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pH calculator with step-by-step solution

Pick what you have: a strong or weak acid or base with its concentration – or a pH, pOH or concentration to convert. You get pH, pOH, [H₃O⁺], [OH⁻] and every step of the working.

mol/L
Table values at 25 °C (CRC Handbook)
custom value only – e.g. 4.76 or Ka 1.75e-5
Your inputs are saved in this browser only.

Result

pH
2.88
The solution is
acidic (pH < 7)
Hydronium ions [H₃O⁺]
0.00131 mol/L
pOH
11.12
Hydroxide ions [OH⁻]
7.636 · 10⁻¹² mol/L
For comparison: approximation
pH ≈ 2.88
Degree of dissociation α
1.31 %
Step-by-step solution
  • Ion product of water at 25 °C: Kw = [H₃O⁺] · [OH⁻] = 1.0 × 10⁻¹⁴, so pH + pOH = 14.
  • Acid constant: pKa = 4.760 → Ka = 10^(−pKa) = 1.738 · 10⁻⁵
  • Equilibrium expression with x = [H₃O⁺] = [A⁻] and [HA] = c − x: Ka = x² / (c − x) → x² + Ka · x − Ka · c = 0 → x = (−Ka + √(Ka² + 4 · Ka · c)) / 2 = (−1.738 · 10⁻⁵ + √(1.738 · 10⁻⁵² + 4 · 1.738 · 10⁻⁵ · 0.1)) / 2 = 0.00131 mol/L
  • pH = −log [H₃O⁺] = −log 0.00131 = 2.883
  • pOH = 14 − pH = 14 − 2.883 = 11.117
  • [OH⁻] = Kw / [H₃O⁺] = 10⁻¹⁴ / 0.00131 = 7.636 · 10⁻¹² mol/L
  • For comparison, the approximation for x ≪ c: pH ≈ ½ · (pKa − log c) = ½ · (4.760 − log 0.1) = 2.880

How it is calculated

How to calculate pH

pH is the negative base-10 logarithm of the hydronium ion concentration: pH = −log [H₃O⁺] (strictly, of its activity – in dilute solutions the two are practically equal). Likewise pOH = −log [OH⁻]. Water itself ionizes to a tiny extent (autoionization); at 25 °C the ion product Kw = [H₃O⁺] · [OH⁻] = 1.0 × 10⁻¹⁴, which gives pH + pOH = 14. Kw changes with temperature – every calculation here assumes 25 °C.

How to calculate pH

  1. Under “What do you want to calculate?” choose strong or weak acid, strong or weak base, or convert.
  2. Enter the concentration in mol/L; for weak acids and bases pick a substance from the list or enter your own pKa/Ka or pKb/Kb.
  3. Read pH, pOH, [H₃O⁺] and [OH⁻] with the full working.

Strong acids and bases

Strong acids such as HCl, HNO₃ or HClO₄ give up their proton completely, so [H₃O⁺] = c. Example: 0.01 M hydrochloric acid has pH = −log 0.01 = 2. Strong bases such as NaOH give [OH⁻] = c: 0.001 M sodium hydroxide has pOH = 3 and therefore pH = 14 − 3 = 11. Ca(OH)₂ releases two OH⁻ per formula unit.

Very dilute solutions: below about 10⁻⁶ mol/L, water supplies a significant share of H₃O⁺. The exact result is [H₃O⁺] = (c + √(c² + 4 · Kw)) / 2, which is why 10⁻⁸ M HCl has pH 6.98 – not 8.

Sulfuric acid (H₂SO₄): the “2 – simplified” option treats both protons as fully released. That is only approximate, because the second step (HSO₄⁻, pKa ≈ 2) dissociates only partly. The calculator flags this.

