# What is the Ka?

Finds the acid ionization constant Ka of a weak acid from the pH and the starting concentration, from the equilibrium concentrations, or from pKa, with pKa and the percent ionization; or pKa from Ka.

- Page: https://www.acalculator.org/chemistry/ka-calculator
- JSON spec: https://www.acalculator.org/chemistry/ka-calculator.json
- Version: 04fe58789b0b

## Default answer

Example with the default inputs (Start from pH, pH 2.09, Starting concentration C₀ (mol/L) 0.125): Ka = 5.65315 × 10⁻⁴ (pKa = 3.24771).

## Inputs

| Key | Label | Description |
| --- | --- | --- |
| find | Start from | What you know: the pH and starting concentration, the equilibrium concentrations, pKa, or Ka. |
| ph | pH | The pH of the acid solution, from 0 to below 7. |
| c | Starting concentration C₀ (mol/L) | The acid concentration as made up, before it ionizes, in moles per liter. |
| h | [H₃O⁺] at equilibrium (mol/L) | The hydronium ion concentration at equilibrium. |
| a | [A⁻] at equilibrium (mol/L) | The conjugate base concentration at equilibrium. |
| ha | [HA] at equilibrium (mol/L) | The concentration of acid left un-ionized at equilibrium. |
| pka | pKa | The pKa of the acid, −log Ka. |
| ka | Ka | The acid ionization constant, such as 1.8e-5 for acetic acid. |

## Outputs

| Key | Label | Description |
| --- | --- | --- |
| result | Answer | Ka with pKa, or pKa with Ka, to 6 significant figures. |
| kaValue | Ka | The acid ionization constant. |
| pkaValue | pKa | −log₁₀ Ka. |
| percent | Percent ionization (%) | The share of the acid ionized: [H₃O⁺] ÷ C₀ × 100 from a pH, or [A⁻] ÷ ([HA] + [A⁻]) × 100 from concentrations. |
| h3o | [H₃O⁺] (mol/L) | The hydronium ion concentration at equilibrium. |
| haLeft | [HA] at equilibrium (mol/L) | The acid left un-ionized: C₀ − [H₃O⁺], or as typed. |
| start | Starting concentration (mol/L) | C₀ as typed, or [HA] + [A⁻] from the equilibrium concentrations. |
| phValue | pH | −log₁₀[H₃O⁺]. |

## Method

Ka = [H₃O⁺][A⁻] ÷ [HA]; from a pH and starting concentration C₀: [H₃O⁺] = [A⁻] = 10^−pH and [HA] = C₀ − 10^−pH; percent ionization = [H₃O⁺] ÷ C₀ × 100 (from equilibrium concentrations, [A⁻] ÷ ([HA] + [A⁻]) × 100); pKa = −log₁₀ Ka and Ka = 10^−pKa.

## Assumptions

- A weak monoprotic acid HA is the only source of H₃O⁺ and A⁻; water’s own 1 × 10⁻⁷ M of H₃O⁺ is neglected, so the pH must be below 7.
- Concentrations stand in for activities, as in introductory chemistry.

## Worked examples

1. find = ph, ph = 2.09, c = 0.125 gives percent = 6.50264, h3o = 0.00812831, kaValue = 0.000565315. Source: OpenStax, Chemistry 2e, §14.3 Relative Strengths of Acids and Bases (Ka = [H₃O⁺][A⁻] ÷ [HA]; percent ionization = [H₃O⁺]eq ÷ [HA]₀ × 100), https://openstax.org/books/chemistry-2e/pages/14-3-relative-strengths-of-acids-and-bases, Example 14.7 Calculation of Percent Ionization from pH: [H₃O⁺] = 0.0081 M, 6.5% ionized.
2. find = ph, ph = 2.89, c = 0.1 gives percent = 1.28825, kaValue = 0.0000168125. Source: OpenStax, Chemistry 2e, §14.3 Relative Strengths of Acids and Bases (Ka = [H₃O⁺][A⁻] ÷ [HA]; percent ionization = [H₃O⁺]eq ÷ [HA]₀ × 100), https://openstax.org/books/chemistry-2e/pages/14-3-relative-strengths-of-acids-and-bases, Example 14.7 Check Your Learning: 1.3% ionized.
3. find = eq, h = 0.00118, a = 0.00118, ha = 0.0787 gives kaValue = 0.0000176925, result = Ka = 1.76925 × 10⁻⁵ (pKa = 4.75221). Source: OpenStax, Chemistry 2e, §14.3 Relative Strengths of Acids and Bases (Ka = [H₃O⁺][A⁻] ÷ [HA]; percent ionization = [H₃O⁺]eq ÷ [HA]₀ × 100), https://openstax.org/books/chemistry-2e/pages/14-3-relative-strengths-of-acids-and-bases, Example 14.9 Determination of Ka from Equilibrium Concentrations: Ka = 1.77 × 10⁻⁵.
4. find = eq, h = 0.027, a = 0.13, ha = 0.29 gives kaValue = 0.0121034. Source: OpenStax, Chemistry 2e, §14.3 Relative Strengths of Acids and Bases (Ka = [H₃O⁺][A⁻] ÷ [HA]; percent ionization = [H₃O⁺]eq ÷ [HA]₀ × 100), https://openstax.org/books/chemistry-2e/pages/14-3-relative-strengths-of-acids-and-bases, Example 14.9 Check Your Learning: Ka = 1.2 × 10⁻².
5. find = ph, ph = 2.34, c = 0.0516 gives kaValue = 0.000444256, percent = 8.8583. Source: OpenStax, Chemistry 2e, §14.3 Relative Strengths of Acids and Bases (Ka = [H₃O⁺][A⁻] ÷ [HA]; percent ionization = [H₃O⁺]eq ÷ [HA]₀ × 100), https://openstax.org/books/chemistry-2e/pages/14-3-relative-strengths-of-acids-and-bases, Example 14.11 Determination of Ka or Kb from pH.
6. find = ka, ka = 0.000018 gives pkaValue = 4.74473, result = pKa = 4.74473 (Ka = 1.8 × 10⁻⁵). Source: OpenStax, Chemistry 2e, §14.6 Buffers (pKa = −log Ka), https://openstax.org/books/chemistry-2e/pages/14-6-buffers: pKa = −log Ka.
7. find = pka, pka = 5 gives kaValue = 0.00001, result = Ka = 1 × 10⁻⁵ (pKa = 5). Source: OpenStax, Chemistry 2e, §14.6 Buffers (pKa = −log Ka), https://openstax.org/books/chemistry-2e/pages/14-6-buffers: Ka = 10^−pKa.

