Why a weak acid needs an equilibrium calculation
A strong acid hands over every proton it has, so its pH follows from the concentration alone. A weak acid does not. Acetic acid at 0.10 mol/L leaves about 98.7% of its molecules intact and releases only the remaining 1.3% as hydrogen ion, so you cannot get the pH without knowing how weak it is. That number is the acid dissociation constant:
Ka is the equilibrium constant for HA ⇌ H+ + A−. Big Ka means the equilibrium sits to the right and the acid is stronger. Because the values span many orders of magnitude — from about 10−2 for the second proton of sulfuric acid down to 10−10 for phenol — chemists usually quote pKa = −log10Ka instead. A lower pKa is a stronger acid, and each unit of pKa is a factor of ten in Ka.
Two things follow that surprise people. First, the pH of a weak acid is not proportional to its concentration: dilute a weak acid ten-fold and the pH rises by about half a unit, not a full unit. Second, dilution makes a weak acid more ionized as a fraction even as it becomes less acidic overall. Both fall directly out of the algebra below.
Setting up and solving the equilibrium
Start with an ICE table. Before anything reacts you have C mol/L of HA and essentially no H+ or A−. Let x be the amount that dissociates. At equilibrium you have C − x of HA and x each of H+ and A−. Substituting into the Ka expression gives
Rearranged, that is the quadratic x2 + Kax − KaC = 0, whose positive root is the formula at the top of this page. Textbooks often skip the quadratic by assuming x ≪ C, which collapses the denominator to C and gives x ≈ √(KaC). That shortcut is convenient but it is always an overestimate of [H+], because the true denominator C − x is smaller than C. The conventional tolerance is the 5% rule: use it only when the resulting ionization is under 5%, which for a given acid means only above a certain concentration.
There is a second approximation buried in the quadratic itself, and it is the one this calculator removes. The quadratic assumes every hydrogen ion came from the acid. In very dilute solution the water contributes comparably, so the correct statement is the charge balance [H+] = [A−] + [OH−]. Writing [A−] = KaC/(Ka + [H+]) and [OH−] = Kw/[H+] gives one equation in one unknown, solved here numerically. Adding the water term can only push [H+] up, never down, so the exact pH is always at or below the quadratic value.
Worked example: 0.10 M acetic acid, Ka = 1.8e-5
Acetic acid at 25 °C has Ka = 1.8×10−5 (pKa 4.74 to 4.76 depending on the table). Work the 0.10 M case by hand.
- Write the quadratic. x2 = Ka(C − x), so x2 + 1.8×10−5x − 1.8×10−6 = 0.
- Discriminant. Ka2 = 3.24×10−10; 4KaC = 4 × 1.8×10−5 × 0.10 = 7.2×10−6. Sum = 7.200324×10−6.
- Square root. √(7.200324×10−6) = 2.68334×10−3.
- Positive root. x = (−1.8×10−5 + 2.68334×10−3) ÷ 2 = 1.3327×10−3 M.
- pH. −log10(1.3327×10−3) = 2.875.
- Percent ionization. 1.3327×10−3 ÷ 0.10 × 100 = 1.33%. Comfortably under 5%.
- Leftover acid. [HA] = 0.10 − 0.0013327 = 0.09867 M.
Compare the shortcut: √(1.8×10−5 × 0.10) = 1.3416×10−3, giving pH 2.872. It is high by 0.7% in [H+] and low by 0.003 in pH — invisible in practice, which is exactly why the shortcut is taught. Now repeat with 0.10 M hydrofluoric acid, Ka = 6.8×10−4: the exact root is 7.913×10−3 (pH 2.102, 7.9% ionized) while the shortcut gives 8.246×10−3 (pH 2.084). Now the shortcut is off by 4% in concentration, and the 5% rule correctly told you in advance not to use it.
How to read the pH and the percent ionization together
Read the two numbers as a pair. The pH tells you how acidic the solution is right now; the percent ionization tells you how much capacity is held in reserve. A 0.10 M acetic acid solution at pH 2.875 has 98.7% of its acid still undissociated, which is why it can neutralise far more base than a pH 2.87 solution of HCl could.
