Equilibrium · Lesson 9 of 14

Weak Acids and Their Ionization

“Calculate weak-acid composition by combining ionization constants with conservation of material.”

Learning Objectives

• Construct a weak-acid ionization table and define the degree of ionization. • Relate Ka, pKa and relative acid strength. • Solve weak-acid pH calculations with a quadratic or a checked approximation. • Explain how dilution affects percentage ionization and hydrogen-ion concentration differently.

A weak acid establishes a proton-transfer equilibrium

A weak acid does not convert completely into ions when dissolved. Its proton transfer to water is opposed by the reverse transfer from hydronium to its conjugate base. The equilibrium constant tells us how the ionized and unionized forms are related. Strength and concentration must be kept separate: Ka is a property of the acid in a specified medium at a specified temperature, while concentration describes the amount present.

Weak-acid equilibriumLaTeX

HA + H_{2} O ⇌ H_{3} O^{+} + A^{-}, K_{a} \approx \frac{[ H _{3} O ^{+} ] [ A ^{-} ]}{[ HA ]}

Activities relative to standard states give the dimensionless thermodynamic constant. The displayed concentration form is the dilute-solution numerical approximation; water activity is incorporated.

The larger Ka is, the greater the tendency of HA to donate a proton in water. The pKa scale reverses this ordering because it is a negative logarithm: a smaller pKa means a larger Ka and a stronger acid. Comparison requires the same temperature and solvent. Acid strength is not a measure of the number of ionizable protons.

The acid logarithmic scaleLaTeX

p K_{a} = - lo g_{10} K_{a}

For example, Ka =

1 0^{- 4}

gives pKa = 4, whereas Ka =

1 0^{- 8}

gives pKa = 8. The first acid is stronger under the same conditions.

Weak acid	Illustrative Ka at 298 K
HF	$3.5 \times 1 0^{- 4}$
HNO2	$4.5 \times 1 0^{- 4}$
HCOOH	$1.8 \times 1 0^{- 4}$
Niacin	$1.5 \times 1 0^{- 5}$
CH3COOH	$1.74 \times 1 0^{- 5}$
C6H5COOH	$6.5 \times 1 0^{- 5}$
HOCl	$3.0 \times 1 0^{- 8}$
HCN	$4.9 \times 1 0^{- 10}$
Phenol	$1.3 \times 1 0^{- 10}$

Track the fraction that ionizes

Let the initial analytical concentration of HA be c and the fraction that ionizes be alpha. If the acid is the principal source of hydronium and no conjugate base is initially added, the amount ionized is c alpha. The acid left is c times one minus alpha. Water ionization is neglected only when the acid-derived hydronium greatly exceeds the water contribution.

Stage	HA	H3O+	A−
Initial	c	Negligible for this model	0
Change	$- c α$	$+ c α$	$+ c α$
Equilibrium	$c (1 - α)$	$c α$	$c α$

Degree of ionizationLaTeX

K_{a} = \frac{c α ^{2}}{1 - α}, % ionization = 100 α

Here 0 ≤ alpha ≤ 1. These equations assume a monoprotic acid without an initially added common ion and with negligible water contribution.

Equivalently let x be the equilibrium hydronium concentration contributed by the acid. The equation becomes $K_{a} = x^{2} / (c - x)$ and rearranges to $x^{2} + K_{a} x - K_{a} c = 0$ . Solving this quadratic gives a positive root smaller than c. It remains reliable when ionization is too extensive to neglect x in the denominator.

Example — HF with appreciable ionization

Problem
For a supplied HF dataset, $K_{a} = 3.2 \times 1 0^{- 4}$ and $c = 0.020 M$ . Calculate its degree of ionization, species concentrations and pH.

1.Use $x^{2} / (0.020 - x) = 3.2 \times 1 0^{- 4}$ . Rearranging gives $x^{2} + 0.00032 x - 0.0000064 = 0$ .
2.The positive root is $x = (- 0.00032 + 0.0003 2^{2} + 4 (0.0000064)) /2 \approx 0.002375 M$ .
3.Thus $[H_{3} O^{+}] = [F^{-}] \approx 0.002375 M$ and $[HF] \approx 0.017625 M$ .
4.The fraction ionized is $0.002375/0.020 = 0.1187$ , or about 11.9%. The pH is about 2.624. Because the fraction exceeds 5%, simply replacing c − x by c would be a poor approximation.

