Percent Ionization Calculator

Background.

This calculator gives the percent ionization and pH of a weak acid or weak base in water, solved exactly. It does not use the quadratic shortcut, and it does not use the √(Ka/C) approximation — it carries water's own autoionization through the full charge balance and finds the root numerically. It also prints the shortcut alongside the exact answer, together with the shortcut's error, because the most useful thing a student can learn here is where the familiar approximation stops working and by how much.

For a weak acid HA at analytical concentration C, mass and charge balance give [H⁺] = [A⁻] + [OH⁻] with [A⁻] = C·Ka/(Ka + [H⁺]). Substituting leaves one equation in one unknown, h − C·Ka/(Ka + h) − Kw/h = 0, whose left side increases strictly with h so there is exactly one positive root. A weak base gives the identical equation in [OH⁻] with Kb in place of Ka. The solver bisects in log space over twenty-five decades, which matters because the root can sit anywhere from 10⁻¹ to 10⁻¹⁴ mol/L depending on the inputs.

The default case is 0.100 mol/L acetic acid at 25 °C, using the NIST critically evaluated pKa of 4.756. The exact answer is [H⁺] = 1.3156 × 10⁻³ mol/L, pH 2.88088, and 1.3156 % ionization — so 98.6844 % of the acetic acid is still sitting there as intact molecules. The √(Ka/C) shortcut gives 1.32434 %, an overestimate of 0.66 %, which for a homework answer is entirely fine.

Now dilute it a hundred thousand-fold, to 1 µmol/L. The exact answer is 94.8176 % ionization at pH 6.01836. The shortcut returns 418.79 %. That is not a small error, it is a number that cannot exist, and it is the reason this page solves the full equation. Two things break at once in dilute solution: the acid is now mostly ionized so the (1 − α) term the shortcut drops is no longer close to 1, and water's own H⁺ is no longer negligible compared with the acid's. Carrying Kw explicitly fixes the second, and solving rather than approximating fixes the first.

One definitional point that is worth stating above the answer. Percent ionization is often written as [H⁺]/C. This page reports [A⁻]/C instead — the actual ionized fraction. The two agree whenever water contributes negligibly, but the [H⁺]/C form counts water's hydrogen ions as though the acid had produced them, which is exactly what lets some calculators print more than 100 % ionization. Defined as [A⁻]/C, the result is bounded by 100 % as a matter of algebra, since [A⁻] = C·Ka/(Ka + [H⁺]) is always less than C.

The scope: one monoprotic solute in water, activity coefficients taken as one, no added conjugate base, and a closed system. If you have both the acid and its conjugate base present, that is a buffer and the Henderson–Hasselbalch page is the right tool. Above roughly 0.5 mol/L, ionic-strength effects that this page ignores start to matter. Polyprotic acids with overlapping constants need the full speciation, not a single-step treatment. And every constant used here is a 25 °C, zero-ionic-strength value, so a bench measurement in salty solution will not match exactly.

What is percent ionization calculator?

Percent ionization — equivalently the degree of ionization or degree of dissociation, expressed as a percentage — is the fraction of a weak electrolyte that has actually ionized at equilibrium. The IUPAC Gold Book defines the degree of ionization (term D01568) and degree of dissociation (D01566) as the ratio of ionized to total solute; multiplying by 100 gives the quantity on this page.

Strong acids and strong bases are, by definition, essentially 100 % ionized in dilute water, so the question is uninteresting for them. Weak electrolytes are not: 0.100 mol/L acetic acid is only about 1.3 % ionized, meaning roughly 99 out of every 100 acetic acid molecules are intact at any instant. That single number explains most of what distinguishes a weak acid from a strong one — the low conductivity, the modest pH, the buffering ability, and the fact that adding base liberates far more acid than the pH alone would suggest.

The counter-intuitive part is dilution. Ostwald's dilution law, Ka = C·α²/(1 − α), published by Wilhelm Ostwald in 1888 from conductance measurements, says that α rises as C falls. Diluting a weak acid makes a LARGER fraction of it ionize, even though the absolute hydrogen-ion concentration goes down and the pH goes up. Both statements are true at once, and confusing them is the single most common error on this topic. The equilibrium shifts toward the ionized side on dilution because that side has more particles.