Weak acids: exact solution and approximation

A weak acid HA dissociates only partly. With x = [H₃O⁺] = [A⁻], the equilibrium expression Ka = x² / (c − x) leads to the quadratic x² + Ka·x − Ka·c = 0, whose solution is

x = (−Ka + √(Ka² + 4 · Ka · c)) / 2

If x is much smaller than c, this simplifies to the familiar approximation pH = ½ · (pKa − log c). Example: acetic acid, c = 0.1 M, pKa = 4.76: Ka = 10^(−4.76) = 1.738 × 10⁻⁵; x = (−1.738 × 10⁻⁵ + √(3.02 × 10⁻¹⁰ + 6.951 × 10⁻⁶)) / 2 = 1.310 × 10⁻³ M → pH = 2.88. The approximation gives ½ · (4.76 + 1) = 2.88 – it works here because only about 1 % of the acid dissociates. Above roughly 5 % (fairly strong weak acids, very low c) it drifts off, and the calculator shows a warning.

Weak bases

A weak base B works the same way with Kb and [OH⁻]: x = (−Kb + √(Kb² + 4 · Kb · c)) / 2, then pOH = −log x and pH = 14 − pOH. Example: 0.1 M ammonia, pKb = 4.75: exactly [OH⁻] = 1.325 × 10⁻³ M, pOH = 2.878, pH = 11.12. The approximation pOH = ½ · (4.75 + 1) = 2.875 gives pH 11.125, often quoted as 11.13. For a conjugate pair, pKa + pKb = 14 (NH₄⁺: pKa = 9.25).

pKa and pKb values (25 °C)

AcidpKaBasepKb
Phosphoric acid H₃PO₄ (1st step)2.16Dimethylamine3.27
Hydrofluoric acid HF3.20Methylamine3.34
Nitrous acid HNO₂3.25Ethylamine3.35
Formic acid HCOOH3.75Trimethylamine4.20
Benzoic acid4.204Ammonia NH₃4.75
Acetic acid CH₃COOH4.756Hydrazine N₂H₄5.9
Carbonic acid H₂CO₃ (1st step)6.35Pyridine8.77
Hypochlorous acid HOCl7.40Aniline9.4
Hydrocyanic acid HCN9.21

Source: CRC Handbook of Chemistry and Physics, 84th edition (2004), as reproduced in Chemistry LibreTexts tables E1/E2. Other references list slightly different values depending on the measurement (e.g. acetic acid 4.75 or 4.76).

More chemistry helpers: the molar mass calculator for concentrations, the scientific notation converter for values like 1.8 × 10⁻⁵, and element data in the interactive periodic table.

Frequently asked questions

How do you calculate the pH of a strong acid?

For a strong monoprotic acid, [H₃O⁺] equals the concentration, so pH = −log c. 0.01 M HCl has pH = −log 0.01 = 2; 0.001 M HCl has pH 3.

How do I find the pH of a weak acid?

Exactly, with the quadratic formula x = (−Ka + √(Ka² + 4·Ka·c)) / 2 and pH = −log x. As long as less than about 5 % dissociates, the shortcut pH = ½ · (pKa − log c) is fine. 0.1 M acetic acid: pH ≈ 2.88.

How do I convert pH to pOH and concentration?

At 25 °C, pH + pOH = 14, [H₃O⁺] = 10^(−pH) and [OH⁻] = 10^(−pOH). So pH 4 means [H₃O⁺] = 10⁻⁴ M, pOH = 10 and [OH⁻] = 10⁻¹⁰ M.

What is the difference between pKa and Ka?

Ka is the acid dissociation constant, pKa = −log Ka. The lower the pKa, the stronger the acid. Ka = 1.75 × 10⁻⁵ corresponds to pKa = 4.76. The calculator accepts either.

Why doesn’t very dilute HCl have a pH of 8?

Because water itself provides 10⁻⁷ M H₃O⁺. In 10⁻⁸ M HCl most of the hydronium comes from water, and the exact result is pH 6.98. Adding an acid can never make a solution basic.

Is pH + pOH always 14?

Only at 25 °C, because Kw depends on temperature. At higher temperatures Kw is larger and neutral pH falls below 7. This calculator uses Kw = 1.0 × 10⁻¹⁴ (25 °C).

Sources and legal basis

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