Use the percent ionization as your validity check. Under 5%, the simple √(KaC) treatment is fine and any textbook answer will agree with this page to two decimals. Between 5% and about 30%, you need the quadratic and hand-waved answers start to visibly disagree. Above roughly 50%, the species is barely weak at all and the tabulated Ka itself becomes the weak link, since values for such acids differ between sources.
A useful cross-check: when C is much larger than Ka, pH ≈ ½(pKa − log10C). For 0.10 M acetic acid that is ½(4.74 + 1) = 2.87, within 0.005 of the exact answer. That relation also explains the half-unit rule: because of the ½, a ten-fold dilution moves the pH by 0.5, not 1.0. Confirm it in the dilution table below.
If your solution contains both the acid and its salt, this page is the wrong tool — the conjugate base suppresses ionization and the pH is set by the ratio of the two. Use the Henderson-Hasselbalch buffer calculator. For fully dissociated acids, use the pH and pOH calculator instead.
Ka and pKa for common weak acids at 25 degC
| Acid | Ka | pKa | pH of 0.10 M | Ionized at 0.10 M |
|---|---|---|---|---|
| Sulfurous (1st) | 1.5×10−2 | 1.82 | 1.50 | 31.9% |
| Chloroacetic | 1.4×10−3 | 2.85 | 1.95 | 11.2% |
| Hydrofluoric | 6.8×10−4 | 3.17 | 2.10 | 7.91% |
| Formic | 1.8×10−4 | 3.75 | 2.38 | 4.15% |
| Benzoic | 6.3×10−5 | 4.20 | 2.61 | 2.48% |
| Acetic | 1.8×10−5 | 4.74 | 2.88 | 1.33% |
| Carbonic (1st) | 4.5×10−7 | 6.35 | 3.67 | 0.21% |
| Hypochlorous | 3.0×10−8 | 7.52 | 4.26 | 0.055% |
| Phenol | 1.0×10−10 | 10.00 | 5.50 | 0.0032% |
Ka values are representative tabulated constants at 25 °C; sources differ in the second significant figure. Enter any of them above to reproduce the pH column.
Assumptions and where they break
- One ionizable proton. For a polyprotic acid this page treats only the first step. That is usually adequate, because Ka2 is typically 104 to 105 times smaller, but it is a genuine approximation.
- Concentration, not activity. The equilibrium constant is strictly written in activities. In dilute solution the difference is small; in high ionic strength it is not, which is why measured pKa values shift with added salt.
- 25 °C. Ka is temperature-dependent through the van 't Hoff relation. Carboxylic acids change little over a few tens of degrees; amine conjugate acids change substantially.
- No other acid or base present. Adding the conjugate base, a second acid, or a buffer invalidates the single-equilibrium treatment entirely.
- The tabulated Ka is right. Published values for the same acid can differ by 10 to 20% in Ka, which is 0.04 to 0.08 in pKa. Do not report a pH to three decimals on the strength of a two-figure constant.
- Water is included here but not in most textbooks. If your homework answer differs from this page below about 10−6 M, this is why.
Related calculations and when to use them instead
The weak-acid equilibrium is the foundation for three other calculations you are likely to need.
Buffers. Add the conjugate base and the pH stops depending on concentration and starts depending on the ratio of base to acid. That is the Henderson-Hasselbalch relation, which is just this same Ka expression rearranged.
Titration. Adding strong base to a weak acid walks you through three regimes: the pure weak acid computed here, a buffer region, and finally the hydrolysis of the pure conjugate base at equivalence — where the pH is above 7, not equal to it. The acid-base titration calculator handles all three.
The conjugate base itself. A solution of sodium acetate is basic, with Kb = Kw/Ka. The weak base pH calculator takes either Kb or the conjugate Ka directly.
Upstream of all of them, the concentration you feed in has to be right. If you are preparing the solution from a solid or from a stock, work through the molarity calculator or the solution dilution calculator first. And if you want to understand Ka as one instance of a general equilibrium constant, see the equilibrium constant calculator.