What a weak base does in water
A weak base does not supply hydroxide directly the way NaOH does. It takes a proton away from water, and the hydroxide left behind is what makes the solution basic:
B + H2O ⇌ BH+ + OH−
The extent of that reaction is the base ionisation constant Kb = [BH+][OH−]/[B]. Water does not appear in the expression because it is the solvent and its concentration is effectively constant. Ammonia at 0.10 mol/L has Kb = 1.8×10−5, which converts only about 1.3% of the ammonia into ammonium and gives pH 11.12 — strongly basic, but nowhere near the pH 13 that 0.10 M NaOH produces.
The calculation runs through pOH first, then converts. That is not an arbitrary detour: the equilibrium produces hydroxide, so hydroxide is what the algebra gives you. You convert at the end using pH = 14.00 − pOH, which holds at 25 °C because Kw = 1.0×10−14. If your problem gives you the Ka of the conjugate acid instead — tables of amines often do — convert with Kb = Kw/Ka before you start, or select that entry mode above and let the calculator do it.
Building the equilibrium expression
Set up the ICE table exactly as you would for an acid. You begin with C mol/L of B and negligible BH+ and OH−. Let x be the amount that reacts. At equilibrium: [B] = C − x, [BH+] = x, [OH−] = x. Substituting,
which is the quadratic x2 + Kbx − KbC = 0. The familiar shortcut sets C − x ≈ C and gives x ≈ √(KbC). Because the real denominator is smaller than C, that shortcut always returns a hydroxide concentration that is too high, and therefore a pH that is too high. The 5% rule tells you when the error is tolerable: if x/C comes out below 0.05, the shortcut is within roughly 2.5% of the true root.
This calculator does not stop at the quadratic. It solves the charge balance [OH−] = [BH+] + [H+], substituting [BH+] = KbC/(Kb + [OH−]) and [H+] = Kw/[OH−]. Below about 10−6 M hydroxide, water's own contribution stops being negligible and the quadratic drifts off; above that concentration the two agree to more decimal places than any measurement can resolve.
One relationship is worth memorising because it saves constant table lookups: KaKb = Kw for a conjugate pair, so pKa + pKb = 14.00. The stronger a base, the weaker its conjugate acid, and by exactly the reciprocal amount. Use the weak acid pH calculator for the other half of the pair.
Worked example: 0.10 M ammonia, Kb = 1.8e-5
Ammonia in water at 25 °C, Kb = 1.8×10−5.
- Write the quadratic. x2 = Kb(C − x), so x2 + 1.8×10−5x − 1.8×10−6 = 0.
- Discriminant. Kb2 = 3.24×10−10; 4KbC = 7.2×10−6; sum 7.200324×10−6.
- Root. √(7.200324×10−6) = 2.68334×10−3; x = (2.68334×10−3 − 1.8×10−5)/2 = 1.3327×10−3 M OH−.
- pOH. −log10(1.3327×10−3) = 2.875.
- pH. 14.00 − 2.875 = 11.125.
- Percent ionization. 1.3327×10−3/0.10 × 100 = 1.33%, well inside the 5% rule.
- Check [H+]. Kw/[OH−] = 1.0×10−14 ÷ 1.3327×10−3 = 7.50×10−12 M, and −log of that is 11.125. The two routes agree.
Notice the symmetry with acetic acid, which has the same numerical K value: acetic acid at 0.10 M gives pH 2.875 and ammonia gives pH 11.125, and 2.875 + 11.125 = 14.00. That is not a coincidence — it follows from KaKb = Kw when the two constants happen to be equal.
Now a case that breaks the shortcut: 0.15 M methylamine, Kb = 4.4×10−4. The exact root is 7.907×10−3 M, giving pOH 2.102 and pH 11.898, with 5.27% ionization. The shortcut √(4.4×10−4 × 0.15) = 8.124×10−3 is 2.7% high, and the 5% rule flagged it before you did the work.
Reading the result
Start with pOH, because that is what the chemistry produces. A pOH below 3 means a genuinely basic solution; a pOH near 7 means the base is either very weak or very dilute and has barely moved the water. Then read percent ionization as your confidence check on the method: under 5% and every textbook route agrees; over 5% and any answer built on √(KbC) is visibly wrong.
