A weak acid HA only partly gives up its proton in water: HA ⇌ H⁺ + A⁻. The acid dissociation constant Ka = [H⁺][A⁻]/[HA] measures how far this equilibrium lies to the right. Because Ka values span many powers of ten – from about 10² for strong acids to 10⁻⁵⁰ for alkanes – chemists use its negative logarithm, pKa = −log₁₀ Ka. A smaller pKa means a stronger acid: acetic acid (pKa 4.76) is about 100,000 times stronger than phenol (pKa 9.99).
pKa is also the pH at which an acid is exactly half ionised. That follows from the Henderson–Hasselbalch equation, pH = pKa + log₁₀([A⁻]/[HA]): when the two concentrations are equal the log term vanishes. So if you measure the pH of a solution that contains known amounts of an acid and its salt – a buffer – you can calculate the pKa directly.
Use this calculator to convert between Ka and pKa, to determine pKa from a buffer measurement or from the base constant of the conjugate base, and to see what fraction of the acid is ionised at a pH you choose – useful for buffers, drug absorption and titration curves.
How to use the pKa Calculator
- Calculate pKa from
- Ka — Scientific notation works, e.g. 1.75e-5
- pKa
- pKb of the conjugate base
- Kb of the conjugate base
- pH — In buffer mode this is the measured pH; in the other modes it sets the pH for the “fraction ionised” result
- Conjugate base concentration [A⁻]
- Weak acid concentration [HA]
Results update as you type, every field has its own unit menu, and everything is calculated in your browser.
How it works
Ka and pKa are converted with the base-10 logarithm. From a buffer the calculator rearranges the Henderson–Hasselbalch equation; from Kb or pKb it uses Ka × Kb = Kw = 1.0 × 10⁻¹⁴ at 25 °C. The fraction ionised follows from the same equation at the pH you enter.
Worked example
From Ka: acetic acid has Ka = 1.75 × 10⁻⁵, so pKa = −log₁₀(1.75 × 10⁻⁵) = 4.757. Its conjugate base acetate has pKb = 14.00 − 4.757 = 9.243. At pH 7 the fraction ionised is 1 ÷ (1 + 10^(4.757 − 7)) = 99.43% – in neutral water acetic acid is almost entirely acetate.
From a buffer: a buffer containing 0.20 M sodium acetate and 0.15 M acetic acid has a measured pH of 4.90. pKa = 4.90 − log₁₀(0.20/0.15) = 4.90 − 0.125 = 4.775.
pKa values of common acids (water, 25 °C)
| Acid | Formula | pKa |
|---|---|---|
| Hydrochloric acid | HCl | ≈ −6 (strong) |
| Phosphoric acid (1st) | H₃PO₄ | 2.15 |
| Hydrofluoric acid | HF | 3.17 |
| Formic acid | HCOOH | 3.75 |
| Lactic acid | CH₃CH(OH)COOH | 3.86 |
| Benzoic acid | C₆H₅COOH | 4.20 |
| Acetic acid | CH₃COOH | 4.76 |
| Carbonic acid (1st) | H₂CO₃ / CO₂(aq) | 6.35 |
| Dihydrogen phosphate | H₂PO₄⁻ | 7.20 |
| Hypochlorous acid | HOCl | 7.53 |
| Boric acid | B(OH)₃ | 9.24 |
| Ammonium ion | NH₄⁺ | 9.25 |
| Phenol | C₆H₅OH | 9.99 |
| Bicarbonate | HCO₃⁻ | 10.33 |
| Hydrogen phosphate | HPO₄²⁻ | 12.35 |
Values are rounded literature values for dilute aqueous solution at 25 °C; polyprotic acids have one pKa per proton.
Tips
- Type formulas with normal capitals and numbers, e.g. Ca(OH)2 or CuSO4·5H2O – brackets and hydrates work.
- Atomic masses are IUPAC standard atomic weights; radioactive elements use the mass number of their longest-lived isotope.
- Electronegativity values are on the Pauling scale.
- Everything is calculated in your browser.
Common mistakes to avoid
- Writing CO (carbon monoxide) when you mean Co (cobalt) – capitals matter.
- Forgetting to multiply the bracket contents by the subscript after the bracket.
- Mixing up atomic number (protons) and mass number (protons + neutrons).
