Weak acid and base equilibria

A weak acid (or base) only partially ionizes: it reaches an equilibrium with its ions rather than dissociating all the way. That is what makes it weak.

The extent is measured by the acid ionization constant K_a (or K_b for a base), the equilibrium constant for the ionization. A larger K_a means a stronger weak acid — one that ionizes further.

How much of it actually stays intact is not fixed by K_a alone — it depends on the concentration too. At the concentrations you will usually meet (roughly 0.01 M and up, with a small K_a), a weak acid is mostly un-ionized, and the percent ionization is only a percent or two. But percent ionization rises as you dilute the solution: the equilibrium shifts toward the ions to keep K_a constant. Dilute a weak acid far enough and a substantial fraction of it ionizes. So “mostly un-ionized” is a statement about a typical solution, not a law about weak acids.

Acids and bases are not two separate stories. For any conjugate acid–base pair, the two constants are locked together by water’s own equilibrium:

Know one and you know the other — a stronger acid necessarily has a weaker conjugate base.

Because it is an equilibrium, finding the pH of a weak acid requires an equilibrium calculation (an ICE table with Ka), not the direct approach used for strong acids. The same is true of a weak base, except that the ICE table runs on K_b and lands you on [OH⁻] — so you finish through pOH.

Use Ka for strength and for the equilibrium — and Kb the same way for a base.

1. Recognize partial ionization. A weak acid reaches equilibrium partway. At ordinary concentrations that leaves it mostly un-ionized, but check the percent ionization rather than assuming it — dilution pushes it up.
2. Use Ka to compare strength. A larger Ka means a stronger weak acid; a smaller Ka, weaker. Kb ranks weak bases the same way, and Ka × Kb = Kw links a conjugate pair.
3. Set up an equilibrium for pH. A weak acid's pH needs an ICE table with Ka, not direct dissociation.
4. For a base, finish through pOH. A Kb ICE table gives [OH⁻], so take pOH = −log[OH⁻], then pH = 14.00 − pOH at 25 °C.
5. Check the x-is-small approximation. Dropping x from the denominator is only safe when the percent ionization comes out under about 5%. If it does not, solve the quadratic.
6. Do not treat it as strong. Assuming complete ionization overestimates [H⁺] and gives the wrong pH.

Ka is the key measure.

Compare. Acid X has K_a = 1 × 10⁻³; acid Y has K_a = 1 × 10⁻⁵.

Stronger. The larger Ka wins: X (1 × 10⁻³) is the stronger weak acid, ionizing further than Y.

Still weak. Both are weak — both sit far below a strong acid, which has a much larger K_a and essentially complete ionization in water. At ordinary concentrations both X and Y stay mostly un-ionized.

Watch the concentration, though. “Mostly un-ionized” is a claim about the solution, not the acid. Percent ionization climbs as you dilute, so a sufficiently dilute sample of X is not mostly intact. The ratio that matters is C/K_a: when it is large the acid stays mostly un-ionized, when it shrinks the acid ionizes substantially.

Now find a pH. Ranking is the easy half. To get an actual number you need the equilibrium, because neither acid dissociates completely. Take 0.100 M acid Y (K_a = 1 × 10⁻⁵) at 25 °C — the same concentration handed to ammonia in §5, so the two calculations line up side by side. Writing the intact acid as HY:

The Ka expression. Products over reactants, and the acid ionizes to give H⁺ directly:

The ICE table. Start with 0.100 M acid and no products, let x mol/L ionize, and read off the equilibrium row:

Substitute and approximate. C/K_a = 0.100 / (1 × 10⁻⁵) = 1 × 10⁴, comfortably large, so the shortcut is worth trying:

Check the approximation. Earn it before you keep it:

Well under 5%, so dropping x from the denominator was legitimate.

pH. A K_a table hands you [H⁺] itself, so the log is the answer — no conversion step, which is exactly the difference from the base problem in §5:

Sanity check. pH = 3.00 — acidic, but nowhere near the pH 1.00 of 0.100 M HCl, a strong acid at the same concentration that ionizes completely. That gap is the partial ionization, and it is the mirror image of the ammonia-versus-NaOH gap in §5.

Where the shortcut breaks. Now run acid X at that same 0.100 M. Its K_a is 1 × 10⁻³, so C/K_a = 100 — not large — and the approximation returns x = 1.0 × 10⁻² M, which is 10% of the starting concentration. That fails the 5% test, so the shortcut is off the table and the quadratic is mandatory: x = 9.5 × 10⁻³ M, giving pH = 2.02 rather than 2.00. The correction happens to be small here, but that is luck, not permission — you cannot know its size until you check, and it grows as C/K_a shrinks. Same procedure, same 0.100 M, and the stronger acid is the one the shortcut fails on.

Why dilution is the trap. The percent ionization is not a property of acid Y — it is a property of this solution of acid Y. Dilute it and the same acid ionizes further:

One acid, one K_a, ionization climbing from a rounding error to more than a quarter of the sample. “Weak acids stay mostly un-ionized” is true of the top of that row and false at the bottom — which is why the 5% check is a step in the procedure, not a formality.

The problem. Find the pH of a 0.100 M solution of ammonia, NH₃, at 25 °C. K_b = 1.8 × 10⁻⁵.

The equilibrium. Ammonia has no proton to donate — it takes one, from water, leaving hydroxide behind. That is the whole reason a base problem produces [OH⁻] and not [H⁺]:

The Kb expression. Water is the solvent, so it does not appear:

The ICE table. Start with 0.100 M ammonia and no products, let x mol/L react, and read off the equilibrium row:

Substitute and approximate. Putting the equilibrium row into K_b:

That last step is the x-is-small approximation, and it is an assumption, not a rule — you have to earn it. It is worth trying here because C/K_b = 0.100 / (1.8 × 10⁻⁵) ≈ 5.6 × 10³, comfortably large. Solve:

Check the approximation. Now go back and justify it, using the percent ionization:

That is under the usual 5% threshold, so dropping x from the denominator was legitimate and the answer stands. (Had it come out above 5% — which is exactly what happens if you dilute this solution far enough — you would have to solve the quadratic instead. Here the quadratic gives 1.33 × 10⁻³ M, a difference too small to matter.)

pOH. A Kb table hands you hydroxide, so hydroxide is what you take the log of:

pH. Convert at the very end — this is the step students skip, reporting pOH as though it were pH:

Sanity check. pH = 11.13 — basic, as a base should be, but nowhere near the pH 13 you would get from 0.100 M NaOH, a strong base that ionizes completely. That gap is the partial ionization.

The bridge back to acids. Because Ka × Kb = Kw, this one number also tells you about ammonium: K_a(NH₄⁺) = (1.0 × 10⁻¹⁴)/(1.8 × 10⁻⁵) = 5.6 × 10⁻¹⁰. Ammonia is a decent weak base, so its conjugate acid is a very feeble one.