{
  "id": "ka",
  "version": "04fe58789b0b",
  "status": "published",
  "name": "Ka Calculator",
  "question": "What is the Ka?",
  "summary": "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.",
  "category": "chemistry",
  "subcategory": "solutions",
  "url": "https://www.acalculator.org/chemistry/ka-calculator",
  "markdown": "https://www.acalculator.org/chemistry/ka-calculator.md",
  "kind": "function",
  "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."
  ],
  "inputs": {
    "$schema": "https://json-schema.org/draft/2020-12/schema",
    "type": "object",
    "properties": {
      "find": {
        "title": "Start from",
        "description": "What you know: the pH and starting concentration, the equilibrium concentrations, pKa, or Ka.",
        "type": "string",
        "enum": [
          "ph",
          "eq",
          "pka",
          "ka"
        ]
      },
      "ph": {
        "title": "pH",
        "description": "The pH of the acid solution, from 0 to below 7.",
        "type": "number",
        "minimum": 0,
        "exclusiveMaximum": 7
      },
      "c": {
        "title": "Starting concentration C₀ (mol/L)",
        "description": "The acid concentration as made up, before it ionizes, in moles per liter.",
        "type": "number",
        "minimum": 1e-10,
        "maximum": 100
      },
      "h": {
        "title": "[H₃O⁺] at equilibrium (mol/L)",
        "description": "The hydronium ion concentration at equilibrium.",
        "type": "number",
        "minimum": 1e-14,
        "maximum": 100
      },
      "a": {
        "title": "[A⁻] at equilibrium (mol/L)",
        "description": "The conjugate base concentration at equilibrium.",
        "type": "number",
        "minimum": 1e-14,
        "maximum": 100
      },
      "ha": {
        "title": "[HA] at equilibrium (mol/L)",
        "description": "The concentration of acid left un-ionized at equilibrium.",
        "type": "number",
        "minimum": 1e-14,
        "maximum": 100
      },
      "pka": {
        "title": "pKa",
        "description": "The pKa of the acid, −log Ka.",
        "type": "number",
        "minimum": -10,
        "maximum": 50
      },
      "ka": {
        "title": "Ka",
        "description": "The acid ionization constant, such as 1.8e-5 for acetic acid.",
        "type": "number",
        "minimum": 1e-50,
        "maximum": 10000000000
      }
    }
  },
  "outputs": {
    "result": {
      "label": "Answer",
      "description": "Ka with pKa, or pKa with Ka, to 6 significant figures.",
      "format": "text"
    },
    "kaValue": {
      "label": "Ka",
      "description": "The acid ionization constant.",
      "format": "number"
    },
    "pkaValue": {
      "label": "pKa",
      "description": "−log₁₀ Ka.",
      "format": "number"
    },
    "percent": {
      "label": "Percent ionization (%)",
      "description": "The share of the acid ionized: [H₃O⁺] ÷ C₀ × 100 from a pH, or [A⁻] ÷ ([HA] + [A⁻]) × 100 from concentrations.",
      "format": "number"
    },
    "h3o": {
      "label": "[H₃O⁺] (mol/L)",
      "description": "The hydronium ion concentration at equilibrium.",
      "format": "number"
    },
    "haLeft": {
      "label": "[HA] at equilibrium (mol/L)",
      "description": "The acid left un-ionized: C₀ − [H₃O⁺], or as typed.",
      "format": "number"
    },
    "start": {
      "label": "Starting concentration (mol/L)",
      "description": "C₀ as typed, or [HA] + [A⁻] from the equilibrium concentrations.",
      "format": "number"
    },
    "phValue": {
      "label": "pH",
      "description": "−log₁₀[H₃O⁺].",
      "format": "number"
    }
  },
  "defaultAnswer": {
    "inputs": {
      "find": "ph",
      "ph": 2.09,
      "c": 0.125,
      "h": 0.00118,
      "a": 0.00118,
      "ha": 0.0787,
      "pka": 4.74,
      "ka": 0.000018
    },
    "outputs": {
      "result": "Ka = 5.65315 × 10⁻⁴ (pKa = 3.24771)",
      "kaValue": 0.0005653151936586336,
      "pkaValue": 3.2477093420943985,
      "percent": 6.502644129312796,
      "h3o": 0.008128305161640995,
      "haLeft": 0.116871694838359,
      "start": 0.125,
      "phValue": 2.09
    },
    "text": "Ka = 5.65315 × 10⁻⁴ (pKa = 3.24771)."
  },
  "examples": [
    {
      "given": {
        "find": "ph",
        "ph": 2.09,
        "c": 0.125
      },
      "expect": {
        "percent": 6.502644129312794,
        "h3o": 0.008128305161640993,
        "kaValue": 0.0005653151936586331
      },
      "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; Python 3",
      "tolerance": 1e-12
    },
    {
      "given": {
        "find": "ph",
        "ph": 2.89,
        "c": 0.1
      },
      "expect": {
        "percent": 1.288249551693134,
        "kaValue": 0.000016812455456421565
      },
      "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; Python 3",
      "tolerance": 1e-12
    },
    {
      "given": {
        "find": "eq",
        "h": 0.00118,
        "a": 0.00118,
        "ha": 0.0787
      },
      "expect": {
        "kaValue": 0.000017692503176620076,
        "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; Python 3"
    },
    {
      "given": {
        "find": "eq",
        "h": 0.027,
        "a": 0.13,
        "ha": 0.29
      },
      "expect": {
        "kaValue": 0.012103448275862069
      },
      "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; Python 3"
    },
    {
      "given": {
        "find": "ph",
        "ph": 2.34,
        "c": 0.0516
      },
      "expect": {
        "kaValue": 0.0004442558600057919,
        "percent": 8.858298248350291
      },
      "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 (OpenStax rounds [H₃O⁺] to 0.0046 M and prints 4.6 × 10⁻⁴; see content.mdx); Python 3",
      "tolerance": 1e-12
    },
    {
      "given": {
        "find": "ka",
        "ka": 0.000018
      },
      "expect": {
        "pkaValue": 4.7447274948966935,
        "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; Example 14.20 gives −log(1.8 × 10⁻⁵) = 4.74; Python 3"
    },
    {
      "given": {
        "find": "pka",
        "pka": 5
      },
      "expect": {
        "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; hand calculation in content.mdx"
    }
  ],
  "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)"
  ],
  "related": [
    "ph",
    "ksp",
    "equilibrium-constant",
    "titration"
  ],
  "changelog": []
}