pH calculator, acids and bases
pH = −log₁₀[H₃O⁺], pH + pOH = pKw
pH solved exactly, never serving an approximation in silence. For a strong acid, water autoionisation is included: a 10⁻⁸ mol/L solution has a pH of 6.98, never 8. An acid cannot make a solution basic. For a weak acid, the tool solves the exact quadratic instead of the textbook √(Ka·C), reports the degree of dissociation α and shows the gap between the two when it matters. pKw is a field, not a hidden constant: neutrality is pKw/2, and that is 7 only at 25 °C.
2.883
- Calculation
- [H⁺] = (−Ka + √(Ka² + 4·Ka·C)) / 2, exact root of [H⁺]² + Ka[H⁺] − Ka·C = 0
- pH from the textbook approximation
- 2.88 (écart 0.002863)
ACIDIC solution: pH 2.883, below neutrality (7).
Degree of dissociation α = 1.31 %: below 5%, the textbook √(Ka·C) approximation is acceptable, and the tool confirms it matches the exact calculation here.
Neutrality sits at pH 7.00 here, but because pKw is 14.00, that is, at 25 °C. It always equals pKw/2: at 100 °C, pure water has a pH of 6.15 while remaining perfectly neutral.
Scientific dossier
What the tool computes, what it assumes, where it stops being valid, and where its data comes from.
Method & formulaspH = −log₁₀[H₃O⁺], pH + pOH = pKw
pH = −log₁₀[H₃O⁺], pH + pOH = pKw
strong acid: [H⁺] = (C + √(C² + 4·Kw)) / 2
weak acid: [H⁺] = (−Ka + √(Ka² + 4·Ka·C)) / 2
textbook approximations: [H⁺] ≈ C, then [H⁺] ≈ √(Ka·C)
Both textbook formulas are limiting cases, and the tool computes what they are the limit of. For a strong acid, [H⁺] = C ignores the protons water itself provides: negligible at 0.1 mol/L, dominant at 10⁻⁸. For a weak acid, √(Ka·C) assumes dissociation does not drain the undissociated acid reservoir, true while α stays small. False as soon as you dilute. Both exact forms are the positive roots of complete quadratic balances; the tool solves them, also computes the approximation, and shows the gap rather than choosing for you.
- Degree of dissociation α
- · the fraction of acid actually dissociated, [H⁺]/C. IT is what decides whether the textbook approximation holds: the usual criterion is α < 5%. It grows on dilution, Ostwald’s dilution law.
- Neutrality
- · the state where [H⁺] = [OH⁻], that is pH = pKw/2. It is 7.00 at 25 °C only: at 100 °C, pKw ≈ 12.3 and pure water has a pH of 6.15, without being acidic.
- Analytical concentration
- · the one you added by weighing the solute, distinct from the equilibrium concentration: for a weak acid the latter is far smaller than the former.
Validity domainThe model handles a single monoacid (or monobase) in aqueous solution, with no ionic strength: concentrations replace activities, which becomes inexact beyond ~0.
The model handles a single monoacid (or monobase) in aqueous solution, with no ionic strength: concentrations replace activities, which becomes inexact beyond ~0.1 mol/L. Polyprotic acids are not covered, for H₂SO₄ only the first. Complete, dissociation matches the “strong acid” mode. Neither mixtures, nor buffers, nor titrations are covered: those are different equilibria with different balances. The pKa values are from your table: none is imposed, which avoids attributing to the tool a constant you did not choose.
Common pitfall: the acid that would turn basicApplying pH = −log C to a 10⁻⁸ mol/L strong acid gives 8, a basic solution obtained by adding acid to water.
Applying pH = −log C to a 10⁻⁸ mol/L strong acid gives 8, a basic solution obtained by adding acid to water. The absurdity comes from forgetting that water already supplies 10⁻⁷ mol/L of protons: the full balance gives 6.98, very slightly acidic. Which is the only possible answer. The rule to remember: as soon as the concentration approaches 10⁻⁶ mol/L, the textbook formula stops being usable, and the tool flags this explicitly instead of serving the absurd result.