A small-ionization approximation must be checked

If x is much smaller than c, subtracting it makes little difference to the denominator. We may then estimate x by the square root of Ka times c. After calculating it, compare x/c with one. A common practical check is that ionization below about five percent makes this approximation reasonably good for ordinary school calculations; the required accuracy ultimately determines what is acceptable.

Small-ionization estimatesLaTeX

[H_{3} O^{+}] \approx K_{a} c, α \approx \frac{K _{a}}{c}

These estimates require small ionization, negligible common ions, dilute-solution behaviour and an acid contribution large compared with water ionization.

Example — Finding the constant from pH

Problem
A monoprotic acid has concentration 0.100 M and pH 4.50. Determine its equilibrium composition, Ka and pKa in the dilute model.

1.Convert pH first: $x = [H_{3} O^{+}] = 1 0^{- 4.50} = 3.162 \times 1 0^{- 5} M$ .
2.The conjugate-base concentration is approximately x and the remaining acid is $0.100 - x = 0.0999684 M$ .
3.Then $K_{a} = x^{2} / (0.100 - x) = 1.0003 \times 1 0^{- 8}$ , giving pKa approximately 8.00. The ionized percentage is only about 0.0316%, so the denominator approximation would also work well.

Example — A checked approximate calculation

Problem
A hypothetical monoprotic acid has $K_{a} = 2.5 \times 1 0^{- 5}$ and concentration 0.080 M. Find its approximate pH and percentage ionization.

1.Estimate $x = (2.5 \times 1 0^{- 5}) (0.080) = 1.414 \times 1 0^{- 3} M$ .
2.Check $x / c = 0.001414/0.080 = 0.01768$ , about 1.77%, which supports the small-ionization approximation.
3.Then pH is approximately 2.85. The exact quadratic gives about 2.853, close to the approximate result.

Dilution increases the fraction but lowers the concentration

From the small-ionization equations, lowering c increases the fraction ionized but decreases the hydronium concentration. These two statements can both be true because the fraction is measured relative to a shrinking total acid concentration. For example, with Ka fixed at $1 0^{- 5}$ , lowering c from 0.10 M to 0.010 M raises the estimated fraction from 1.0% to 3.16%, while hydronium falls from 0.0010 M to 0.000316 M. Dilution does not change Ka at fixed temperature in the idealised model.

Two different effects of dilution— Approximate results for the same acid with Ka =

1 0^{- 5}

. The panels use different vertical scales and units; compare changes within each panel.

Before calculating, identify the important proton-transfer reaction. A weak acid with a Ka much larger than Kw generally dominates over water when its concentration is not extremely small. But comparing constants alone is insufficient at extreme dilution: concentrations must also be considered. If an approximation produces a hydronium concentration near that of pure water, include water equilibrium and charge balance rather than forcing the simple formula.

Common mistake

Do not calculate percentage ionization as 100Ka. Ka is an equilibrium ratio, while alpha is a fraction determined by both Ka and the initial concentration. Nor should every weak-acid problem automatically use the square-root formula.

Quick check

Which acid is strongest under the same conditions?

Quick check

For a monoprotic acid initially at c and with ionized fraction alpha, the remaining acid concentration is

Quick check

When is replacing c − x by c least justified?

Quick check

Within the small-ionization model, dilution of a weak acid generally

Quick check

If pH = 4.50, hydrogen-ion concentration is approximately

Quick check

The exact simple monoprotic-acid quadratic uses

Practice Problems

Using Ka = $1.74 \times 1 0^{- 5}$ , find degree of ionization and pH for 0.050 M acetic acid, checking your approximation.
A 0.010 M monoprotic acid has pH 4.15. Find its conjugate-base concentration, Ka and pKa.
For 0.10 M bromoacetic acid with alpha = 0.132, calculate hydronium concentration and pKa.
Rank acids with pKa values 3.0, 4.8 and 9.2 by strength.
For Ka = $1 0^{- 5}$ , compare hydronium concentration and percentage ionization at 0.10 M and 0.010 M.
Explain why a weak acid can have a lower pH than a sufficiently dilute strong acid.

Key Takeaways

Ka and pKa measure equilibrium acid strength under specified conditions. A change table gives Ka = c alpha squared divided by one minus alpha for the simple monoprotic model. Approximate only after identifying the dominant equilibria, and check the resulting ionized fraction.

Previous · Lesson 8

Water, pH, and Measuring Acidity

Next · Lesson 10

Weak Bases and Conjugate Constants

A weak acid establishes a proton-transfer equilibrium

Track the fraction that ionizes

A small-ionization approximation must be checked

Dilution increases the fraction but lowers the concentration

Quiz

Practice Problems

Key Takeaways