The relationship to pH is direct but not identical. For a reasonably concentrated weak acid, [H⁺] ≈ [A⁻] and percent ionization ≈ 100·[H⁺]/C, so the two are interchangeable. In dilute solution water supplies a comparable amount of H⁺, the approximation fails, and the two must be distinguished — which is why this page solves for [A⁻] explicitly rather than inferring it from [H⁺].

What percent ionization does not describe: it is not a rate. Proton transfer in water is among the fastest reactions known, essentially diffusion-limited, so the equilibrium is established effectively instantaneously. Percent ionization is a statement about the position of that equilibrium, not about how long it takes to get there.

How to use this calculator.

1. 01Choose weak acid or weak base. The two use the same algebra but a different charge balance, and the calculator returns both pH and pOH in either case.
2. 02Enter the pKa (or pKb) at your temperature. If your table gives Ka or Kb, take the negative base-10 logarithm first. For a base you can also use pKb = pKw − pKa of the conjugate acid.
3. 03Enter the ANALYTICAL concentration — the total amount you dissolved per litre, before any of it ionized. This is not the equilibrium concentration of the un-ionized form.
4. 04Read the exact percent ionization first, then compare it with the shortcut estimate and its error. If the error is under a per cent or two, the shortcut would have been fine for your purposes.
5. 05If the shortcut estimate is above 100 %, you are in the regime where the approximation has failed completely. That happens for very dilute or relatively strong weak acids, and the exact answer beside it is the one to use.
6. 06Check that the ionized and un-ionized percentages sum to 100. They always will; it is a quick way to confirm you have read the right two numbers.
7. 07For a solution that also contains the conjugate base, stop — that is a buffer, and this calculation does not apply. Use the buffer pH or buffer capacity calculators instead.
8. 08If your solution is not at 25 °C, look up the pKa and pKw for your temperature before entering them. The defaults are 25 °C, zero-ionic-strength values.

The formula.

DEFINITION. Percent ionization α = 100 × [A⁻]/C for a weak acid, or 100 × [BH⁺]/C for a weak base, where C is the analytical concentration. Note the numerator: the ionized form of the solute, not [H⁺].

THE EXACT EQUATION. For HA ⇌ H⁺ + A⁻, mass balance gives C = [HA] + [A⁻] and the equilibrium expression gives [A⁻] = C·Ka/(Ka + [H⁺]). Charge balance in a solution containing only the acid and water is [H⁺] = [A⁻] + [OH⁻]. Substituting and writing h for [H⁺]:

h − C·Ka/(Ka + h) − Kw/h = 0

Every term on the left increases with h, so the function is strictly increasing and has exactly one positive root. For a weak base the charge balance is [OH⁻] = [BH⁺] + [H⁺] and the identical equation holds in [OH⁻] with Kb replacing Ka.

HOW IT IS SOLVED. By bisection on p = −log₁₀z over the interval [−5, 25], 200 iterations. Working in the log domain matters: the root can lie anywhere across roughly twenty-five decades, and a linear bisection would spend its entire budget resolving the first one. Because the function is monotonic, the bracket is guaranteed and there is no possibility of converging on a spurious root.

WORKED. 0.100 mol/L acetic acid, pKa 4.756, pKw 14.00. Ka = 1.7538805 × 10⁻⁵. The root is [H⁺] = 1.315601 × 10⁻³ mol/L, so pH = 2.88088 and pOH = 11.11912. Then [A⁻] = 0.1 × Ka/(Ka + [H⁺]) = 1.3156012 × 10⁻³ mol/L = 1315.60 µmol/L, and α = 1.3156 %, leaving 98.6844 % un-ionized.

CROSS-CHECK AGAINST OSTWALD'S DILUTION LAW. Ostwald's 1888 result, Ka = C·α²/(1 − α), was derived from conductance measurements and is completely independent of the electrochemical data behind the NIST pKa. Substituting the exact α back in gives C·α²/(1 − α) = 1.75388049 × 10⁻⁵ against Ka = 1.75388050 × 10⁻⁵ — agreement to eight significant figures, with the residual being the solver's tolerance rather than a modelling difference.

