pH Calculator

The pH formula is pH = -log10[H+], where [H+] is the hydrogen ion concentration in moles per liter. A solution with [H+] = 4.5 x 10^-4 mol/L has a pH of -log10(4.5 x 10^-4) = 3.35. The reverse is [H+] = 10^-pH, so pH 3.35 means 10^-3.35 = 4.5 x 10^-4 mol/L. Texts that write [H3O+] mean the same quantity: a free proton in water is attached to a water molecule, and the two notations count the same ions.

pOH is the same idea for hydroxide: pOH = -log10[OH-]. Water itself splits slightly into H+ and OH-, and the product of the two is fixed: [H+][OH-] = Kw = 1.0 x 10^-14 at 25 C. Taking logs gives the identity everybody memorizes, pH + pOH = 14. So [OH-] = 2.0 x 10^-3 mol/L is pOH 2.70 and pH 11.30. The 14 is not a law of nature; it is -log10 of Kw at 25 C, and Kw rises with temperature. Warm pure water has a pH below 7 and is still exactly neutral, because neutral means [H+] = [OH-], not pH 7. This calculator assumes 25 C throughout.

Strong acids and bases dissociate completely, so 0.010 mol/L HCl gives [H+] = 0.010 and pH 2.00. The common mistake is with very dilute solutions. The shortcut pH = -log10 C says 10^-8 mol/L HCl has pH 8: an acid making water basic, which is impossible. Water's own 10^-7 mol/L of H+ is no longer negligible at that dilution, and the exact treatment solves [H+] = C + Kw/[H+], a quadratic with the answer [H+] = (C + sqrt(C^2 + 4Kw)) / 2. For 10^-8 mol/L that is pH 6.98: slightly acidic, as it must be. The page does this automatically and shows how far the shortcut would have been off. A base with two hydroxides per formula unit, like Ba(OH)2, supplies twice its molarity in OH-.

Weak acids only partly ionize. For an acid HA at concentration C with dissociation constant Ka, the equilibrium is Ka = x^2 / (C - x), where x = [H+]. Rearranged, x^2 + Ka x - Ka C = 0, which the quadratic formula solves exactly. The classroom shortcut drops the x from C - x, giving x = sqrt(Ka C), and it is acceptable when x comes out under 5% of C (the 5% rule). For 0.10 mol/L of an acid with pKa 4.76, x = 1.3 x 10^-3, about 1.3% of C, so the shortcut holds and pH = 2.88. At 10^-4 mol/L the same acid is over a third ionized and the shortcut is wrong; the table shows where it breaks. pKa is -log10 Ka, so either can be entered.

Buffers are a weak acid mixed with its conjugate base, and the Henderson-Hasselbalch equation gives their pH: pH = pKa + log10([A-]/[HA]). Equal amounts give pH = pKa; ten times as much base gives pKa + 1. It assumes the amounts you mixed are the amounts present at equilibrium, which is close when both are much larger than [H+] and [OH-]. It fails for dilute buffers and for pKa values far from 7, so the page also solves the full equilibrium and tells you when the two disagree. A buffer works best within one pH unit of its pKa, a ratio between 0.1 and 10.

Two limits apply to every mode. Concentration stands in for activity, which is the textbook convention and close for dilute solutions but drifts in concentrated ones, so a pH meter will not agree with a calculated pH of 0 or 14. And the answer carries as many decimal places as the concentration has significant figures: 4.5 x 10^-4 is known to two figures, so its pH is 3.35, not 3.346787. The page computes chemistry; it is not guidance on what pH anything should be.