Compare the pH against what a strong base of the same concentration would give. 0.10 M NaOH is pH 13.00; 0.10 M ammonia is pH 11.12. That gap of nearly two units is the whole practical difference between a strong and a weak base, and it is why ammonia is used as a cleaning agent and household bleach solutions are not interchangeable with lye.
Watch the concentration dependence too. The approximation pOH ≈ ½(pKb − log10C) holds whenever C ≫ Kb, so a ten-fold dilution moves the pOH by about 0.5 units, not 1.0. The dilution table this calculator builds shows the effect directly for the base you entered.
If you have both the base and its conjugate acid in solution — ammonia plus ammonium chloride, for instance — this is not the right calculation. The pH is then fixed by the ratio of the two, and you want the Henderson-Hasselbalch buffer calculator. For a fully dissociated base such as NaOH or Ba(OH)2, use the pH and pOH calculator.
Kb, pKb and pH for common weak bases at 25 degC
| Base | Kb | pKb | pKa of conjugate | pH of 0.10 M | Ionized |
|---|---|---|---|---|---|
| Dimethylamine | 5.4×10−4 | 3.27 | 10.73 | 11.85 | 7.08% |
| Methylamine | 4.4×10−4 | 3.36 | 10.64 | 11.81 | 6.42% |
| Trimethylamine | 6.3×10−5 | 4.20 | 9.80 | 11.39 | 2.48% |
| Ammonia | 1.8×10−5 | 4.74 | 9.26 | 11.12 | 1.33% |
| Hydrazine | 1.7×10−6 | 5.77 | 8.23 | 10.61 | 0.411% |
| Pyridine | 1.7×10−9 | 8.77 | 5.23 | 9.12 | 0.0130% |
| Acetate ion | 5.6×10−10 | 9.25 | 4.75 | 8.87 | 0.00748% |
| Aniline | 4.3×10−10 | 9.37 | 4.63 | 8.82 | 0.00656% |
Kb values are representative tabulated constants at 25 °C; sources differ in the second significant figure. The pH and ionization columns are computed from the Kb in the same row for C = 0.10 M.
Assumptions and common errors
- Reporting pOH as pH. The equilibrium gives hydroxide, so the raw answer is pOH. Forgetting the final 14.00 − pOH step turns a basic solution into an acidic one.
- Confusing Kb with the conjugate Ka. They are reciprocally related through Kw, not equal. Ammonia's Kb is 1.8×10−5; the ammonium ion's Ka is 5.6×10−10. Select the matching entry mode above.
- Using the shortcut past 5% ionization. For methylamine at 0.15 M it inflates [OH−] by 2.7%; for stronger amines the error grows quickly.
- Single equilibrium only. This treatment assumes nothing else in solution reacts. Adding the conjugate acid, a second base, or a salt of a different weak acid invalidates it.
- Concentration in place of activity. Above roughly 0.1 M the two diverge and the calculated pH becomes an estimate.
- Polyfunctional bases treated as monofunctional. Carbonate, phosphate and ethylenediamine have two or more protonation steps; entering only the first Kb underestimates the pH.
- Assuming 25 °C. Amine ionisation is noticeably temperature-dependent, which is why buffer bottles carry a temperature coefficient. Enter a constant measured at your working temperature if precision matters.
Where this fits with the other acid-base tools
Four calculations cover almost all routine acid-base work, and they hand off to one another in a predictable order.
Strong electrolytes need no equilibrium at all — the concentration gives the ion directly. A single weak species, acid or base, needs the quadratic on this page. A conjugate pair together needs Henderson-Hasselbalch. A reaction in progress needs a titration treatment, because the composition changes with every millilitre added.
That last case is where weak bases most often show up in practice. Titrating a weak base with a strong acid gives an equivalence point below pH 7, because what remains at equivalence is the conjugate acid BH+, which hydrolyses. The acid-base titration calculator handles the stoichiometry and the curve. If you need the underlying constant expressed in the general form, the equilibrium constant calculator covers K for any reaction, and the molarity calculator is where you should start if the concentration itself is still to be worked out.