Lesson 2 · 25 min

Weak acids and bases: partial ionization

At the same 0.1 M, acetic acid has one seventy-fifth of the [H₃O⁺] of hydrochloric acid. Ka tells you how much ionizes — and why dilution makes a weak acid ionize more.

In this lesson

Compare weak acids with $K_\mathrm{a}$ and $\mathrm{p}K_\mathrm{a}$
Calculate the pH of a weak acid and use the 5 % rule to judge the shortcut
Explain why dilution raises the percent ionization of a weak acid
Handle weak bases and salt solutions with $K_\mathrm{a}K_\mathrm{b} = K_\mathrm{w}$

At the same 0.10 M, hydrochloric acid has pH 1.00 but acetic acid has pH 2.88 — about one seventy-sixth of the $\ce{H3O+}$ . Acetic acid molecules can give up protons; they just keep taking them back, so only about 1.3 % are ionized at any moment. That is a weak acid.

Predict first

Dilute 0.10 M acetic acid (pH 2.88) a hundredfold to 0.0010 M. What happens to the pH?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

Rises by 2, to about 4.9 Rises by about 1, to about 3.9 Barely changes Actually falls

The acid dissociation constant Ka

K_\mathrm{a} = \frac{[\mathrm{A^-}][\mathrm{H_3O^+}]}{[\mathrm{HA}]},\qquad \mathrm{p}K_\mathrm{a} = -\lg K_\mathrm{a}

Water’s concentration barely changes; it is folded into the constant and does not appear.

$K_\mathrm{a}$: acid dissociation constant: larger means stronger
$\mathrm{p}K_\mathrm{a}$: smaller means stronger; 4.76 for acetic acid
$[\mathrm{HA}]$: un-ionized acid at equilibrium

Start from concentration $C$ and let $x$ ionize. At equilibrium $[\ce{H3O+}] = [\ce{A-}] = x$ and $[\ce{HA}] = C - x$ :

K_\mathrm{a} = \frac{x^2}{C - x}

If very little ionizes ( $x \ll C$ ), the denominator is about $C$ , which gives the most-used textbook shortcut

[\mathrm{H_3O^+}] \approx \sqrt{K_\mathrm{a} C}

As a rule of thumb, while the percent ionization $x/C$ stays below 5 % the shortcut is off by less than about 0.01 pH. Otherwise solve the quadratic.

\alpha = \frac{x}{C} \approx \sqrt{\frac{K_\mathrm{a}}{C}}

The fraction ionised grows as one over the square root of the concentration (Ostwald’s dilution law): dilute 100-fold and α rises about tenfold, so

[\ce{H3O+}] = \alpha C

falls only tenfold and the pH rises by only about 1. Once α passes 5 % the approximation runs high: 13 % for 0.0010 M acetic acid against 12 % exact.

$\alpha$: fraction ionised (degree of ionisation)
$C$: initial concentration of the weak acid

Exact solution vs. the √(Ka·C) shortcut

Weak-acid ionization explorer

Weak acid Initial concentration · 0.05 M

$\ce{HF + H2O <=> F- + H3O+}$ K_a = 6.76 × 10⁻⁴

Exact pH2.26

Shortcut pH2.23

Percent ionized11.0%

Shortcut error0.025

Shortcut fails: more than 5 % ionized — solve the quadratic (or exactly, as here).

What 50 acid molecules are doing right now

intact ionized

Dilution makes a weak acid ionize more

Switch acids and find the concentration where the shortcut starts to fail. The number of ionized molecules among the 50 is the exact result, rounded.

Worked example

Hydrofluoric acid: when the shortcut fails

Find the pH of 0.050 M HF ( $\mathrm{p}K_\mathrm{a} = 3.17$ ).

Weak bases and salt solutions

A weak base takes protons from water: $\ce{NH3} + \ce{H2O} \mathrel{\ce{<=>}} \ce{NH4+} + \ce{OH-}$ , with constant $K_\mathrm{b}$ . The two constants of a conjugate pair are not independent: adding the acid’s ionization to the base’s reaction gives exactly water’s self-ionization, so

K_\mathrm{a}K_\mathrm{b} = K_\mathrm{w},\qquad \mathrm{p}K_\mathrm{a} + \mathrm{p}K_\mathrm{b} = 14.00

$\ce{NH4+}$ has $\mathrm{p}K_\mathrm{a} = 9.25$ , so $\ce{NH3}$ has $\mathrm{p}K_\mathrm{b} = 4.75$ . Likewise a salt solution is acidic or basic depending on whether its ions trade protons with water: $\ce{NH4Cl}$ is acidic, $\ce{CH3COONa}$ is basic, and $\ce{NaCl}$ is neutral (neither $\ce{Na+}$ nor $\ce{Cl-}$ exchanges protons appreciably).

Concept check

Which of these 0.1 M solutions is basic?

\ce{NH4Cl}

\ce{NaCl}

\ce{CH3COONa}

\ce{HCOOH}

Try the calculation

Calculate the pH of 0.10 M aqueous ammonia at 25 °C ( $\mathrm{p}K_\mathrm{a}$ of $\ce{NH4+}$ = 9.25).

Key takeaways

Weak acids ionize partly; a larger $K_\mathrm{a}$ (smaller $\mathrm{p}K_\mathrm{a}$ ) means a stronger acid.
$[\ce{H3O+}] \approx \sqrt{K_\mathrm{a}C}$ holds only below about 5 % ionization; otherwise solve the quadratic.
Dilution drives a weak acid to ionize more: a hundredfold dilution raises the pH by only about one unit.
Conjugate pairs obey $K_\mathrm{a}K_\mathrm{b} = K_\mathrm{w}$ ; a salt’s pH is set by how its ions trade protons with water.

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Data and references

D. C. Harris, Quantitative Chemical Analysis, 9th ed. (2016), Ch. 9; Appendix G (pKa values at 25 °C).
P. Atkins, J. de Paula, J. Keeler, Atkins’ Physical Chemistry, 11th ed., Oxford University Press (2018), Topic 6C.