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Library / Foundations: Chemistry / FCH-24

Foundations: Chemistry

pH of a weak acid

Estimates the pH of a weak acid from its Ka and concentration.

Ref
FCH-24
Rev
v1
Field
Foundations: Chemistry
Result
pH

For HA ⇌ H⁺ + A⁻ with [H⁺] = [A⁻] and [HA] ≈ initial concentration c₀ (valid when the acid is only slightly dissociated): [H⁺] = √(Ka c₀). Then pH = −log₁₀[H⁺].

A weak acid only partly splits into ions, so it makes far fewer hydrogen ions than a strong acid of the same concentration.

Calculator

Solve for
mol/dm³
1e-120.01
mol/dm³
1e-45
  • [H⁺]0.001342 mol/dm³
  • Fraction dissociated1.342 %

What drives the result

Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.

pH () current value 2.872
KaAcid dissociation constantKey driver
1.56.5
c₀Initial concentration
2.024.37
  • Ethanoic acid at 0.100 mol/dm³ has a pH of about 2.87, compared with 1.00 for the same concentration of hydrochloric acid.
  • The approximation fails when Ka is large or the acid very dilute.
Chart data as tables

Ranked variables

VariableTypical rangeLowest resultHighest resultSwing ()Local stepChange in result
Ka Acid dissociation constant1e-12 to 0.01 mol/dm³1.56.55+10%−0.0207
c₀ Initial concentration1e-4 to 5 mol/dm³2.0234.3722.349+10%−0.0207

Sweep of Initial concentration

Initial concentration (mol/dm³)pH
1e-44.372
3.867e-44.079
0.0014953.785
0.0057823.491
0.022363.198
0.086472.904
0.33442.61
1.2932.317
52.023

Sweep one variable

pH ()Hover, touch, or use the arrow keys on the chart
22.533.544.5 1e−40.0010.010.11 Initial concentration (mol/dm³), log scale now 2.872

Variables and how each one contributes

pH

pH

Result ·

What the equation gives you. Everything else feeds it.

Ka

Acid dissociation constant

mol/dm³ · typical 1e-12 to 0.01

Ka of the acid (ethanoic acid 1.8×10⁻⁵). A larger Ka means a stronger acid and a lower pH.

c₀

Initial concentration

mol/dm³ · typical 1e-4 to 5

Concentration of the acid before dissociation. More concentrated acid has a lower pH.

Worked example with the values above

  1. [H⁺] = √(Ka c₀) = √(1.8×10−5 × 0.1) = 0.001342 mol/dm³.
  2. Take −log₁₀.
  3. pH = 2.872.

Assumptions and limits

  • Dissociation is small compared with c₀.
  • Water’s own H⁺ is negligible.

Common mistakes

  • Using c₀ as [H⁺].
  • Forgetting the square root.

Statistics and uncertainty

Ka is typically known to two significant figures, which affects pH in the second decimal place.

Where it is used

  • Predicting acidity of weak acids.
  • Buffer and cleaning solution formulation.

Maths you need

  • Equilibrium constants: Kc
  • pH: −log[H⁺]

Related formulas

Practise this formula