acalculator

What is the Ka?

Enter the pH of a weak acid solution and the acid’s starting concentration, or its equilibrium concentrations. The calculator gives the acid ionization constant Ka, pKa and the percent ionization, and it converts between Ka and pKa.

Your numbers

Start from
Answer
Ka = 5.65315 × 10⁻⁴ (pKa = 3.24771)

Ka = 5.65315 × 10⁻⁴ (pKa = 3.24771).

Ka
0.000565315
pKa
3.24771
Percent ionization (%)
6.50264
[H₃O⁺] (mol/L)
0.00812831
[HA] at equilibrium (mol/L)
0.116872
Starting concentration (mol/L)
0.125
pH
2.09

Answer: Ka = 5.65315 × 10⁻⁴ (pKa = 3.24771). Ka = 5.65315 × 10⁻⁴ (pKa = 3.24771).

How to calculate

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.

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).

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.

  • 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.

Machine-readable copies: Markdown, JSON.

Worked examples

Each example is checked against the calculator on every build.

  1. Start from pH, pH 2.09, Starting concentration C₀ (mol/L) 0.125 gives Percent ionization (%) 6.50264, [H₃O⁺] (mol/L) 0.00812831, Ka 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. Start from pH, pH 2.89, Starting concentration C₀ (mol/L) 0.1 gives Percent ionization (%) 1.28825, Ka 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. Start from Concentrations, [H₃O⁺] at equilibrium (mol/L) 0.00118, [A⁻] at equilibrium (mol/L) 0.00118, [HA] at equilibrium (mol/L) 0.0787 gives Ka 0.0000176925, Answer 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. Start from Concentrations, [H₃O⁺] at equilibrium (mol/L) 0.027, [A⁻] at equilibrium (mol/L) 0.13, [HA] at equilibrium (mol/L) 0.29 gives Ka 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. Start from pH, pH 2.34, Starting concentration C₀ (mol/L) 0.0516 gives Ka 0.000444256, Percent ionization (%) 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. Start from Ka, Ka 0.000018 gives pKa 4.74473, Answer 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. Start from pKa, pKa 5 gives Ka 0.00001, Answer 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

How it works

Acid ionization constant (OpenStax §14.3). For a weak acid HA + H₂O ⇌ H₃O⁺ + A⁻,

Ka = [H₃O⁺][A⁻] ÷ [HA]

with the concentrations at equilibrium, in mol/L. Pick what you know:

  • pH and starting concentration C₀: [H₃O⁺] = [A⁻] = 10^−pH and [HA] = C₀ − 10^−pH, so Ka = (10^−pH)² ÷ (C₀ − 10^−pH). Percent ionization = 10^−pH ÷ C₀ × 100.
  • Equilibrium concentrations: type [H₃O⁺], [A⁻] and [HA]. Ka = [H₃O⁺][A⁻] ÷ [HA]. The starting concentration is taken as [HA] + [A⁻] (all the A⁻ came from the acid), so percent ionization = [A⁻] ÷ ([HA] + [A⁻]) × 100, and pH = −log₁₀[H₃O⁺].
  • pKa: Ka = 10^−pKa (OpenStax §14.6).
  • Ka: pKa = −log₁₀ Ka.

pKa = −log₁₀ Ka in every case.

Exact values. Typed numbers are exact decimals. 10^−pH and 10^−pKa are exact when the pH or pKa is a whole number (pH 3 gives exactly 0.001 M); otherwise they are the nearest double-precision number, and the rest is worked exactly from it. pKa and the pH from [H₃O⁺] are logarithms, worked in double precision.

Rules the page checks.

  • pH from 0 to below 7; C₀ from 10⁻¹⁰ to 100 mol/L.
  • 10^−pH must be below C₀ (exactly when the pH is a whole number); if it is not, the page gives no answer, because a weak acid of that concentration cannot give that pH.
  • Equilibrium concentrations from 10⁻¹⁴ to 100 mol/L; pKa from −10 to 50; Ka from 10⁻⁵⁰ to 10¹⁰.

How results are written. The sentence is “Ka = 5.65315 × 10⁻⁴ (pKa = 3.24771)”, or in the Ka mode “pKa = 4.74473 (Ka = 1.8 × 10⁻⁵)”. Ka has 6 significant figures with trailing zeros dropped, written m × 10ⁿ when it is below 0.001 or 1,000,000 and up, and plainly between (Ka = 0.0121034). The form is judged on the value before rounding, so 999,999.9 is written 1,000,000, and 9.999995 × 10⁻⁵ is written 1 × 10⁻⁴. pKa has 6 significant figures, trailing zeros dropped (pKa = 5), with the minus sign − for a negative pKa, and m × 10ⁿ below 0.000001 (pKa = −4.34294 × 10⁻⁸ for Ka = 1.0000001). Rows show Ka, pKa, the percent ionization (%), [H₃O⁺], [HA] at equilibrium, the starting concentration and the pH (the pH mode shows the typed pH); a row writes a number below 0.000001 or from 10¹⁵ as m × 10ⁿ (a percent ionization of 1 × 10⁻¹⁴).

Assumptions. One weak monoprotic acid is the only source of H₃O⁺ and A⁻ (water’s own H₃O⁺ is neglected), and concentrations stand in for activities.

Worked examples by hand

0.125 M nitrous acid at pH 2.09 (OpenStax Example 14.7). [H₃O⁺] = 10^−2.09 = 0.00812831 M. Percent ionization = 0.00812831 ÷ 0.125 × 100 = 6.50264% (OpenStax: 6.5%). [HNO₂] = 0.125 − 0.00812831 = 0.116872 M, so Ka = 0.00812831² ÷ 0.116872 = 5.65315 × 10⁻⁴ and pKa = 3.24771.

0.10 M acetic acid at pH 2.89 (Check Your Learning). [H₃O⁺] = 10^−2.89 = 0.00128825 M: 1.28825% ionized (OpenStax: 1.3%); Ka = 0.00128825² ÷ 0.0987118 = 1.68125 × 10⁻⁵.

Acetic acid at equilibrium (Example 14.9). Ka = 0.00118 × 0.00118 ÷ 0.0787 = 1.76925 × 10⁻⁵ (OpenStax: 1.77 × 10⁻⁵); pKa = 4.75221.

HSO₄⁻ (Check Your Learning). Ka = 0.027 × 0.13 ÷ 0.29 = 0.0121034 (OpenStax: 1.2 × 10⁻²).

0.0516 M nitrous acid at pH 2.34 (Example 14.11). [H₃O⁺] = 10^−2.34 = 0.00457088 M and [HNO₂] = 0.0516 − 0.00457088 = 0.0470291 M, so Ka = 0.00457088² ÷ 0.0470291 = 4.44256 × 10⁻⁴. OpenStax rounds [H₃O⁺] to 0.0046 M and prints 4.6 × 10⁻⁴; its own numbers, (0.0046)² ÷ 0.0470, give 4.5 × 10⁻⁴.

pKa from Ka. −log₁₀(1.8 × 10⁻⁵) = 4.74473 (OpenStax: 4.74). Ka from pKa 5: 10⁻⁵ = 1 × 10⁻⁵.

Other questions people ask

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.