengths of Acids and Bases
• OpenStax Chemistry 2e: pH and pOH