Strong vs. Weak Acid Dissociation

8.3 Weak Acid Equilibria

Strong vs. Weak Acid Dissociation

Side-by-side animated beakers at equal concentration: strong acid fully splits, weak acid shows partial dissociation with live particle counts, pH, and % dissociation.

KapH% DissociationQuadratic Solve

Topic 8.3

Weak Acid and Base Equilibria

Explain the relationship among pH, pOH, and concentrations of all species in a solution of a monoprotic weak acid or weak base.

Weak acids react with water to produce hydronium ions, but only a small percentage of the molecules ionize. The concentration of H₃O⁺ is therefore much less than the initial acid concentration, and the vast majority of acid molecules remain un-ionized.

A weak acid solution is an equilibrium between an un-ionized acid and its conjugate base:

HA(aq) + H₂O(l) ⇌ H₃O⁺(aq) + A⁻(aq) K_a = [H₃O⁺][A⁻]/[HA], pK_a = −log K_a

The pH can be determined from the initial acid concentration and the pK_a. Weak bases behave identically:

B(aq) + H₂O(l) ⇌ HB⁺(aq) + OH⁻(aq) K_b = [OH⁻][HB⁺]/[B], pK_b = −log K_b

Percent ionization can be calculated from the pK_a and initial concentration, or directly from the initial and equilibrium concentrations:

% ionization = ([H₃O⁺]_eq / [HA]₀) × 100%

Counterintuitively, percent ionization increases as a weak acid is diluted, even though pH rises. Dilution shifts the ionization equilibrium toward the side with more particles.

EK 8.3.A.6 gives the conjugate relationship, both forms printed on the equation sheet:

K_w = K_a × K_b and pK_w = pK_a + pK_b

This is Unit 7's rule for adding reactions (7.6): the acid ionization and its conjugate base ionization sum to the autoionization of water, so their K values multiply to K_w.

The calculation. Build an ICE table, substitute into K_a, and apply the "x is small" approximation — which is almost always valid here, since typical K_a values are around 10⁻⁵. Verify with the 5% rule.

Key points

A weak acid needs an ICE table; a strong acid does not.
K_a × K_b = K_w for a conjugate pair — the stronger the acid, the weaker its conjugate base.
Percent ionization increases on dilution, even though pH rises.
Smaller pK_a means stronger acid (pK_a and strength run opposite).

Equations

$K_{a} = \frac{[ H _{3} O ^{+} ] [ A ^{-} ]}{[ HA ]}$
on the exam sheet
$K_{b} = \frac{[ O H ^{-} ] [ H B ^{+} ]}{[ B ]}$
on the exam sheet
$p K_{a} = - lo g K_{a}, p K_{b} = - lo g K_{b}$
on the exam sheet
$K_{w} = K_{a} \times K_{b}, p K_{w} = p K_{a} + p K_{b}$
on the exam sheet
$% ionization = \frac{[ H _{3} O ^{+} ] _{e q}}{[ HA ] _{0}} \times 100%$
not on the sheet

Common mistakes

Setting [H₃O⁺] equal to the initial acid concentration. That is the strong-acid shortcut and it is badly wrong for weak acids.
Using K_a for the conjugate base. Convert with K_b = K_w/K_a first.
Skipping the 5% check.
Confusing strength with concentration. A 6 M acetic acid solution is concentrated but still weak.
Thinking dilution lowers percent ionization. It raises it.

Worked example

Calculate the pH and percent ionization of 0.150 M acetic acid, K_a = 1.8 × 10⁻⁵. Then find K_b for the acetate ion.

Equilibrium and ICE table: CH₃COOH + H₂O ⇌ H₃O⁺ + CH₃COO⁻

	CH₃COOH	H₃O⁺	CH₃COO⁻
I	0.150	~0	0
C	−x	+x	+x
E	0.150−x	x	x

Substitute with the approximation (K_a is small, so 0.150 − x ≈ 0.150):
1.8 × 10⁻⁵ = x² / 0.150
x² = 2.70 × 10⁻⁶ → x = 1.64 × 10⁻³ M

Check the 5% rule: (1.64 × 10⁻³ / 0.150) × 100% = 1.1% < 5% ✓

pH = −log(1.64 × 10⁻³) = 2.78

Percent ionization = 1.1% — the solution is 98.9% intact acetic acid molecules.

K_b for acetate:
K_b = K_w/K_a = (1.0 × 10⁻¹⁴)/(1.8 × 10⁻⁵) = 5.6 × 10⁻¹⁰

A very small K_b, confirming that acetate is a weak base — as expected for the conjugate of a moderately weak acid.

Full notes for topic 8.3 →