WHY THE SHORTCUT OVERESTIMATES. Dropping the (1 − α) denominator in Ostwald's law gives Ka ≈ Cα², that is α ≈ √(Ka/C). Since (1 − α) < 1, the shortcut always returns too large an α. At 0.100 mol/L acetic acid it gives 1.32434 % against the exact 1.3156 %, an overestimate of 0.66 %. At 1 µmol/L it gives 418.79 % against the exact 94.8176 % — an impossible value, because nothing constrains √(Ka/C) to stay below 1.

DILUTION DIRECTION. α rises as C falls, while pH also rises. Both are correct simultaneously: the equilibrium shifts toward the side with more particles as the solution is diluted, so a larger fraction ionizes, but the absolute hydrogen-ion concentration still falls.

CONVENTIONS. 25 °C, 0.1 MPa, dilute aqueous, zero ionic strength, closed system, one monoprotic solute. α, pH, pOH and the pK values are dimensionless; concentration is in mol/L; the ionized species concentration is in µmol/L; [H⁺] is printed in scientific notation because a numeric field would round it to zero.

ROUNDING STAGE. Nothing is rounded at an intermediate step, including inside the 200-iteration bisection. Rounding happens only when a result is returned, at ten decimal places, with a twelve-significant-digit fallback so a tiny but genuinely non-zero value is never reported as exactly zero.

INVALID DOMAIN. A concentration of zero is rejected rather than reported as zero ionization: the ratio [A⁻]/C is 0/0 and therefore undefined, and the calculator says so under the field. A pK outside −5 to 20, a concentration above 50 mol/L and a pKw of zero or less are likewise rejected with messages naming the field.

A worked example.

Frequently asked questions.

References& sources.

1. [1]IUPAC, Compendium of Chemical Terminology (the Gold Book), entries "degree of ionization" (term identifier D01568, doi:10.1351/goldbook.D01568) and "degree of dissociation" (D01566, doi:10.1351/goldbook.D01566) — the ratio of ionized to total solute that this page reports ×100. Independent, open access; identifiers confirmed 2026-07-29 through the public index (the Gold Book server refuses automated retrieval).
2. [2]Goldberg, R. N.; Kishore, N.; Lennen, R. M. "Thermodynamic Quantities for the Ionization Reactions of Buffers." Journal of Physical and Chemical Reference Data 31(2), 231–370 (2002). NIST Standard Reference Data. Table 7.2 (Acetate): selected pK = 4.756 at 298.15 K and I = 0, evaluation grade AAA — the default pKa. The same table lists 4.763 from MacInnes and Shedlovsky's conductance measurements, the source-conflict figure quoted in the FAQ. Table 7.7 (Ammonia): pK = 9.245, grade AAA, giving the default pKb of 4.755. Table 7.51 gives phosphate's 2.148 / 7.198 / 12.35. Open access; retrieved 2026-07-29.
3. [3]SECOND, INDEPENDENT AUTHORITY (BUILD-BRIEF §9.1). Ostwald, W. "Über die Dissociationstheorie der Elektrolyte." Zeitschrift für physikalische Chemie 2, 36–37 (1888) — Ostwald's dilution law, Ka = C·α²/(1 − α), derived from conductance measurements and therefore entirely independent of the electrochemical and calorimetric data behind the NIST pK values. The exact solver on this page reproduces it to eight significant figures wherever water's autoionization is negligible; the test suite checks the identity at five different pK and concentration combinations. PRINT / BIBLIOGRAPHIC — the 1888 volume has no machine-readable public copy and is cited as such rather than linked as though it were live.
4. [4]International Association for the Properties of Water and Steam, IAPWS R11-24, "Revised Release on the Ionization Constant of H₂O" (June 2024), Table 3: pKw = 14.95 at 0 °C, 13.99 at 25 °C, 13.26 at 50 °C, 12.25 at 100 °C at 0.1 MPa / saturation. Source of the pKw values offered on the input field and quoted in the temperature guidance. Open access; retrieved 2026-07-29.