## FAQ

### How do I calculate Ka from pH?

Turn the pH into [H₃O⁺] = 10^−pH. For a weak acid HA alone in water, [A⁻] equals [H₃O⁺] and the acid left is the starting concentration minus [H₃O⁺]. Then Ka = [H₃O⁺]² ÷ (C₀ − [H₃O⁺]). For 0.125 M nitrous acid at pH 2.09: [H₃O⁺] = 0.00812831 M and Ka = 0.00812831² ÷ 0.116872 = 5.65 × 10⁻⁴.

### How do I calculate pKa?

pKa = −log Ka. Acetic acid has Ka = 1.8 × 10⁻⁵, so pKa = −log(1.8 × 10⁻⁵) = 4.74. Going back, Ka = 10^−pKa. A smaller pKa means a stronger acid.

### How do I calculate percent ionization?

Divide the equilibrium [H₃O⁺] by the starting acid concentration and multiply by 100. OpenStax’s 0.125 M nitrous acid at pH 2.09 has [H₃O⁺] = 0.0081 M, so it is 0.0081 ÷ 0.125 × 100 = 6.5% ionized.

### How do I find Ka from equilibrium concentrations?

Put them into Ka = [H₃O⁺][A⁻] ÷ [HA]. Vinegar’s acetic acid at equilibrium has [CH₃CO₂H] = 0.0787 M and [H₃O⁺] = [CH₃CO₂⁻] = 0.00118 M: Ka = 0.00118² ÷ 0.0787 = 1.77 × 10⁻⁵.

### Why must the pH be below 7?

An acid solution has more H₃O⁺ than pure water, whose own 1 × 10⁻⁷ M gives pH 7. The formula leaves water’s share out, which is fair only when the acid supplies far more H₃O⁺ than water does.

### Why does my answer differ from the textbook’s?

Textbooks round [H₃O⁺] before dividing. OpenStax’s Example 14.11 (0.0516 M nitrous acid, pH 2.34) rounds [H₃O⁺] to 0.0046 M and prints Ka = 4.6 × 10⁻⁴; (0.0046)² ÷ 0.0470 is 4.5 × 10⁻⁴, and with the unrounded 10^−2.34 = 0.00457088 M the page gives 4.44256 × 10⁻⁴.

## Sources

- OpenStax, Chemistry 2e, §14.3 Relative Strengths of Acids and Bases: Ka = [H₃O⁺][A⁻] ÷ [HA], percent ionization = [H₃O⁺]eq ÷ [HA]₀ × 100, Example 14.7 (0.125 M HNO₂ at pH 2.09: 6.5%) and its Check Your Learning (0.10 M acetic acid at pH 2.89: 1.3%), Example 14.9 (acetic acid: 1.77 × 10⁻⁵) and its Check Your Learning (HSO₄⁻: 1.2 × 10⁻²), Example 14.11 (0.0516 M HNO₂ at pH 2.34). https://openstax.org/books/chemistry-2e/pages/14-3-relative-strengths-of-acids-and-bases (retrieved 2026-10-19)
- OpenStax, Chemistry 2e, §14.6 Buffers: pKa = −log Ka; Example 14.20 (−log(1.8 × 10⁻⁵) = 4.74). https://openstax.org/books/chemistry-2e/pages/14-6-buffers (retrieved 2026-10-19)