Inputs explained
| Field | What to enter |
|---|---|
| Calculate pKa from | Choose one: Ka (acid dissociation constant), pKa → Ka (reverse), pH and buffer concentrations (Henderson–Hasselbalch), pKb of the conjugate base … |
| Ka | A number. Scientific notation works, e.g. 1.75e-5 |
| pKa | A number |
| pKb of the conjugate base | A number |
| Kb of the conjugate base | A number |
| pH | A number. In buffer mode this is the measured pH; in the other modes it sets the pH for the “fraction ionised” result |
| Conjugate base concentration [A⁻] | Units: mol/L (M), mmol/L (mM), µmol/L (µM), nmol/L (nM), mol/m³ |
| Weak acid concentration [HA] | Units: mol/L (M), mmol/L (mM), µmol/L (µM), nmol/L (nM), mol/m³ |
What each result means
- Ka = 10^(-[pKa])
- pKb of the conjugate base (25 °C) = 14 – [pKa]
- Kb of the conjugate base (25 °C) = 10^([pKa] – 14)
- Fraction ionised (as A⁻) at the pH entered = (1÷(1 + 10^([pKa] – [pH])))×100
- Ratio [A⁻]/[HA] at that pH = 10^([pH] – [pKa])
Formula
pKa = −log₁₀(Ka) · Ka = 10^(−pKa) · pKa = pH − log₁₀([A⁻]/[HA]) · pKa + pKb = pKw = 14.00 (25 °C) · fraction ionised = 1 ÷ (1 + 10^(pKa − pH))Worked example
Inputs — Calculate pKa from: Ka (acid dissociation constant); Ka: 1.75e-05; pH: 4.9.
Results — pKa = 4.757; Ka = 1.7500e-5; pKb of the conjugate base (25 °C) = 9.243; Kb of the conjugate base (25 °C) = 5.7143e-10; Fraction ionised (as A⁻) at the pH entered = 58.16 %; Ratio [A⁻]/[HA] at that pH = 1.3901; Acid strength = Weak acid (pKa 4–10).
Key terms
- Acid dissociation constant (Ka)
- Equilibrium constant [H⁺][A⁻]/[HA] for the ionisation of an acid in water.
- pKa
- The negative base-10 logarithm of Ka; lower values mean stronger acids.
- Conjugate base (A⁻)
- The species left after an acid loses a proton.
- Henderson–Hasselbalch equation
- pH = pKa + log₁₀([A⁻]/[HA]), linking the pH of a buffer to its composition.
- Buffer
- A solution of a weak acid and its conjugate base that resists changes in pH.
Frequently asked questions
How do I convert Ka to pKa?
Take the negative base-10 logarithm: pKa = −log₁₀(Ka). For Ka = 1.75 × 10⁻⁵ (acetic acid), pKa = 4.757.
How do I convert pKa to Ka?
Ka = 10^(−pKa). A pKa of 9.25 gives Ka = 5.6 × 10⁻¹⁰.
How do I find pKa from pH?
Measure the pH of a solution with known [HA] and [A⁻] and use pKa = pH − log₁₀([A⁻]/[HA]). At pH 4.90 with 0.20 M acetate and 0.15 M acetic acid, pKa = 4.90 − 0.125 = 4.775.
Does a lower pKa mean a stronger acid?
Yes. Each unit lower means a Ka ten times larger. Formic acid (3.75) is about ten times stronger than acetic acid (4.76).
What is the relationship between pKa and pKb?
For a conjugate acid–base pair in water at 25 °C, pKa + pKb = 14.00. Ammonia has pKb 4.75, so the ammonium ion has pKa 9.25.
What does pKa = pH mean?
At that pH the acid is exactly half ionised: [A⁻] = [HA]. Buffers work best within about ±1 pH unit of the pKa.
Why does the "14" change at other temperatures?
pKw depends on temperature – it is 14.00 at 25 °C but about 13.26 at 50 °C – so the pKa + pKb sum used here is valid at 25 °C only.
Is pKa affected by concentration?
The thermodynamic pKa is a constant at a given temperature; the apparent value measured in concentrated buffers shifts slightly because of activity effects.
What formula does the pKa Calculator use?
pKa = −log₁₀(Ka) · Ka = 10^(−pKa) · pKa = pH − log₁₀([A⁻]/[HA]) · pKa + pKb = pKw = 14.00 (25 °C) · fraction ionised = 1 ÷ (1 + 10^(pKa − pH)).